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Stewart 9e Section 2.9: Linear Approximations and Differentials
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1
Linearization
Error Rate 100%
Wrong
Find the linearization \(L(x)\) of the function at \(a\): \(f(x) = x^3 - x^2 + 3\), \(a = -2\).
My Answer
-2t^3/sqrt(1-t^4) dt
Answer
\(L(x) = 16 x + 23\)
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Explanation
\(f'(x) = 3 x^2 - 2 x\). At \(a = -2\): \(f(-2) = -8 - 4 + 3 = -9\) and \(f'(-2) = 12 + 4 = 16\). Hence \(L(x) = f(-2) + f'(-2)(x - (-2)) = -9 + 16(x + 2) = 16 x + 23\).
2
Linearization
Error Rate 100%
Wrong
Find the linearization \(L(x)\) of the function at \(a\): \(f(x) = \cos(2 x)\), \(a = \dfrac{\pi}{6}\).
(No answer submitted)
Answer
\(L(x) = \dfrac{1}{2} - \sqrt{3} \left(x - \dfrac{\pi}{6}\right)\)
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Explanation
\(f\left(\dfrac{\pi}{6}\right) = \cos\left(\dfrac{\pi}{3}\right) = \dfrac{1}{2}\). \(f'(x) = -2 \sin(2 x)\), so \(f'\left(\dfrac{\pi}{6}\right) = -2 \sin\left(\dfrac{\pi}{3}\right) = -2 \left(\dfrac{\sqrt{3}}{2}\right) = -\sqrt{3}\). Hence \(L(x) = \dfrac{1}{2} - \sqrt{3}\left(x - \dfrac{\pi}{6}\right)\).
3
Linearization
Error Rate 100%
Wrong
Find the linearization \(L(x)\) of the function at \(a\): \(f(x) = \sqrt[3]{x}\), \(a = 8\).
(No answer submitted)
Answer
\(L(x) = 2 + (x - 8)/12\)
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Explanation
\(f(8) = \sqrt[3]{8} = 2\). \(f'(x) = \left(\dfrac{1}{3}\right) x^{-\dfrac{2}{3}}\), so \(f'(8) = \left(\dfrac{1}{3}\right)\left(\dfrac{1}{4}\right) = \dfrac{1}{12}\). Hence \(L(x) = 2 + \left(\dfrac{1}{12}\right)(x - 8)\).
4
Linearization
Error Rate 100%
Wrong
Find the linearization \(L(x)\) of the function at \(a\): \(f(x) = \dfrac{2}{\sqrt{x^2 - 5}}\), \(a = 3\).
(No answer submitted)
Answer
\(L(x) = 1 - \left(\dfrac{3}{4}\right)(x - 3)\)
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Explanation
\(f(3) = \dfrac{2}{\sqrt{4}} = 1\). Writing \(f(x) = 2(x^2 - 5)^{-\dfrac{1}{2}}\), by the chain rule \(f'(x) = 2 \cdot \left(-\dfrac{1}{2}\right)(x^2 - 5)^{-\dfrac{3}{2}}(2 x) = -2 x/(x^2 - 5)^{\dfrac{3}{2}}\). At \(x = 3\): \(f'(3) = -6/(4)^{\dfrac{3}{2}} = -\dfrac{6}{8} = -\dfrac{3}{4}\). Hence \(L(x) = 1 - \left(\dfrac{3}{4}\right)(x - 3)\).
5
Linear Approximation
Error Rate 100%
Wrong
Find the linear approximation of the function \(f(x) = \sqrt{1 - x}\) at \(a = 0\) and use it to approximate the numbers \(\sqrt{0.9}\) and \(\sqrt{0.99}\). Illustrate by graphing \(f\) and the tangent line.
(No answer submitted)
Answer
\(L(x) = 1 - \dfrac{x}{2}\); \(\sqrt{0.9} \approx 0.95\), \(\sqrt{0.99} \approx 0.995\).
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Explanation
\(f(0) = 1\) and \(f'(x) = -1/(2 \sqrt{1 - x})\), so \(f'(0) = -\dfrac{1}{2}\). Hence \(L(x) = 1 - \dfrac{x}{2}\). Now \(\sqrt{0.9} = \sqrt{1 - 0.1} \approx 1 - 0.\dfrac{1}{2} = 0.95\) and \(\sqrt{0.99} = \sqrt{1 - 0.01} \approx 1 - 0.\dfrac{01}{2} = 0.995\).
6
Linear Approximation
Error Rate 100%
Wrong
Find the linear approximation of the function \(g(x) = \sqrt[3]{1 + x}\) at \(a = 0\) and use it to approximate the numbers \(\sqrt[3]{0.95}\) and \(\sqrt[3]{1.1}\). Illustrate by graphing \(g\) and the tangent line.
(No answer submitted)
Answer
\(L(x) = 1 + \dfrac{x}{3}\); \(\sqrt[3]{0.95} \approx 0.9833\), \(\sqrt[3]{1.1} \approx 1.0333\).
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Explanation
\(g(0) = 1\) and \(g'(x) = \left(\dfrac{1}{3}\right)(1 + x)^{-\dfrac{2}{3}}\), so \(g'(0) = \dfrac{1}{3}\). Hence \(L(x) = 1 + \dfrac{x}{3}\). Now \(\sqrt[3]{0.95} = \sqrt[3]{1 + (-0.05)} \approx 1 - 0.\dfrac{05}{3} \approx 0.9833\) and \(\sqrt[3]{1.1} = \sqrt[3]{1 + 0.1} \approx 1 + 0.\dfrac{1}{3} \approx 1.0333\).
7
Linear Approximation Accuracy
Error Rate 100%
Wrong
Verify the given linear approximation at \(a = 0\): \(\sqrt[4]{1 + 2 x} \approx 1 + \left(\dfrac{1}{2}\right) x\). Then determine the values of \(x\) for which the linear approximation is accurate to within \(0.1\).
(No answer submitted)
Answer
Verified. Accurate within \(0.1\) for approximately \(-0.69 < x < 1.09\).
