Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #461 : Derivative Review

Define \displaystyle f (x,y) = e ^{x^{2}+ y^{2}}.

Find \displaystyle f _{xy} (x,y).

Possible Answers:

\displaystyle f _{xy} (x,y)= 4xy e ^{x^{2}+y^{2}}

\displaystyle f _{xy} (x,y)= xy e ^{x^{2}+y^{2}}

\displaystyle f _{xy} (x,y)=2 xy e ^{x^{2}+y^{2}}

\displaystyle f _{xy} (x,y)= xe ^{x^{2}+y^{2}} + y e ^{x^{2}+y^{2}}

\displaystyle f _{xy} (x,y)=2xe ^{x^{2}+y^{2}} + 2y e ^{x^{2}+y^{2}}

Correct answer:

\displaystyle f _{xy} (x,y)= 4xy e ^{x^{2}+y^{2}}

Explanation:

First, find \displaystyle f_{x} = \frac{\partial f}{\partial x} as follows:

\displaystyle f (x,y) = e ^{x^{2}+ y^{2}} = e ^{x^{2} } e ^{y^{2} }

\displaystyle \frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \left (e ^{x^{2} } e ^{y^{2} } \right )

\displaystyle = 2x e ^{x^{2}} e ^{y^{2}}

Therefore, \displaystyle f_{x} (x,y ) = 2x e ^{x^{2}} e ^{y^{2}}.

Now, find \displaystyle f_{xy} = \frac{\partial }{\partial y} f_{x} as follows:

\displaystyle f_{x} (x,y )= 2x e ^{x^{2}} e ^{y^{2}}

\displaystyle \frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y} \left ( 2x e ^{x^{2}} e ^{y^{2}} \right )

\displaystyle =2x e ^{x^{2}} \frac{\partial }{\partial y} \left ( e ^{y^{2}} \right )

\displaystyle =2x e ^{x^{2}} \frac{\partial }{\partial y} \left ( y^{2} \right ) e ^{y^{2}}

\displaystyle =2x e ^{x^{2}} \cdot 2y \cdot e ^{y^{2}}

\displaystyle =4xy e ^{x^{2}} e ^{y^{2}} = 4xy e ^{x^{2}+y^{2}}

Example Question #461 : Derivatives

Define \displaystyle f (x,y) = \cos (x^{2} + y^{2}).

Find \displaystyle f _{xy} (x,y).

Possible Answers:

\displaystyle f _{xy} (x,y) = 4x \cos (x^{2} + y^{2}) + 4y \cos (x^{2} + y^{2})

\displaystyle f _{xy} (x,y) =- 4x \cos (x^{2} + y^{2}) + 4y \cos (x^{2} + y^{2})

\displaystyle f _{xy} (x,y) =- 4xy \cos (x^{2} + y^{2})

\displaystyle f _{xy} (x,y) = 4xy \cos (x^{2} + y^{2})

\displaystyle f _{xy} (x,y) = 4x \cos (x^{2} + y^{2}) - 4y \cos (x^{2} + y^{2})

Correct answer:

\displaystyle f _{xy} (x,y) =- 4xy \cos (x^{2} + y^{2})

Explanation:

First, find \displaystyle f_{x} = \frac{\partial f}{\partial x} as follows:

\displaystyle f (x,y) = \cos (x^{2} + y^{2})

\displaystyle \frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \cos (x^{2} + y^{2})

\displaystyle = \frac{\partial }{\partial x}(x^{2} + y^{2}) \cdot \left [ - \sin (x^{2} + y^{2}) \right ]

\displaystyle = (2x + 0) \cdot \left [ - \sin (x^{2} + y^{2}) \right ]

\displaystyle = - 2x\sin (x^{2} + y^{2})

Therefore, \displaystyle f_{x} (x,y ) = - 2x\sin (x^{2} + y^{2}).

