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Stewart 9th Section 2.6: Implicit Differentiation
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1
Implicit Diff - Basic
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오답
(a) Find \(y'\) by implicit differentiation.
(b) Solve the equation explicitly for \(y\) and differentiate to get \(y'\) in terms of \(x\).
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a).
\(5x^2 - y^3 = 7\)
(미작성)
정답
(a) \(y' = \dfrac{10x}{3y^2}\)
(b) \(y = (5x^2 - 7)^{\dfrac{1}{3}}\), \(y' = \dfrac{10x}{3}(5x^2 - 7)^{-\dfrac{2}{3}}\)
(c) Substituting \(y = (5x^2 - 7)^{\dfrac{1}{3}}\) into \(\dfrac{10x}{3y^2}\) gives the same result.
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2
Implicit Diff - Basic
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오답
(a) Find \(y'\) by implicit differentiation.
(b) Solve the equation explicitly for \(y\) and differentiate to get \(y'\) in terms of \(x\).
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a).
\(6x^4 + y^5 = 2x\)
(미작성)
정답
(a) \(y' = \dfrac{2 - 24x^3}{5y^4}\)
(b) \(y = (2x - 6x^4)^{\dfrac{1}{5}}\), \(y' = \dfrac{2 - 24x^3}{5}(2x - 6x^4)^{-\dfrac{4}{5}}\)
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3
Implicit Diff - Basic
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오답
(a) Find \(y'\) by implicit differentiation.
(b) Solve the equation explicitly for \(y\) and differentiate to get \(y'\) in terms of \(x\).
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a).
\(\sqrt{x} + \sqrt{y} = 1\)
(미작성)
정답
(a) \(y' = -\dfrac{\sqrt{y}}{\sqrt{x}} = -\sqrt{\dfrac{y}{x}}\)
(b) \(y = (1 - \sqrt{x})^2\), \(y' = -\dfrac{1 - \sqrt{x}}{\sqrt{x}}\)
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4
Implicit Diff - Basic
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오답
(a) Find \(y'\) by implicit differentiation.
(b) Solve the equation explicitly for \(y\) and differentiate to get \(y'\) in terms of \(x\).
(c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a).
\(\dfrac{2}{x} - \dfrac{1}{y} = 4\)
(미작성)
정답
(a) \(y' = -\dfrac{y^2}{x^2} \cdot \dfrac{2}{1} = -\dfrac{2y^2}{x^2}\)
(b) \(y = \dfrac{x}{4x - 2}\), differentiate using quotient rule.
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5
Implicit Diff - Intermediate
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오답
\( x^2 - 4x y + y^2 = 4 \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{4y - 2x}{2y - 4x} = \dfrac{2y - x}{y - 2x}\)
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6
Implicit Diff - Intermediate
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오답
\( 2x^2 + x y - y^2 = 2 \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{-(4x + y)}{x - 2y} = \dfrac{4x + y}{2y - x}\)
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7
Implicit Diff - Intermediate
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오답
\( x^4 + x^2 y^2 + y^3 = 5 \)
(미작성)
정답
\(\dfrac{d y}{d x} = -\dfrac{4x^3 + 2x y^2}{2x^2 y + 3y^2}\)
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8
Implicit Diff - Intermediate
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오답
\( x^3 - x y^2 + y^3 = 1 \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{y^2 - 3x^2}{3y^2 - 2x y}\)
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9
Implicit Diff - Intermediate
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오답
\( \dfrac{x^2}{x + y} = y^2 + 1 \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{2x(x + y) - x^2}{(x + y)^2 \cdot (-1) + 2y(x+y)^2} \). Use quotient rule on the left side, then solve for \(y'\).