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Explanation
Let \(f(x) = (1 + 2 x)^{\dfrac{1}{4}}\). Then \(f(0) = 1\) and \(f'(x) = \left(\dfrac{1}{4}\right)(1 + 2 x)^{-\dfrac{3}{4}}(2) = 1/(2 (1 + 2 x)^{\dfrac{3}{4}})\), so \(f'(0) = \dfrac{1}{2}\). Hence \(L(x) = 1 + \dfrac{x}{2}\), verifying the approximation. Graphing \(y = (1 + 2 x)^{\dfrac{1}{4}}\) and \(y = 1 + \dfrac{x}{2} \pm 0.1\) shows the linearization is within \(0.1\) of the function for approximately \(-0.69 < x < 1.09\).
8
Linear Approximation Accuracy
Error Rate 100%
Wrong
Verify the given linear approximation at \(a = 0\): \((1 + x)^{-3} \approx 1 - 3 x\). Then determine the values of \(x\) for which the linear approximation is accurate to within \(0.1\).
(No answer submitted)
Answer
Verified. Accurate within \(0.1\) for approximately \(-0.116 < x < 0.144\).
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Explanation
Let \(f(x) = (1 + x)^{-3}\). Then \(f(0) = 1\) and \(f'(x) = -3 (1 + x)^{-4}\), so \(f'(0) = -3\). Hence \(L(x) = 1 - 3 x\), verifying the approximation. Solving \(|(1 + x)^{-3} - (1 - 3 x)| < 0.1\) graphically gives approximately \(-0.116 < x < 0.144\).
9
Linear Approximation Accuracy
Error Rate 100%
Wrong
Verify the given linear approximation at \(a = 0\): \(1/(1 + 2 x)^4 \approx 1 - 8 x\). Then determine the values of \(x\) for which the linear approximation is accurate to within \(0.1\).
(No answer submitted)
Answer
Verified. Accurate within \(0.1\) for approximately \(-0.045 < x < 0.055\).
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Explanation
Let \(f(x) = (1 + 2 x)^{-4}\). Then \(f(0) = 1\) and \(f'(x) = -8 (1 + 2 x)^{-5}\), so \(f'(0) = -8\). Hence \(L(x) = 1 - 8 x\), verifying the approximation. Graphical analysis of \(|(1 + 2 x)^{-4} - (1 - 8 x)| < 0.1\) gives approximately \(-0.045 < x < 0.055\).
10
Linear Approximation Accuracy
Error Rate 100%
Wrong
Verify the given linear approximation at \(a = 0\): \(\tan x \approx x\). Then determine the values of \(x\) for which the linear approximation is accurate to within \(0.1\).
(No answer submitted)
Answer
Verified. Accurate within \(0.1\) for approximately \(-0.63 < x < 0.63\).
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Explanation
Let \(f(x) = \tan x\). Then \(f(0) = 0\) and \(f'(x) = \sec^2 x\), so \(f'(0) = 1\). Hence \(L(x) = x\), verifying the approximation. Solving \(|\tan x - x| < 0.1\) graphically gives approximately \(-0.63 < x < 0.63\).
11
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = (x^2 - 3)^{-2}\).
(No answer submitted)
Answer
\(d y = -4 x/(x^2 - 3)^3 d x\)
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Explanation
Using the chain rule, \(\dfrac{d y}{d x} = -2 (x^2 - 3)^{-3} (2 x) = -4 x/(x^2 - 3)^3\). Hence \(d y = -4 x/(x^2 - 3)^3 d x\).
12
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = \sqrt{1 - t^4}\).
(No answer submitted)
Answer
\(d y = -2 t^3/\sqrt{1 - t^4} d t\)
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Explanation
Writing \(y = (1 - t^4)^{\dfrac{1}{2}}\), the chain rule gives \(\dfrac{d y}{d t} = \left(\dfrac{1}{2}\right)(1 - t^4)^{-\dfrac{1}{2}}(-4 t^3) = -2 t^3/\sqrt{1 - t^4}\). Hence \(d y = -2 t^3/\sqrt{1 - t^4} d t\).
13
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = \dfrac{1 + 2 u}{1 + 3 u}\).
(No answer submitted)
Answer
\(d y = -1/(1 + 3 u)^2 d u\)
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Explanation
By the quotient rule, \(\dfrac{d y}{d u} = \dfrac{2 (1 + 3 u) - (1 + 2 u)(3)}{1 + 3 u}^2 = \dfrac{2 + 6 u - 3 - 6 u}{1 + 3 u}^2 = -1/(1 + 3 u)^2\). Hence \(d y = -1/(1 + 3 u)^2 d u\).
14
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = \theta^2 \sin(2 \theta)\).
(No answer submitted)
Answer
\(d y = (2 \theta \sin(2 \theta) + 2 \theta^2 \cos(2 \theta)) d \theta\)
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Explanation
By the product and chain rules, \(\dfrac{d y}{d \theta} = 2 \theta \sin(2 \theta) + \theta^2 (2 \cos(2 \theta)) = 2 \theta \sin(2 \theta) + 2 \theta^2 \cos(2 \theta) = 2 \theta (\sin(2 \theta) + \theta \cos(2 \theta))\). Hence \(d y = 2 \theta (\sin(2 \theta) + \theta \cos(2 \theta)) d \theta\).
15
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = 1/(x^2 - 3 x)\).
(No answer submitted)
Answer
\(d y = \dfrac{3 - 2 x}{x^2 - 3 x}^2 d x\)
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Explanation
Writing \(y = (x^2 - 3 x)^{-1}\), the chain rule gives \(\dfrac{d y}{d x} = -(x^2 - 3 x)^{-2}(2 x - 3) = \dfrac{3 - 2 x}{x^2 - 3 x}^2\). Hence \(d y = \dfrac{3 - 2 x}{x^2 - 3 x}^2 d x\).
16
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = \sqrt{1 + \cos \theta}\).
(No answer submitted)
Answer
\(d y = -\sin \theta/(2 \sqrt{1 + \cos \theta}) d \theta\)
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Explanation
Writing \(y = (1 + \cos \theta)^{\dfrac{1}{2}}\), the chain rule gives \(\dfrac{d y}{d \theta} = \left(\dfrac{1}{2}\right)(1 + \cos \theta)^{-\dfrac{1}{2}}(-\sin \theta) = -\sin \theta/(2 \sqrt{1 + \cos \theta})\). Hence \(d y = -\sin \theta/(2 \sqrt{1 + \cos \theta}) d \theta\).
17
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = \sqrt{t \cos t}\).