Now, find \displaystyle f_{xy} = \frac{\partial }{\partial y} f_{x} as follows:

\displaystyle f_{x} (x,y ) = - 2x\sin (x^{2} + y^{2})

\displaystyle \frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left [ - 2x\sin (x^{2} + y^{2}) \right ]

\displaystyle =- 2x \frac{\partial }{\partial y}\left [ \sin (x^{2} + y^{2}) \right ]

\displaystyle =- 2x\cdot \frac{\partial }{\partial y}(x^{2} + y^{2}) \cdot \cos (x^{2} + y^{2})

\displaystyle =- 2x\cdot (0 + 2y) \cdot \cos (x^{2} + y^{2})

\displaystyle =- 2x\cdot ( 2y) \cdot \cos (x^{2} + y^{2})

\displaystyle =- 4xy \cos (x^{2} + y^{2})

Example Question #1591 : Calculus Ii

Define \displaystyle f (x,y) = \log xy.

Find \displaystyle f _{xy} (x,y).

Possible Answers:

\displaystyle f _{xy} (x,y) = \frac{x+y}{x y}

\displaystyle f _{xy} (x,y) = 1

\displaystyle f _{xy} (x,y) = \frac{1}{x+y}

\displaystyle f _{xy} (x,y) = 0

\displaystyle f _{xy} (x,y) = \frac{1}{xy}

Correct answer:

\displaystyle f _{xy} (x,y) = 0

Explanation:

First, find \displaystyle f_{x} = \frac{\partial f}{\partial x} as follows:

\displaystyle f (x,y) = \log xy = \log x + \log y = \frac{ \ln x }{\ln 10}+ \frac{ \ln y }{\ln 10}

\displaystyle f (x,y) = \frac{ 1}{\ln 10}\left ( \ln x+ \ln y \right )

\displaystyle \frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x}\left [ \frac{ 1}{\ln 10}\left ( \ln x+ \ln y \right ) \right ]

\displaystyle = \frac{ 1}{\ln 10} \left ( \frac{1}{x}+0 \right )

\displaystyle = \frac{ 1}{\ln 10} \cdot \frac{1}{x}

Therefore, \displaystyle f_{x} (x,y )= \frac{ 1}{\ln 10} \cdot \frac{1}{x}.

Now, find \displaystyle f_{xy} = \frac{\partial }{\partial y} f_{x} as follows:

\displaystyle f_{x} (x,y )= \frac{ 1}{\ln 10} \cdot \frac{1}{x}

\displaystyle \frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left ( \frac{ 1}{\ln 10} \cdot \frac{1}{x} \right )

\displaystyle =0

Example Question #462 : Derivatives

Let the initial approximation of a solution of the equation

\displaystyle e^{x} = \sin x

be \displaystyle x_{1} = -2.

Use one iteration of Newton's method to find an approximation for \displaystyle x_{2}. Give your answer to the nearest thousandth.

Possible Answers:

\displaystyle -4.095

\displaystyle -3.795

\displaystyle -4.195

\displaystyle -3.695

\displaystyle -3.895

Correct answer:

\displaystyle -3.895

Explanation:

Rewrite the equation to be solved for \displaystyle x as \displaystyle e^{x} - \sin x = 0.

Let \displaystyle f(x ) = e^{x} - \sin x

\displaystyle f'(x ) = \frac{\mathrm{d} }{\mathrm{d} x} \left (e^{x} - \sin x \right )

The problem amounts to finding a zero of \displaystyle f(x ). By Newton's method, the second approximation can be derived from the first using the equation

\displaystyle x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}.

Since \displaystyle x_{1} = -2

\displaystyle f(-2 ) = e^{-2} - \sin (-2) \approx 0.1353 - (-0.9093 ) \approx 1.0446 

and

\displaystyle f'(-2 ) = e^{-2} - \cos (-2) \approx 0.1353 - \left ( -0.4161 \right )\approx 0.5514

Use these to find the approximation:

\displaystyle x_{2} =-2 - \frac{f(-2)}{f'(-2)} \approx -2 - \frac{1.0446}{0.5514}\approx -2 -1.8945 \approx -3.895

Example Question #467 : Derivative Review

Define \displaystyle f(x) = e^x- 4x.

Give the minimum value of \displaystyle f on the set of all real numbers.