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10
Implicit Diff - Intermediate
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오답
\( y^5 + x^2 y^3 = 1 + x^4 y \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{4x^3 y - 2x y^3}{5y^4 + 3x^2 y^2 - x^4}\)
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11
Implicit Diff - Intermediate
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오답
\( \sin x + \cos y = 2x - 3y \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{2 - \cos x}{3 - \sin y}\)
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12
Implicit Diff - Intermediate
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오답
\( y \sin(x^2) = x \sin(y^2) \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{\sin(y^2) - 2x y \cos(x^2)}{\sin(x^2) - 2x y \cos(y^2)}\)
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13
Implicit Diff - Intermediate
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오답
\( \sin(x + y) = \cos x + \cos y \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{\sin x + \cos(x + y)}{\sin y - \cos(x + y)} = -\dfrac{\cos(x + y) + \sin x}{\cos(x + y) - \sin y}\)
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14
Implicit Diff - Intermediate
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오답
\( \tan(x - y) = 2x y^3 + 1 \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{\sec^2(x - y) - 2y^3}{\sec^2(x - y) + 6x y^2}\). Differentiate noting \(\dfrac{d}{d x}[\tan(x-y)] = \sec^2(x-y)(1 - y')\).
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15
Implicit Diff - Intermediate
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오답
\( \tan\left(\dfrac{x}{y}\right) = x + y \)
(미작성)
정답
Use \(\dfrac{d}{d x}[\tan\left(\dfrac{x}{y}\right)] = \sec^2\left(\dfrac{x}{y}\right) \cdot \dfrac{y - x y'}{y^2}\) and set equal to \(1 + y'\). Solve for \(y'\).
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16
Implicit Diff - Intermediate
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오답
\( \sin(x y) = \cos(x + y) \)
(미작성)
정답
\(\dfrac{d y}{d x} = -\dfrac{y \cos(x y) + \sin(x + y)}{x \cos(x y) + \sin(x + y)}\)
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17
Implicit Diff - Intermediate
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오답
\( \sqrt{x + y} = x^4 + y^4 \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{4x^3 - \dfrac{1}{2\sqrt{x + y}}}{\dfrac{1}{2\sqrt{x + y}} - 4y^3}\)
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18
Implicit Diff - Intermediate
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오답
\( \sin x \cos y = x^2 - 5y \)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{2x - \cos x \cos y}{5 - \sin x \sin y}\)
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19
Implicit Diff - Intermediate
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오답
\( \sqrt{x y} = 1 + x^2 y \)
(미작성)
정답
Differentiate: \(\dfrac{y + x y'}{2\sqrt{x y}} = 2x y + x^2 y'\). Solve for \(y'\).
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20
Implicit Diff - Intermediate
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오답
\( x y = \sqrt{x^2 + y^2} \)
(미작성)
정답
Differentiate: \(y + x y' = \dfrac{2x + 2y y'}{2\sqrt{x^2 + y^2}} = \dfrac{x + y y'}{\sqrt{x^2 + y^2}}\). Solve for \(y'\).
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21
Implicit Diff - Given Values
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오답
If \(f(x) + x^2 [f(x)]^3 = 10\) and \(f(1) = 2\), find \(f'(1)\).
(미작성)
정답
Differentiate: \(f'(x) + 2x[f(x)]^3 + 3x^2[f(x)]^2 f'(x) = 0\). At \(x = 1\): \(f'(1) + 2(8) + 3(4)f'(1) = 0\), so \(13 f'(1) = -16\), \(f'(1) = -\dfrac{16}{13}\).
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22
Implicit Diff - Given Values
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오답
If \(g(x) + x \sin g(x) = x^2\), find \(g'(0)\).
(미작성)
정답
At \(x = 0\): \(g(0) = 0\). Differentiate: \(g'(x) + \sin g(x) + x \cos g(x) \cdot g'(x) = 2x\). At \(x = 0\): \(g'(0) + 0 + 0 = 0\), so \(g'(0) = 0\).
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23
Implicit Diff - dx/dy
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오답
Regard \(y\) as the independent variable and \(x\) as the dependent variable and use implicit differentiation to find \(\dfrac{d x}{d y}\).
\(x^4 y^2 - x^3 y + 2x y^3 = 0\)
(미작성)
정답
Differentiate with respect to \(y\): \(4x^3 x' y^2 + 2x^4 y - 3x^2 x' y - x^3 + 2x' y^3 + 6x y^2 = 0\). Solve for \(x' = \dfrac{d x}{d y}\).
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24
Implicit Diff - dx/dy
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오답
Regard \(y\) as the independent variable and \(x\) as the dependent variable and use implicit differentiation to find \(\dfrac{d x}{d y}\).
\(y \sec x = x \tan y\)
(미작성)
정답
Differentiate with respect to \(y\): \(\sec x + y \sec x \tan x \cdot x' = x' \tan y + x \sec^2 y\). Solve for \(x' = \dfrac{d x}{d y}\).