(No answer submitted)
Answer
\(d y = \dfrac{\cos t - t \sin t}{2 \sqrt{t \cos t}} d t\)
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Explanation
Writing \(y = (t \cos t)^{\dfrac{1}{2}}\) and using the chain and product rules: \(\dfrac{d y}{d t} = \left(\dfrac{1}{2}\right)(t \cos t)^{-\dfrac{1}{2}} (\cos t - t \sin t) = \dfrac{\cos t - t \sin t}{2 \sqrt{t \cos t}}\). Hence \(d y = \dfrac{\cos t - t \sin t}{2 \sqrt{t \cos t}} d t\).
18
Differentials
Error Rate 100%
Wrong
Find the differential of the function: \(y = \left(\dfrac{1}{x}\right) \sin x\).
(No answer submitted)
Answer
\(d y = (x \cos x - \sin x)/x^2 d x\)
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Explanation
Writing \(y = \sin \dfrac{x}{x}\), by the quotient rule \(\dfrac{d y}{d x} = (x \cos x - \sin x)/x^2\). Hence \(d y = (x \cos x - \sin x)/x^2 d x\).
19
Differentials - Evaluation
Error Rate 100%
Wrong
Let \(y = \tan x\). (a) Find the differential \(d y\). (b) Evaluate \(d y\) for \(x = \dfrac{\pi}{4}\) and \(d x = -0.1\).
(No answer submitted)
Answer
(a) \(d y = \sec^2 x d x\). (b) \(d y = -0.2\).
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Explanation
(a) \(\dfrac{d y}{d x} = \sec^2 x\), so \(d y = \sec^2 x d x\). (b) \(\sec\left(\dfrac{\pi}{4}\right) = \sqrt{2}\), so \(\sec^2\left(\dfrac{\pi}{4}\right) = 2\). Hence \(d y = 2(-0.1) = -0.2\).
20
Differentials - Evaluation
Error Rate 100%
Wrong
Let \(y = \cos(\pi x)\). (a) Find the differential \(d y\). (b) Evaluate \(d y\) for \(x = \dfrac{1}{2}\) and \(d x = -0.02\).
(No answer submitted)
Answer
(a) \(d y = -\pi \sin(\pi x) d x\). (b) \(d y = 0.02 \pi \approx 0.0628\).
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Explanation
(a) \(\dfrac{d y}{d x} = -\pi \sin(\pi x)\), so \(d y = -\pi \sin(\pi x) d x\). (b) \(\sin\left(\dfrac{\pi}{2}\right) = 1\), so \(d y = -\pi (1)(-0.02) = 0.02 \pi \approx 0.0628\).
21
Differentials - Evaluation
Error Rate 100%
Wrong
Let \(y = \sqrt{3 + x^2}\). (a) Find the differential \(d y\). (b) Evaluate \(d y\) for \(x = 1\) and \(d x = -0.1\).
(No answer submitted)
Answer
(a) \(d y = \dfrac{x}{\sqrt{3 + x^2}} d x\). (b) \(d y = -0.05\).
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Explanation
(a) \(\dfrac{d y}{d x} = \dfrac{x}{\sqrt{3 + x^2}}\), so \(d y = \dfrac{x}{\sqrt{3 + x^2}} d x\). (b) At \(x = 1\): \(\sqrt{4} = 2\), so \(d y = \left(\dfrac{1}{2}\right)(-0.1) = -0.05\).
22
Differentials - Evaluation
Error Rate 100%
Wrong
Let \(y = \dfrac{x + 1}{x - 1}\). (a) Find the differential \(d y\). (b) Evaluate \(d y\) for \(x = 2\) and \(d x = 0.05\).
(No answer submitted)
Answer
(a) \(d y = -2/(x - 1)^2 d x\). (b) \(d y = -0.1\).
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Explanation
(a) By the quotient rule, \(\dfrac{d y}{d x} = \dfrac{(x - 1) - (x + 1)}{x - 1}^2 = -2/(x - 1)^2\). So \(d y = -2/(x - 1)^2 d x\). (b) At \(x = 2\): \((x - 1)^2 = 1\), so \(d y = -2 (0.05) = -0.1\).
23
Increment vs Differential
Error Rate 100%
Wrong
Compute \(\Delta y\) and \(d y\) for \(y = x^2 - 4 x\), \(x = 3\), \(\Delta x = 0.5\). Then sketch a diagram showing the line segments with lengths \(d x\), \(d y\), and \(\Delta y\).
(No answer submitted)
Answer
\(\Delta y = 1.25\), \(d y = 1\).
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Explanation
\(f(3) = 9 - 12 = -3\) and \(f(3.5) = 12.25 - 14 = -1.75\), so \(\Delta y = -1.75 - (-3) = 1.25\). The differential is \(d y = (2 x - 4) d x = (6 - 4)(0.5) = 1\). The diagram shows \(d x\) along the \(x\)-axis and the tangent rise \(d y = 1\) vs. the actual curve rise \(\Delta y = 1.25\).
24
Increment vs Differential
Error Rate 100%
Wrong
Compute \(\Delta y\) and \(d y\) for \(y = x - x^3\), \(x = 0\), \(\Delta x = -0.3\). Then sketch a diagram showing the line segments with lengths \(d x\), \(d y\), and \(\Delta y\).
(No answer submitted)
Answer
\(\Delta y = -0.273\), \(d y = -0.3\).
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Explanation
\(f(0) = 0\) and \(f(-0.3) = -0.3 - (-0.027) = -0.273\), so \(\Delta y = -0.273\). The differential is \(d y = (1 - 3 x^2) d x = (1)(-0.3) = -0.3\).
25
Increment vs Differential
Error Rate 100%
Wrong
Compute \(\Delta y\) and \(d y\) for \(y = \sqrt{x - 2}\), \(x = 3\), \(\Delta x = 0.8\). Then sketch a diagram showing the line segments with lengths \(d x\), \(d y\), and \(\Delta y\).
(No answer submitted)
Answer
\(\Delta y \approx 0.3416\), \(d y = 0.4\).
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Explanation
\(f(3) = \sqrt{1} = 1\) and \(f(3.8) = \sqrt{1.8} \approx 1.3416\), so \(\Delta y \approx 0.3416\). The differential is \(d y = 1/(2 \sqrt{x - 2}) d x = \left(\dfrac{1}{2}\right)(0.8) = 0.4\).
26
Increment vs Differential
Error Rate 100%
Wrong
Compute \(\Delta y\) and \(d y\) for \(y = x^3\), \(x = 1\), \(\Delta x = 0.5\). Then sketch a diagram showing the line segments with lengths \(d x\), \(d y\), and \(\Delta y\).