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25
Implicit Diff - Tangent Lines
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오답
\(y \sin 2x = x \cos 2y\), \(\quad \left(\dfrac{\pi}{2}, \dfrac{\pi}{4}\right)\)
(미작성)
정답
Differentiate: \(y' \sin 2x + 2y \cos 2x = \cos 2y - 2x \sin 2y \cdot y'\). At \(\left(\dfrac{\pi}{2}, \dfrac{\pi}{4}\right)\): evaluate and solve for \(y'\), then write tangent line equation.
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26
Implicit Diff - Tangent Lines
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오답
\(\tan(x + y) + \sec(x - y) = 2\), \(\quad \left(\dfrac{\pi}{8}, \dfrac{\pi}{8}\right)\)
(미작성)
정답
At \(\left(\dfrac{\pi}{8}, \dfrac{\pi}{8}\right)\): \(\tan\left(\dfrac{\pi}{4}\right) + \sec(0) = 1 + 1 = 2\) checks. Differentiate and evaluate to find the tangent line.
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27
Implicit Diff - Tangent Lines
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오답
\(x^{\dfrac{2}{3}} + y^{\dfrac{2}{3}} = 4\), \(\quad (-3\sqrt{3}, 1)\) (astroid)
(미작성)
정답
\(\dfrac{d y}{d x} = -\left(\dfrac{y}{x}\right)^{\dfrac{1}{3}}\). At \((-3\sqrt{3}, 1)\): \(y' = -\left(\dfrac{1}{-3\sqrt{3}}\right)^{\dfrac{1}{3}}\). Write tangent line.
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28
Implicit Diff - Tangent Lines
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오답
\(y^2(6 - x) = x^3\), \(\quad (2, \sqrt{2})\) (cissoid of Diocles)
(미작성)
정답
Differentiate: \(2y y'(6 - x) + y^2(-1) = 3x^2\). At \((2, \sqrt{2})\): solve for \(y'\) and write tangent line equation.
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29
Implicit Diff - Tangent Lines
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오답
\(x^2 - x y - y^2 = 1\), \(\quad (2, 1)\) (hyperbola)
(미작성)
정답
\(\dfrac{d y}{d x} = \dfrac{2x - y}{x + 2y}\). At \((2, 1)\): \(y' = \dfrac{4 - 1}{2 + 2} = \dfrac{3}{4}\). Tangent: \(y - 1 = \dfrac{3}{4}(x - 2)\).
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30
Implicit Diff - Tangent Lines
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오답
\(x^2 + 2x y + 4y^2 = 12\), \(\quad (2, 1)\) (ellipse)
(미작성)
정답
\(\dfrac{d y}{d x} = -\dfrac{2x + 2y}{2x + 8y} = -\dfrac{x + y}{x + 4y}\). At \((2, 1)\): \(y' = -\dfrac{3}{6} = -\dfrac{1}{2}\). Tangent: \(y - 1 = -\dfrac{1}{2}(x - 2)\).
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31
Implicit Diff - Tangent Lines
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오답
\(x^2 + y^2 = (2x^2 + 2y^2 - x)^2\), \(\quad \left(0, \dfrac{1}{2}\right)\) (cardioid)
(미작성)
정답
Differentiate both sides using chain rule. At \(\left(0, \dfrac{1}{2}\right)\): the curve satisfies \(\dfrac{1}{4} = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\). Find \(y'\) and write tangent line.
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32
Implicit Diff - Tangent Lines
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오답
\(x^2 y^2 = (y + 1)^2(4 - y^2)\), \(\quad (2\sqrt{3}, 1)\) (conchoid of Nicomedes)
(미작성)
정답
Differentiate both sides implicitly. At \((2\sqrt{3}, 1)\): evaluate and solve for \(y'\), then write tangent line equation.
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33
Implicit Diff - Tangent Lines
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오답
\(2(x^2 + y^2)^2 = 25(x^2 - y^2)\), \(\quad (3, 1)\) (lemniscate)
(미작성)
정답
Differentiate: \(4(x^2 + y^2)(2x + 2y y') = 25(2x - 2y y')\). At \((3, 1)\): \(4(10)(6 + 2y') = 25(6 - 2y')\). Solve for \(y'\) and write tangent line.
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34
Implicit Diff - Tangent Lines
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오답
\(y^2(y^2 - 4) = x^2(x^2 - 5)\), \(\quad (0, -2)\) (devil's curve)
(미작성)
정답
Differentiate: \(2y y'(y^2 - 4) + y^2(2y y') = 2x(x^2 - 5) + x^2(2x)\), i.e., \(2y y'(2y^2 - 4) = 2x(2x^2 - 5)\). At \((0, -2)\): \(y' = 0\). Tangent: \(y = -2\).
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35
Implicit Diff - Tangent Lines
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오답
(a) The curve with equation \(y^2 = 5x^4 - x^2\) is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point \((1, 2)\).
(b) Illustrate part (a) by graphing the curve and the tangent line on a common screen.
(미작성)
정답
(a) Differentiate: \(2y y' = 20x^3 - 2x\). At \((1, 2)\): \(4y' = 18\), so \(y' = \dfrac{9}{2}\). Tangent: \(y - 2 = \dfrac{9}{2}(x - 1)\).
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36
Implicit Diff - Tangent Lines
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오답
(a) The curve with equation \(y^2 = x^3 + 3x^2\) is called a Tschirnhausen cubic. Find an equation of the tangent line to this curve at the point \((1, -2)\).
(b) At what points does this curve have horizontal tangents?
(c) Illustrate parts (a) and (b) by graphing the curve and the tangent lines on a common screen.
(미작성)
정답
(a) \(2y y' = 3x^2 + 6x\). At \((1, -2)\): \(-4y' = 9\), \(y' = -\dfrac{9}{4}\). Tangent: \(y + 2 = -\dfrac{9}{4}(x - 1)\).
(b) Horizontal tangents where \(3x^2 + 6x = 0\), i.e., \(x = 0\) or \(x = -2\). At \(x = -2\): \(y^2 = -8 + 12 = 4\), \(y = \pm 2\). At \(x = 0\): \(y = 0\).
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37
Implicit Diff - Second Derivative
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오답
Find \(y''\) by implicit differentiation. Simplify where possible.
\(x^2 + 4y^2 = 4\)
(미작성)
정답
\(y' = -\dfrac{x}{4y}\). \(y'' = -\dfrac{4y - x \cdot 4y'}{16y^2} = -\dfrac{4y + \dfrac{x^2}{y}}{16y^2} = -\dfrac{4y^2 + x^2}{16y^3} = -\dfrac{4}{16y^3} = -\dfrac{1}{4y^3}\).
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38
Implicit Diff - Second Derivative
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오답
Find \(y''\) by implicit differentiation. Simplify where possible.
\(x^2 + x y + y^2 = 3\)
(미작성)
정답
\(y' = -\dfrac{2x + y}{x + 2y}\). Differentiate again using quotient rule and substitute \(y'\) to find \(y''\). Use \(x^2 + x y + y^2 = 3\) to simplify: \(y'' = -\dfrac{18}{(x + 2y)^3}\).
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39
Implicit Diff - Second Derivative
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오답
Find \(y''\) by implicit differentiation. Simplify where possible.
\(\sin y + \cos x = 1\)
(미작성)
정답
\(y' = \dfrac{\sin x}{\cos y}\). \(y'' = \dfrac{\cos x \cos y - \sin x (-\sin y) y'}{\cos^2 y} = \dfrac{\cos x \cos y + \sin^2 x \sin y}{\cos^3 y}\). Can use \(\sin y = 1 - \cos x\) to simplify further.
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40
Implicit Diff - Second Derivative
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오답
Find \(y''\) by implicit differentiation. Simplify where possible.
\(x^3 - y^3 = 7\)
(미작성)
정답
\(y' = \dfrac{x^2}{y^2}\). \(y'' = \dfrac{2x y^2 - x^2 \cdot 2y y'}{y^4} = \dfrac{2x y - 2x^2 y'}{y^3} = \dfrac{2x y - \dfrac{2x^4}{y^2}}{y^3} = -\dfrac{2x(x^3 - y^3)}{y^5} = -\dfrac{14x}{y^5}\).