(No answer submitted)
Answer
\(\Delta y = 2.375\), \(d y = 1.5\).
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Explanation
\(f(1) = 1\) and \(f(1.5) = 3.375\), so \(\Delta y = 2.375\). The differential is \(d y = 3 x^2 d x = 3 (1)(0.5) = 1.5\).
27
Comparing Delta y and dy
Error Rate 100%
Wrong
Compare the values of \(\Delta y\) and \(d y\) for \(f(x) = x^4 - x + 1\) if \(x\) changes from 1 to 1.05. What if \(x\) changes from 1 to 1.01? Does the approximation \(\Delta y \approx d y\) become better as \(\Delta x\) gets smaller?
(No answer submitted)
Answer
\(x: 1 \rightarrow 1.05\): \(\Delta y \approx 0.16551\), \(d y = 0.15\). \(x: 1 \rightarrow 1.01\): \(\Delta y \approx 0.03060\), \(d y = 0.03\). Yes, the approximation improves.
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Explanation
\(f(1) = 1\) and \(d y = (4 x^3 - 1) d x = 3 d x\) at \(x = 1\). (a) \(f(1.05) = (1.05)^4 - 1.05 + 1 \approx 1.16551\), \(\Delta y \approx 0.16551\); \(d y = 3 (0.05) = 0.15\). (b) \(f(1.01) \approx 1.03060\), \(\Delta y \approx 0.03060\); \(d y = 3 (0.01) = 0.03\). The relative discrepancy \(|\Delta y - d y|/|\Delta y|\) is smaller for the smaller \(\Delta x\), so the approximation improves.
28
Comparing Delta y and dy
Error Rate 100%
Wrong
Compare the values of \(\Delta y\) and \(d y\) for \(f(x) = (x^3 + 3)^2\) if \(x\) changes from 1 to 1.05. What if \(x\) changes from 1 to 1.01? Does the approximation \(\Delta y \approx d y\) become better as \(\Delta x\) gets smaller?
(No answer submitted)
Answer
\(x: 1 \rightarrow 1.05\): \(\Delta y \approx 1.2858\), \(d y = 1.2\). \(x: 1 \rightarrow 1.01\): \(\Delta y \approx 0.2433\), \(d y = 0.24\). Yes.
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Explanation
\(f(1) = 16\) and \(d y = 2 (x^3 + 3)(3 x^2) d x = 24 d x\) at \(x = 1\). (a) \(f(1.05) = (1.157625 + 3)^2 \approx 17.2858\), \(\Delta y \approx 1.2858\); \(d y = 24 (0.05) = 1.2\). (b) \(f(1.01) \approx 16.2433\), \(\Delta y \approx 0.2433\); \(d y = 24 (0.01) = 0.24\). The approximation improves as \(\Delta x\) decreases.
29
Comparing Delta y and dy
Error Rate 100%
Wrong
Compare the values of \(\Delta y\) and \(d y\) for \(f(x) = \sqrt{5 - x}\) if \(x\) changes from 1 to 1.05. What if \(x\) changes from 1 to 1.01? Does the approximation \(\Delta y \approx d y\) become better as \(\Delta x\) gets smaller?
(No answer submitted)
Answer
\(x: 1 \rightarrow 1.05\): \(\Delta y \approx -0.01254\), \(d y = -0.0125\). \(x: 1 \rightarrow 1.01\): \(\Delta y \approx -0.00250\), \(d y = -0.0025\). Yes.
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Explanation
\(f(1) = 2\) and \(d y = -1/(2 \sqrt{5 - x}) d x = -\left(\dfrac{1}{4}\right) d x\) at \(x = 1\). (a) \(f(1.05) = \sqrt{3.95} \approx 1.98746\), \(\Delta y \approx -0.01254\); \(d y = -\left(\dfrac{1}{4}\right)(0.05) = -0.0125\). (b) \(f(1.01) = \sqrt{3.99} \approx 1.99750\), \(\Delta y \approx -0.00250\); \(d y = -\left(\dfrac{1}{4}\right)(0.01) = -0.0025\). The approximation improves.
30
Comparing Delta y and dy
Error Rate 100%
Wrong
Compare the values of \(\Delta y\) and \(d y\) for \(f(x) = 1/(x^2 + 1)\) if \(x\) changes from 1 to 1.05. What if \(x\) changes from 1 to 1.01? Does the approximation \(\Delta y \approx d y\) become better as \(\Delta x\) gets smaller?
(No answer submitted)
Answer
\(x: 1 \rightarrow 1.05\): \(\Delta y \approx -0.02438\), \(d y = -0.025\). \(x: 1 \rightarrow 1.01\): \(\Delta y \approx -0.00498\), \(d y = -0.005\). Yes.
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Explanation
\(f(1) = \dfrac{1}{2}\) and \(d y = -2 x/(x^2 + 1)^2 d x = -\left(\dfrac{1}{2}\right) d x\) at \(x = 1\). (a) \(f(1.05) = \dfrac{1}{2}.1025 \approx 0.47562\), \(\Delta y \approx -0.02438\); \(d y = -\left(\dfrac{1}{2}\right)(0.05) = -0.025\). (b) \(f(1.01) \approx 0.49502\), \(\Delta y \approx -0.00498\); \(d y = -\left(\dfrac{1}{2}\right)(0.01) = -0.005\). The approximation improves.
31
Estimation - Linear Approximation
Error Rate 100%
Wrong
Use a linear approximation (or differentials) to estimate the given number: \((1.999)^4\).
(No answer submitted)
Answer
\((1.999)^4 \approx 15.968\)
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Explanation
Let \(f(x) = x^4\) at \(a = 2\). Then \(f(2) = 16\) and \(f'(2) = 4 (2)^3 = 32\), so \(L(x) = 16 + 32(x - 2)\). Hence \((1.999)^4 = f(1.999) \approx 16 + 32(-0.001) = 15.968\).
32
Estimation - Linear Approximation
Error Rate 100%
Wrong
Use a linear approximation (or differentials) to estimate the given number: \(\dfrac{1}{4}.002\).
(No answer submitted)
Answer
\(\dfrac{1}{4}.002 \approx 0.249875\)
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Explanation
Let \(f(x) = \dfrac{1}{x}\) at \(a = 4\). Then \(f(4) = \dfrac{1}{4}\) and \(f'(4) = -\dfrac{1}{16}\), so \(L(x) = \dfrac{1}{4} - \left(\dfrac{1}{16}\right)(x - 4)\). Hence \(\dfrac{1}{4}.002 \approx \dfrac{1}{4} - \left(\dfrac{1}{16}\right)(0.002) = 0.25 - 0.000125 = 0.249875\).
33
Estimation - Linear Approximation
Error Rate 100%
Wrong
Use a linear approximation (or differentials) to estimate the given number: \(\sqrt[3]{1001}\).
(No answer submitted)
Answer
\(\sqrt[3]{1001} \approx 10.00333\)
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Explanation
Let \(f(x) = x^{\dfrac{1}{3}}\) at \(a = 1000\). Then \(f(1000) = 10\) and \(f'(1000) = \left(\dfrac{1}{3}\right)(1000)^{-\dfrac{2}{3}} = \dfrac{1}{300}\), so \(L(x) = 10 + (x - 1000)/300\). Hence \(\sqrt[3]{1001} \approx 10 + \dfrac{1}{300} \approx 10.00333\).
34
Estimation - Linear Approximation
Error Rate 100%
Wrong
Use a linear approximation (or differentials) to estimate the given number: \(\sqrt{100.5}\).
(No answer submitted)
Answer
\(\sqrt{100.5} \approx 10.025\)
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Explanation
Let \(f(x) = \sqrt{x}\) at \(a = 100\). Then \(f(100) = 10\) and \(f'(100) = \dfrac{1}{20}\), so \(L(x) = 10 + (x - 100)/20\). Hence \(\sqrt{100.5} \approx 10 + 0.\dfrac{5}{20} = 10.025\).
35
Estimation - Linear Approximation
Error Rate 100%
Wrong
Use a linear approximation (or differentials) to estimate the given number: \(\tan 2^{\circ}\).
(No answer submitted)
Answer
\(\tan 2^{\circ} \approx \dfrac{\pi}{90} \approx 0.0349\)
Compare with the answer above and grade yourself:
Explanation
Convert to radians: \(2^{\circ} = \dfrac{\pi}{90}\) rad. Let \(f(x) = \tan x\) at \(a = 0\). Then \(f(0) = 0\) and \(f'(0) = \sec^2 0 = 1\), so \(L(x) = x\). Hence \(\tan\left(\dfrac{\pi}{90}\right) \approx \dfrac{\pi}{90} \approx 0.0349\).
36
Estimation - Linear Approximation
Error Rate 100%
Wrong
Use a linear approximation (or differentials) to estimate the given number: \(\cos 29^{\circ}\).
(No answer submitted)
Answer
\(\cos 29^{\circ} \approx \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{360} \approx 0.8747\)
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Explanation
Note \(29^{\circ} = 30^{\circ} - 1^{\circ}\), so \(29^{\circ} = \dfrac{\pi}{6} - \dfrac{\pi}{180}\) rad. Let \(f(x) = \cos x\) at \(a = \dfrac{\pi}{6}\). Then \(f\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\) and \(f'\left(\dfrac{\pi}{6}\right) = -\sin\left(\dfrac{\pi}{6}\right) = -\dfrac{1}{2}\), so \(L(x) = \dfrac{\sqrt{3}}{2} - \left(\dfrac{1}{2}\right)\left(x - \dfrac{\pi}{6}\right)\). Hence \(\cos 29^{\circ} \approx \dfrac{\sqrt{3}}{2} - \left(\dfrac{1}{2}\right)\left(-\dfrac{\pi}{180}\right) = \dfrac{\sqrt{3}}{2} + \dfrac{\pi}{360} \approx 0.8660 + 0.00873 \approx 0.8747\).
37
Explanation - Linear Approximation
Error Rate 100%
Wrong
Explain, in terms of linear approximations or differentials, why the approximation \(\sec 0.08 \approx 1\) is reasonable.
(No answer submitted)
Answer
The linearization of \(f(x) = \sec x\) at \(a = 0\) is \(L(x) = 1\), so \(\sec 0.08 \approx 1\).
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Explanation
Let \(f(x) = \sec x\). Then \(f(0) = 1\) and \(f'(x) = \sec x \tan x\), so \(f'(0) = 1 \cdot 0 = 0\). The linearization at \(a = 0\) is \(L(x) = 1 + 0 \cdot x = 1\). Since \(0.08\) is close to \(0\), \(\sec 0.08 \approx L(0.08) = 1\).
38
Explanation - Linear Approximation
Error Rate 100%
Wrong
Explain, in terms of linear approximations or differentials, why the approximation \(\sqrt{4.02} \approx 2.005\) is reasonable.
(No answer submitted)
Answer
The linearization of \(f(x) = \sqrt{x}\) at \(a = 4\) gives \(L(4.02) = 2 + 0.\dfrac{02}{4} = 2.005\).
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Explanation
Let \(f(x) = \sqrt{x}\). Then \(f(4) = 2\) and \(f'(x) = 1/(2 \sqrt{x})\), so \(f'(4) = \dfrac{1}{4}\). The linearization at \(a = 4\) is \(L(x) = 2 + \left(\dfrac{1}{4}\right)(x - 4)\). Since \(4.02\) is close to \(4\), \(\sqrt{4.02} \approx L(4.02) = 2 + (0.02)/4 = 2.005\).
39
Error Estimation - Cube
Error Rate 100%
Wrong
The edge of a cube was found to be 30 cm with a possible error in measurement of 0.1 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing (a) the volume of the cube and (b) the surface area of the cube.
(No answer submitted)
Answer
(a) Max error \(\approx 270 \text{cm}^3\); relative error \(= 0.01\); percentage error \(= 1%\). (b) Max error \(= 36 \text{cm}^2\); relative error \(\approx 0.00667\); percentage error \(\approx 0.67%\).
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Explanation
Let \(x\) be the edge length. (a) \(V = x^3\), so \(d V = 3 x^2 d x\). At \(x = 30\), \(d x = 0.1\): \(d V = 3 (900)(0.1) = 270 \text{cm}^3\). Relative error: \((d V)/V = 3 (d x)/x = 3(0.1)/30 = 0.01\), or \(1%\). (b) \(S = 6 x^2\), so \(d S = 12 x d x = 12 (30)(0.1) = 36 \text{cm}^2\). Relative error: \((d S)/S = 2 (d x)/x = 2(0.1)/30 \approx 0.00667\), or about \(0.67%\).