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41
Implicit Diff - Second Derivative
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오답
If \(x y + y^3 = 1\), find the value of \(y''\) at the point where \(x = 0\).
(미작성)
정답
At \(x = 0\): \(y^3 = 1\), so \(y = 1\). \(y' = -\dfrac{y}{x + 3y^2}\). At \((0, 1)\): \(y' = -\dfrac{1}{3}\). Differentiate again to find \(y''\) at \((0, 1)\).
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42
Implicit Diff - Second Derivative
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오답
If \(x^2 + x y + y^3 = 1\), find the value of \(y'''\) at the point where \(x = 1\).
(미작성)
정답
At \(x = 1\): \(1 + y + y^3 = 1\), so \(y + y^3 = 0\), giving \(y = 0\). Find \(y'\), \(y''\), and \(y'''\) successively at \((1, 0)\).
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43
Implicit Diff - Curves
오답률 100%
오답
Fanciful shapes can be created by using the implicit plotting capabilities of computer algebra systems.
(a) Graph the curve with equation \(y(y^2 - 1)(y - 2) = x(x - 1)(x - 2)\). At how many points does this curve have horizontal tangents? Estimate the \(x\)-coordinates of these points.
(b) Find equations of the tangent lines at the points \((0, 1)\) and \((0, 2)\).
(c) Find the exact \(x\)-coordinates of the points in part (a).
(d) Create even more fanciful curves by modifying the equation in part (a).
(미작성)
정답
(b) At \((0, 1)\): implicit differentiation gives \(y'(y^2 - 1)(y - 2) + ...\) evaluated. At \((0, 2)\): similarly.
(c) Horizontal tangents where \(\dfrac{d}{d x}[x(x-1)(x-2)] \neq 0\) but the \(y\)-factor equation gives \(y' = 0\).
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44
Implicit Diff - Curves
오답률 100%
오답
(a) The curve with equation \(2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2\) has been likened to a bouncing wagon. Use a computer algebra system to graph this curve and discover why.
(b) At how many points does this curve have horizontal tangent lines? Find the \(x\)-coordinates of these points.
(미작성)
정답
(b) Horizontal tangents where \(\dfrac{d y}{d x} = 0\), i.e., \(4x^3 - 6x^2 + 2x = 0\), giving \(2x(2x^2 - 3x + 1) = 0\), so \(x = 0\), \(x = \dfrac{1}{2}\), \(x = 1\).
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45
Implicit Diff - Curves
오답률 100%
오답
Find the points on the lemniscate in Exercise 33 where the tangent is horizontal.
(미작성)
정답
From \(2(x^2 + y^2)^2 = 25(x^2 - y^2)\), set \(y' = 0\): this requires \(4(x^2 + y^2)(2x) = 25(2x)\), so \(x^2 + y^2 = \dfrac{25}{4}\) (when \(x \neq 0\)). Combine with original equation to find the points.
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46
Implicit Diff - Ellipse/Hyperbola
오답률 100%
오답
Show that the tangent line to the ellipse
\( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1 \)
at the point \((x_0, y_0)\) has the equation
\( \dfrac{x_0 x}{a^2} + \dfrac{y_0 y}{b^2} = 1 \)
(미작성)
정답
Implicit differentiation gives \(y' = -\dfrac{b^2 x_0}{a^2 y_0}\). Tangent at \((x_0, y_0)\): \(y - y_0 = -\dfrac{b^2 x_0}{a^2 y_0}(x - x_0)\). Multiply through by \(\dfrac{y_0}{b^2}\) and use \(\dfrac{x_0^2}{a^2} + \dfrac{y_0^2}{b^2} = 1\) to simplify.
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47
Implicit Diff - Ellipse/Hyperbola
오답률 100%
오답
Find an equation of the tangent line to the hyperbola
\( \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \)
at the point \((x_0, y_0)\).
(미작성)
정답
By the same method as Exercise 46: \(\dfrac{x_0 x}{a^2} - \dfrac{y_0 y}{b^2} = 1\).
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48
Implicit Diff - Ellipse/Hyperbola
오답률 100%
오답
Show that the sum of the \(x\)- and \(y\)-intercepts of any tangent line to the curve \(\sqrt{x} + \sqrt{y} = \sqrt{c}\) is equal to \(c\).