40
Error Estimation - Disk
Error Rate 100%
Wrong
The radius of a circular disk is given as 24 cm with a maximum error in measurement of 0.2 cm. (a) Use differentials to estimate the maximum error in the calculated area of the disk. (b) What is the relative error and the percentage error?
(No answer submitted)
Answer
(a) Max error \(\approx 9.6 \pi \approx 30.16 \text{cm}^2\). (b) Relative error \(\approx 0.0167\); percentage error \(\approx 1.67%\).
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Explanation
\(A = \pi r^2\), so \(d A = 2 \pi r d r\). At \(r = 24\), \(d r = 0.2\): \(d A = 2 \pi (24)(0.2) = 9.6 \pi \approx 30.16 \text{cm}^2\). Relative error: \((d A)/A = 2 (d r)/r = 2(0.2)/24 \approx 0.0167\), or about \(1.67%\).
41
Error Estimation - Sphere
Error Rate 100%
Wrong
The circumference of a sphere was measured to be 84 cm with a possible error of 0.5 cm. (a) Use differentials to estimate the maximum error in the calculated surface area. What is the relative error? (b) Use differentials to estimate the maximum error in the calculated volume. What is the relative error?
(No answer submitted)
Answer
(a) Max error \(\approx \dfrac{84}{\pi} \approx 26.74 \text{cm}^2\); relative error \(= \dfrac{1}{84} \approx 0.0119\). (b) Max error \(= \dfrac{1764}{p}i^2 \approx 178.79 \text{cm}^3\); relative error \(\approx 0.0179\).
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Explanation
Let \(C\) denote circumference, so \(r = C/(2 \pi)\) and \(d r = \dfrac{d C}{2 \pi}\). (a) \(S = 4 \pi r^2 = C^2/\pi\). Then \(d S = \left(2 \dfrac{C}{\pi}\right) d C = (2 (84)/\pi)(0.5) = \dfrac{84}{\pi} \approx 26.74 \text{cm}^2\). Relative error: \((d S)/S = 2 (d C)/C = 2 (0.5)/84 = \dfrac{1}{84} \approx 0.0119\). (b) \(V = \left(\dfrac{4}{3}\right) \pi r^3 = C^3/(6 \pi^2)\). Then \(d V = (C^2/(2 \pi^2)) d C = (84^2/(2 \pi^2))(0.5) = \dfrac{1764}{p}i^2 \approx 178.79 \text{cm}^3\). Relative error: \((d V)/V = 3 (d C)/C = 3 (0.5)/84 = 1.\dfrac{5}{84} \approx 0.0179\).
42
Error Estimation - Paint Volume
Error Rate 100%
Wrong
Use differentials to estimate the amount of paint needed to apply a coat of paint 0.05 cm thick to a hemispherical dome with diameter 50 m.
(No answer submitted)
Answer
Approximately \(0.625 \pi \approx 1.96 \text{m}^3\).
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Explanation
The dome has radius \(r = 25\) m. The hemisphere's volume is \(V = \left(\dfrac{2}{3}\right) \pi r^3\), so \(d V = 2 \pi r^2 d r\). The paint thickness is \(d r = 0.05 \text{cm} = 0.0005 \text{m}\). Hence \(d V = 2 \pi (25)^2 (0.0005) = 0.625 \pi \approx 1.96 \text{m}^3\).
43
Differentials - Cylindrical Shell
Error Rate 100%
Wrong
(a) Use differentials to find a formula for the approximate volume of a thin cylindrical shell with height \(h\), inner radius \(r\), and thickness \(\Delta r\). (b) What is the error involved in using the formula from part (a)?
(No answer submitted)
Answer
(a) \(V \approx 2 \pi r h \Delta r\). (b) Error \(= \pi h (\Delta r)^2\).
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Explanation
(a) The cylinder volume is \(V = \pi r^2 h\). Treating \(h\) as constant and using the differential, \(d V = 2 \pi r h d r\). So the shell's volume is approximately \(2 \pi r h \Delta r\). (b) The exact shell volume is \(\pi h ((r + \Delta r)^2 - r^2) = \pi h (2 r \Delta r + (\Delta r)^2)\). Subtracting the approximation \(2 \pi r h \Delta r\) leaves an error of \(\pi h (\Delta r)^2\).
44
Error Estimation - Right Triangle
Error Rate 100%
Wrong
One side of a right triangle is known to be 20 cm long and the opposite angle is measured as \(30^{\circ}\), with a possible error of \(\pm 1^{\circ}\). (a) Use differentials to estimate the error in computing the length of the hypotenuse. (b) What is the percentage error?
(No answer submitted)
Answer
(a) Approximately \(\pm (2 \pi \sqrt{3})/9 \approx \pm 1.21 \text{cm}\). (b) Approximately \(3%\).
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Explanation
Let \(\theta\) denote the opposite angle and \(h\) the hypotenuse. Since \(\sin \theta = \dfrac{20}{h}\), we have \(h = \dfrac{20}{\sin} \theta\), so \(\dfrac{d h}{d \theta} = -20 \cos \dfrac{\theta}{si}n^2 \theta\). At \(\theta = \dfrac{\pi}{6}\): \(\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}\) and \(\sin^2\left(\dfrac{\pi}{6}\right) = \dfrac{1}{4}\). With \(d \theta = \pm 1^{\circ} = \pm \dfrac{\pi}{180}\) rad: \(d h = -20 \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{4}} \left(\pm \dfrac{\pi}{180}\right) = \mp (2 \pi \sqrt{3})/9 \approx \mp 1.21 \text{cm}\). (b) At \(\theta = \dfrac{\pi}{6}\), \(h = 20/\left(\dfrac{1}{2}\right) = 40 \text{cm}\). Percentage error \(\approx 1.\dfrac{21}{40} \approx 0.03\), or about \(3%\).
45
Application - Ohm's Law
Error Rate 100%
Wrong
If a current \(I\) passes through a resistor with resistance \(R\), Ohm's Law states that the voltage drop is \(V = R I\). If \(V\) is constant and \(R\) is measured with a certain error, use differentials to show that the relative error in calculating \(I\) is approximately the same (in magnitude) as the relative error in \(R\).
(No answer submitted)
Answer
\(|(d I)/I| = |(d R)/R|\).