(미작성)
정답
At \((x_0, y_0)\): \(y' = -\sqrt{\dfrac{y_0}{x_0}}\). Tangent: \(y - y_0 = -\sqrt{\dfrac{y_0}{x_0}}(x - x_0)\). Find \(x\)-intercept (set \(y = 0\)) and \(y\)-intercept (set \(x = 0\)). Their sum equals \((\sqrt{x_0} + \sqrt{y_0})^2 = c\).
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49
Implicit Diff - Geometry
오답률 100%
오답
Show that the tangent line to the circle \(x^2 + y^2 = r^2\) at the point \(P(x_0, y_0)\) is perpendicular to the radius \(O P\).
(미작성)
정답
Slope of \(O P\) is \(\dfrac{y_0}{x_0}\). Implicit differentiation: \(y' = -\dfrac{x_0}{y_0}\). Product of slopes: \(\dfrac{y_0}{x_0} \cdot \left(-\dfrac{x_0}{y_0}\right) = -1\). Therefore perpendicular.
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50
Implicit Diff - Geometry
오답률 100%
오답
The Power Rule can be proved using implicit differentiation for the case where \(n\) is a rational number, \(n = \dfrac{p}{q}\), and \(y = f(x) = x^n\) is assumed beforehand to be a differentiable function. If \(y = x^{\dfrac{p}{q}}\), then \(y^q = x^p\). Use implicit differentiation to show that
\( y' = \dfrac{p}{q} x^{\dfrac{p}{q} - 1} \)
(미작성)
정답
Differentiate \(y^q = x^p\): \(q y^{q-1} y' = p x^{p-1}\). So \(y' = \dfrac{p x^{p-1}}{q y^{q-1}} = \dfrac{p x^{p-1}}{q \left(x^{\dfrac{p}{q}}\right)^{q-1}} = \dfrac{p x^{p-1}}{q x^{p - \dfrac{p}{q}}} = \dfrac{p}{q} x^{\dfrac{p}{q} - 1}\).
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51
Implicit Diff - Orthogonal Trajectories
오답률 100%
오답
Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Show that the given families of curves are orthogonal trajectories of each other; that is, every curve in one family is orthogonal to every curve in the other family. Sketch both families of curves on the same axes.
\(x^2 + y^2 = r^2\), \(\quad a x + b y = 0\)
(미작성)
정답
First family (circles): \(y' = -\dfrac{x}{y}\). Second family (lines through origin): slope \(= -\dfrac{a}{b}\), but \(a x + b y = 0\) means \(\dfrac{y}{x} = -\dfrac{a}{b}\), so slope of line through origin to \((x, y)\) is \(\dfrac{y}{x}\). Product of slopes: \(\left(-\dfrac{x}{y}\right)\left(\dfrac{y}{x}\right) = -1\).
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52
Implicit Diff - Orthogonal Trajectories
오답률 100%
오답
Show that the given families of curves are orthogonal trajectories of each other.
\(x^2 + y^2 = a x\), \(\quad x^2 + y^2 = b y\)
(미작성)
정답
First family: \(2x + 2y y' = a\), but \(a = \dfrac{x^2 + y^2}{x}\), so \(y'_1 = \dfrac{a - 2x}{2y} = \dfrac{y^2 - x^2}{2x y}\). Similarly \(y'_2 = \dfrac{2x y}{x^2 - y^2}\). Product = \(-1\).
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53
Implicit Diff - Orthogonal Trajectories
오답률 100%
오답
Show that the given families of curves are orthogonal trajectories of each other.
\(y = c x^2\), \(\quad x^2 + 2y^2 = k\)
(미작성)
정답
First family: \(y' = 2c x\). Since \(c = \dfrac{y}{x^2}\), \(y'_1 = \dfrac{2y}{x}\). Second family: \(2x + 4y y' = 0\), \(y'_2 = -\dfrac{x}{2y}\). Product: \(\dfrac{2y}{x} \cdot \left(-\dfrac{x}{2y}\right) = -1\).
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54
Implicit Diff - Orthogonal Trajectories
오답률 100%
오답
Show that the given families of curves are orthogonal trajectories of each other.