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Explanation
Since \(V\) is constant, \(I = \dfrac{V}{R}\). Differentiating, \(\dfrac{d I}{d R} = -V/R^2\), so \(d I = -(V/R^2) d R\). Then \((d I)/I = \dfrac{-(V/R^2) d R}{\dfrac{V}{R}} = -(d R)/R\). Taking absolute values, \(|(d I)/I| = |(d R)/R|\), which means the relative error in \(I\) matches the relative error in \(R\) in magnitude.
46
Application - Poiseuille's Law
Error Rate 100%
Wrong
When blood flows along a blood vessel, the flux \(F\) (the volume of blood per unit time that flows past a given point) is proportional to the fourth power of the radius \(R\) of the blood vessel: \(F = k R^4\). Show that the relative change in \(F\) is about four times the relative change in \(R\). How will a \(5%\) increase in the radius affect the flow of blood?
(No answer submitted)
Answer
\((d F)/F = 4 (d R)/R\); a \(5%\) increase in \(R\) produces about a \(20%\) increase in \(F\).
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Explanation
Since \(F = k R^4\), \(\dfrac{d F}{d R} = 4 k R^3\), so \(d F = 4 k R^3 d R\). Then \((d F)/F = \dfrac{4 k R^3 d R}{k R^4} = 4 (d R)/R\). Hence the relative change in \(F\) is approximately four times the relative change in \(R\). A \(5%\) increase in \(R\) \(((d R)/R = 0.05)\) gives \((d F)/F \approx 4 (0.05) = 0.20\), or a \(20%\) increase in flow.
47
Differential Rules - Proof
Error Rate 100%
Wrong
Establish the following rules for working with differentials (where \(c\) denotes a constant and \(u\) and \(v\) are functions of \(x\)): (a) \(d c = 0\), (b) \(d(c u) = c d u\), (c) \(d(u + v) = d u + d v\), (d) \(d(u v) = u d v + v d u\), (e) \(d\left(\dfrac{u}{v}\right) = (v d u - u d v)/v^2\), (f) \(d(x^n) = n x^{n-1} d x\).
(No answer submitted)
Answer
Each rule follows directly from the corresponding derivative rule.
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Explanation
(a) If \(f(x) = c\), then \(f'(x) = 0\), so \(d c = 0 d x = 0\). (b) \((c u)' = c u'\), so \(d(c u) = c u' d x = c d u\). (c) \((u + v)' = u' + v'\), so \(d(u + v) = (u' + v') d x = u' d x + v' d x = d u + d v\). (d) By the product rule, \((u v)' = u' v + u v'\), so \(d(u v) = (u' v + u v') d x = v (u' d x) + u (v' d x) = v d u + u d v\). (e) By the quotient rule, \(\left(\dfrac{u}{v}\right)' = (u' v - u v')/v^2\), so \(d\left(\dfrac{u}{v}\right) = (u' v - u v')/v^2 d x = (v d u - u d v)/v^2\). (f) \((x^n)' = n x^{n-1}\), so \(d(x^n) = n x^{n-1} d x\).
48
Application - Pendulum
Error Rate 100%
Wrong
In physics textbooks, the period \(T\) of a pendulum of length \(L\) is often given as \(T \approx 2 \pi \sqrt{\dfrac{L}{g}}\), provided that the pendulum swings through a relatively small arc. In the course of deriving this formula, the equation \(a_T = -g \sin \theta\) for the tangential acceleration of the bob of the pendulum is obtained, and then \(\sin \theta\) is replaced by \(\theta\) with the remark that for small angles, \(\theta\) (in radians) is very close to \(\sin \theta\). (a) Verify the linear approximation at 0 for the sine function: \(\sin \theta \approx \theta\). (b) If \(\theta = \dfrac{\pi}{18}\) (equivalent to \(10^{\circ}\)) and we approximate \(\sin \theta\) by \(\theta\), what is the percentage error? (c) Use a graph to determine the values of \(\theta\) for which \(\sin \theta\) and \(\theta\) differ by less than \(2%\). What are the values in degrees?
(No answer submitted)
Answer
(a) Verified: \(L(\theta) = \theta\). (b) Approximately \(0.51%\). (c) Approximately \(|\theta| < 0.344 \text{rad}\), equivalent to \(|\theta| < 19.7^{\circ}\).
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Explanation
(a) Let \(f(\theta) = \sin \theta\). Then \(f(0) = 0\) and \(f'(0) = \cos 0 = 1\), so \(L(\theta) = \theta\), verifying \(\sin \theta \approx \theta\) for \(\theta\) near \(0\). (b) \(\sin\left(\dfrac{\pi}{18}\right) \approx 0.17365\) and \(\dfrac{\pi}{18} \approx 0.17453\). Percentage error \(= |\dfrac{\pi}{18} - \sin\left(\dfrac{\pi}{18}\right)|/\sin\left(\dfrac{\pi}{18}\right) \approx 0.00507\), or about \(0.51%\). (c) Graphing \(y = (\theta - \sin \theta)/\sin \theta\) and locating where \(|y| < 0.02\) gives approximately \(|\theta| < 0.344 \text{rad}\), which corresponds to \(|\theta| < 0.344 \cdot \left(\dfrac{180}{\pi}\right) \approx 19.7^{\circ}\).
49
Linear Approximation from Graph
Error Rate 100%
Wrong
Suppose that the only information we have about a function \(f\) is that \(f(1) = 5\) and the graph of its derivative is as shown. (a) Use a linear approximation to estimate \(f(0.9)\) and \(f(1.1)\). (b) Are your estimates in part (a) too large or too small? Explain.
(No answer submitted)
Answer
(a) From the graph \(f'(1) \approx 2\): \(f(0.9) \approx 4.8\) and \(f(1.1) \approx 5.2\). (b) Estimates are too large because \(f'\) is decreasing near \(x = 1\) (so \(f\) is concave down), making the tangent line lie above the curve.
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Explanation
(a) Reading \(f'(1) \approx 2\) from the graph, the linearization is \(L(x) = f(1) + f'(1)(x - 1) = 5 + 2 (x - 1)\). So \(f(0.9) \approx L(0.9) = 5 + 2 (-0.1) = 4.8\) and \(f(1.1) \approx L(1.1) = 5 + 2 (0.1) = 5.2\). (b) Since the graph of \(f'\) is decreasing at \(x = 1\), we have \(f''(1) < 0\), so \(f\) is concave down near \(x = 1\). The tangent line therefore lies above the graph of \(f\), making both estimates too large.