\(y = a x^3\), \(\quad x^2 + 3y^2 = b\)
(미작성)
정답
First family: \(y' = 3a x^2 = \dfrac{3y}{x}\). Second family: \(2x + 6y y' = 0\), \(y' = -\dfrac{x}{3y}\). Product: \(\dfrac{3y}{x} \cdot \left(-\dfrac{x}{3y}\right) = -1\).
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55
Implicit Diff - Orthogonal Trajectories
오답률 100%
오답
Show that the families of ellipses and hyperbolas
\( \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, \quad \dfrac{x^2}{A^2} - \dfrac{y^2}{B^2} = 1 \)
are orthogonal trajectories of each other if \(a^2 < A^2\) and \(a^2 - b^2 = A^2 + B^2\) (so that the conics have the same foci).
(미작성)
정답
Ellipse: \(y' = -\dfrac{b^2 x}{a^2 y}\). Hyperbola: \(y' = \dfrac{B^2 x}{A^2 y}\). Product: \(-\dfrac{b^2 B^2}{a^2 A^2}\). Using \(a^2 - b^2 = A^2 + B^2\) and the equations at intersection to show the product equals \(-1\).
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56
Implicit Diff - Orthogonal Trajectories
오답률 100%
오답
Find the value of the number \(a\) such that the families of curves \(y = (x + c)^{-1}\) and \(y = a(x + k)^{\dfrac{1}{3}}\) are orthogonal trajectories of each other.
(미작성)
정답
First family: \(y' = -(x + c)^{-2} = -y^2\). Second family: \(y' = \dfrac{a}{3}(x + k)^{-\dfrac{2}{3}} = \dfrac{y}{3(x + k)} = \dfrac{a^3}{3y^2}\) (using \(y^3 = a^3(x + k)\)). For orthogonality: \((-y^2) \cdot \dfrac{a^3}{3y^2} = -1\), so \(a^3 = 3\), \(a = \sqrt[3]{3}\).
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57
Implicit Diff - Applied/Proof
오답률 100%
오답
The Van der Waals equation for \(n\) moles of a gas is
\( \left(P + \dfrac{n^2 a}{V^2}\right)(V - n b) = n R T \)
where \(P\) is the pressure, \(V\) is the volume, and \(T\) is the temperature of the gas. The constant \(R\) is the universal gas constant and \(a\) and \(b\) are positive constants that are characteristic of a particular gas.
(a) Use implicit differentiation to find \(\dfrac{d V}{d P}\).
(b) Find \(\dfrac{d V}{d P}\) for carbon dioxide at a specific volume and pressure.
(미작성)
정답
(a) Differentiate with respect to \(P\): \((V - n b) + \left(P + \dfrac{n^2 a}{V^2}\right)\left(\dfrac{d V}{d P}\right) + \dfrac{-2n^2 a}{V^3} \cdot \dfrac{d V}{d P} \cdot (V - n b) = 0\) (at constant \(T\)). Solve for \(\dfrac{d V}{d P}\).
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58
Implicit Diff - Applied/Proof
오답률 100%
오답
(a) The equation \(x^2 + x y + y^2 + 1 = 0\) is an implicit curve. Use implicit differentiation to find \(y'\).
(b) Plot the curve in part (a). What do you see? Prove it.
(c) Comment on the result of part (a) in light of part (b).
(미작성)
정답
(a) \(y' = -\dfrac{2x + y}{x + 2y}\).
(b) The curve has no real points. \(x^2 + x y + y^2 + 1 = \left(x + \dfrac{y}{2}\right)^2 + \dfrac{3y^2}{4} + 1 \geq 1 > 0\) for all real \((x, y)\).
(c) The derivative formula is meaningless since there are no points on the curve.
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59
Implicit Diff - Applied/Proof
오답률 100%
오답
The equation \(x^2 - x y + y^2 = 3\) represents a rotated ellipse, that is, an ellipse whose axes are not parallel to the coordinate axes. Find the points at which this ellipse crosses the \(x\)-axis and show that the tangent lines at these points are parallel.
(미작성)
정답
On \(x\)-axis: \(y = 0\), so \(x^2 = 3\), \(x = \pm \sqrt{3}\). \(y' = \dfrac{y - 2x}{2y - x}\). At \((\sqrt{3}, 0)\): \(y' = \dfrac{-2\sqrt{3}}{-\sqrt{3}} = 2\). At \((-\sqrt{3}, 0)\): \(y' = \dfrac{2\sqrt{3}}{\sqrt{3}} = 2\). Both slopes equal 2, so tangent lines are parallel.