50
Linear Approximation - Concavity Check
Error Rate 100%
Wrong
Suppose that we don't have a formula for \(g(x)\) but we know that \(g(2) = -4\) and \(g'(x) = \sqrt{x^2 + 5}\) for all \(x\). (a) Use a linear approximation to estimate \(g(1.95)\) and \(g(2.05)\). (b) Are your estimates in part (a) too large or too small? Explain.
(No answer submitted)
Answer
(a) \(g(1.95) \approx -4.15\) and \(g(2.05) \approx -3.85\). (b) Estimates are too small because \(g\) is concave up near \(x = 2\), so the tangent line lies below the curve.
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Explanation
(a) \(g'(2) = \sqrt{4 + 5} = 3\), so \(L(x) = -4 + 3 (x - 2)\). Hence \(g(1.95) \approx L(1.95) = -4 + 3 (-0.05) = -4.15\) and \(g(2.05) \approx L(2.05) = -4 + 3 (0.05) = -3.85\). (b) Computing \(g''(x) = \dfrac{x}{\sqrt{x^2 + 5}}\), we have \(g''(2) = \dfrac{2}{3} > 0\), so \(g\) is concave up near \(x = 2\). The tangent line lies below the curve, so both estimates underestimate the true values.
51
Example - Linearization and Linear Approximation
Error Rate 100%
Wrong
Find the linearization \(L(x)\) of the function \(f(x) = \sqrt{x + 3}\) at \(a = 1\) and use it to approximate the numbers \(\sqrt{3.98}\) and \(\sqrt{4.05}\). Are these approximations overestimates or underestimates?
(No answer submitted)
Answer
\(L(x) = \dfrac{7}{4} + \dfrac{x}{4}\); \(\sqrt{3.98} \approx 1.995\), \(\sqrt{4.05} \approx 2.0125\); both are overestimates.
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Explanation
We have \(f'(x) = 1/(2 \sqrt{x + 3})\), so \(f(1) = 2\) and \(f'(1) = \dfrac{1}{4}\). The linearization is \(L(x) = f(1) + f'(1)(x - 1) = 2 + \left(\dfrac{1}{4}\right)(x - 1) = \dfrac{7}{4} + \dfrac{x}{4}\). The corresponding linear approximation is \(\sqrt{x + 3} \approx \dfrac{7}{4} + \dfrac{x}{4}\) when \(x\) is near 1. So \(\sqrt{3.98} = \sqrt{0.98 + 3} \approx \dfrac{7}{4} + 0.\dfrac{98}{4} = 1.995\) and \(\sqrt{4.05} = \sqrt{1.05 + 3} \approx \dfrac{7}{4} + 1.\dfrac{05}{4} = 2.0125\). Both are overestimates because the tangent line lies above the curve \(y = \sqrt{x + 3}\).
52
Example - Accuracy of Linear Approximation
Error Rate 100%
Wrong
For what values of \(x\) is the linear approximation \(\sqrt{x + 3} \approx \dfrac{7}{4} + \dfrac{x}{4}\) accurate to within \(0.5\)? What about accuracy to within \(0.1\)?
(No answer submitted)
Answer
Accurate within \(0.5\) for \(-2.6 < x < 8.6\); accurate within \(0.1\) for \(-1.1 < x < 3.9\).
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Explanation
Accuracy to within \(0.5\) means \(|\sqrt{x + 3} - \left(\dfrac{7}{4} + \dfrac{x}{4}\right)| < 0.5\), equivalently \(\sqrt{x + 3} - 0.5 < \dfrac{7}{4} + \dfrac{x}{4} < \sqrt{x + 3} + 0.5\). Graphing \(y = \sqrt{x + 3} \pm 0.5\) together with the tangent line \(y = \dfrac{7}{4} + \dfrac{x}{4}\) shows that the line lies between the two curves when approximately \(-2.66 < x < 8.66\). Rounding inward gives \(-2.6 < x < 8.6\). Similarly, requiring accuracy within \(0.1\) gives approximately \(-1.1 < x < 3.9\).
53
Example - Differentials vs Increment
Error Rate 100%
Wrong
Compare the values of \(\Delta y\) and \(d y\) if \(y = f(x) = x^3 + x^2 - 2x + 1\) and \(x\) changes (a) from 2 to 2.05 and (b) from 2 to 2.01.
(No answer submitted)
Answer
(a) \(\Delta y = 0.717625\), \(d y = 0.7\). (b) \(\Delta y = 0.140701\), \(d y = 0.14\). The approximation \(\Delta y \approx d y\) improves as \(\Delta x\) decreases.
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Explanation
We have \(f(2) = 8 + 4 - 4 + 1 = 9\) and \(d y = f'(x) d x = (3 x^2 + 2 x - 2) d x\). At \(x = 2\), \(d y = [3(4) + 4 - 2] d x = 14 d x\). (a) \(f(2.05) = (2.05)^3 + (2.05)^2 - 2(2.05) + 1 = 9.717625\), so \(\Delta y = f(2.05) - f(2) = 0.717625\). With \(d x = 0.05\): \(d y = 14(0.05) = 0.7\). (b) \(f(2.01) = (2.01)^3 + (2.01)^2 - 2(2.01) + 1 = 9.140701\), so \(\Delta y = 0.140701\). With \(d x = 0.01\): \(d y = 14(0.01) = 0.14\). Note \(d y\) is much easier to compute than \(\Delta y\), and the approximation \(\Delta y \approx d y\) becomes better as \(\Delta x\) becomes smaller.
54
Example - Error Estimation with Differentials
Error Rate 100%
Wrong
The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?
(No answer submitted)
Answer
Maximum error \(\approx 277 \text{cm}^3\) (relative error \(\approx 0.7%\)).
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Explanation
The sphere has volume \(V = \left(\dfrac{4}{3}\right) \pi r^3\), so \(d V = 4 \pi r^2 d r\). With \(r = 21\) and \(d r = 0.05\): \(d V = 4 \pi (21)^2 (0.05) = 4 \pi (441)(0.05) \approx 277 \text{cm}^3\). The relative error is \((\Delta V)/V \approx (d V)/V = \dfrac{4 \pi r^2 d r}{\left(\dfrac{4}{3}\right) \pi r^3} = 3 (d r)/r = 3(0.05)/21 \approx 0.0071\), so the percentage error in the volume is about \(0.7%\) (compared to \(0.24%\) in the radius).