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60
Implicit Diff - Applied/Proof
오답률 100%
오답
(a) Where does the normal line to the ellipse \(x^2 - x y + y^2 = 3\) at the point \((-1, 1)\) intersect the ellipse a second time?
(b) Illustrate part (a) by graphing the ellipse and the normal line.
(미작성)
정답
(a) At \((-1, 1)\): \(y' = \dfrac{1 - 2(-1)}{2(1) - (-1)} = \dfrac{3}{3} = 1\). Normal slope \(= -1\). Normal line: \(y - 1 = -(x + 1)\), i.e., \(y = -x\). Substitute into \(x^2 - x y + y^2 = 3\): \(x^2 + x^2 + x^2 = 3\), \(x^2 = 1\), \(x = \pm 1\). Second intersection at \((1, -1)\).
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61
Implicit Diff - Applied/Proof
오답률 100%
오답
Find all points on the curve \(x^2 y^2 + x y = 2\) where the slope of the tangent line is \(-1\).
(미작성)
정답
Differentiate: \(2x y^2 + 2x^2 y y' + y + x y' = 0\). Set \(y' = -1\): \(2x y^2 - 2x^2 y + y - x = 0\). Factor and solve simultaneously with \(x^2 y^2 + x y = 2\).
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62
Implicit Diff - Applied/Proof
오답률 100%
오답
Find equations of both tangent lines to the ellipse \(x^2 + 4y^2 = 36\) that pass through the point \((12, 3)\).
(미작성)
정답
Tangent at \((x_0, y_0)\) on ellipse: \(x_0 x + 4 y_0 y = 36\). This passes through \((12, 3)\): \(12 x_0 + 12 y_0 = 36\), i.e., \(x_0 + y_0 = 3\). Combine with \(x_0^2 + 4 y_0^2 = 36\) to find the tangent points.
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63
Implicit Diff - Applied/Proof
오답률 100%
오답
Use implicit differentiation to find \(\dfrac{d y}{d x}\) for the equation \(\dfrac{x}{y} = y^2 + 1\) (where \(y \neq 0\)). Then show that you get the same answer when you perform implicit differentiation on the equivalent equation \(x = y^3 + y\) (where \(y \neq 0\)).
(미작성)
정답
From \(\dfrac{x}{y} = y^2 + 1\): \(\dfrac{y - x y'}{y^2} = 2y y'\). Solve for \(y' = \dfrac{y}{x + 2y^3} = \dfrac{1}{3y^2 + 1}\) (using \(x = y^3 + y\)).
From \(x = y^3 + y\): \(1 = 3y^2 y' + y'\), so \(y' = \dfrac{1}{3y^2 + 1}\). Same result.
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64
Implicit Diff - Applied/Proof
오답률 100%
오답
The Bessel function of order 0, \(y = J(x)\), satisfies the differential equation \(x y'' + y' + x y = 0\) for all values of \(x\) and its value at 0 is \(J(0) = 1\).
(a) Find \(J'(0)\).
(b) Use implicit differentiation of the differential equation to find \(J''(0)\).
(미작성)
정답
(a) At \(x = 0\): \(0 \cdot y''(0) + y'(0) + 0 = 0\), so \(J'(0) = 0\).
(b) Differentiate: \(y'' + x y''' + y'' + y + x y' = 0\). At \(x = 0\): \(2 J''(0) + J(0) = 0\), so \(J''(0) = -\dfrac{1}{2}\).
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65
Implicit Diff - Applied/Proof
오답률 100%
오답
A lamp is located three units to the right of the \(y\)-axis and a shadow is created by the elliptical region \(x^2 + 4y^2 \leq 5\). If the point \((-5, 0)\) is on the edge of the shadow, how far above the \(x\)-axis is the lamp?
(미작성)
정답
The lamp is at \((3, h)\). The shadow edge passes through \((-5, 0)\) tangent to the ellipse \(x^2 + 4y^2 = 5\). Find the tangent line from \((-5, 0)\) to the ellipse, then determine \(h\) so the line from \((3, h)\) through the tangent point reaches \((-5, 0)\).
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