Possible Answers:

\displaystyle 4 - 4 \ln 4

\displaystyle 2- 4 \ln 2

The function has no miminum value.

\displaystyle e^4 - 16

\displaystyle e^2 - 4

Correct answer:

\displaystyle 4 - 4 \ln 4

Explanation:

This function is continuous and differentiable everywhere. First we find the value(s) of \displaystyle x for which \displaystyle f'(x) =0.

\displaystyle f(x) = e^x- 4x

\displaystyle f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left ( e^x- 4x \right )

\displaystyle f'(x) = e^x- 4

 

\displaystyle f'(x) =0

\displaystyle e^x- 4 =0

\displaystyle e^x= 4

\displaystyle \ln e^x= \ln 4

\displaystyle x= \ln 4

 

Therefore, this is the only possible minimum. We determine whether it is a minimum by evaluating \displaystyle f''(\ln 4):

\displaystyle f'(x) = e^x- 4

\displaystyle f''(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^x- 4 \right )

\displaystyle f''(x) = e^x

\displaystyle f''(\ln 4) = e^{\ln 4} = 4 > 0

Since \displaystyle f''(\ln 4) > 0\displaystyle f has its minimum value at \displaystyle x= \ln 4; it is

\displaystyle f( \ln 4) = e^{ \ln 4}- 4 \cdot \ln 4 = 4 - 4 \ln 4.

Example Question #1592 : Calculus Ii

Consider the equation

\displaystyle y =\ln \left ( \sin x + \cos y \right ).

Which of the following is equal to \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} ?

Possible Answers:

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \sin x}{ \sin x + \cos x + \cos y }

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y + \cos y }

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \cos x + \cos y }

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \sin x}{ \sin x + \sin y + \cos y }

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \sin x + \cos x}{ \sin y + \cos y }

Correct answer:

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y + \cos y }

Explanation:

\displaystyle y =\ln \left ( \sin x + \cos y \right )

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( \sin x + \cos y \right )

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x}= \frac{\frac{\mathrm{d} }{\mathrm{d} x} \left ( \sin x + \cos y \right )}{\sin x + \cos y}

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} \left ( \sin x + \cos y \right )= \frac{\mathrm{d} }{\mathrm{d} x} \left ( \sin x + \cos y \right )

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} \left (\sin x + \sin y+ \cos y \right )= \cos x

\displaystyle \frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y+ \cos y }

Example Question #1593 : Calculus Ii

Define \displaystyle f(x) = 4x + 8 \sin x.

Give the minimum value of \displaystyle f(x) on the interval  \displaystyle [-\pi, \pi].

Possible Answers:

\displaystyle -\frac{4}{3} \pi - 4

\displaystyle 0

\displaystyle -8

\displaystyle -\frac{8}{3} \pi - 4 \sqrt {3}

\displaystyle -4 \pi

Correct answer:

\displaystyle -\frac{8}{3} \pi - 4 \sqrt {3}

Explanation:

\displaystyle f(x) = 4x + 8 \sin x

\displaystyle f'(x) = 4 + 8 \cos x

We first look for \displaystyle x such that \displaystyle f'(x) = 0:

\displaystyle 4 + 8 \cos x = 0

\displaystyle 8 \cos x = -4

\displaystyle \cos x = -\frac{1}{2}

The two values on the interval \displaystyle [-\pi, \pi] for which this holds true are \displaystyle -\frac{2}{3} \pi , \frac{2}{3} \pi, so we evaluate \displaystyle f for the values \displaystyle - \pi, -\frac{2}{3} \pi , \frac{2}{3} \pi, \pi:

\displaystyle f(-\pi ) = 4(-\pi ) + 8 \sin (-\pi ) = -4 \pi + 8 \cdot 0 = -4 \pi \approx -12. 5

     \displaystyle = -\frac{8}{3} \pi + 8 \cdot \left (- \frac{\sqrt{3}}{2} \right ) = -\frac{8}{3} \pi - 4 \sqrt {3} \approx -15.3

     \displaystyle = \frac{8}{3} \pi + 8 \cdot \left ( \frac{\sqrt{3}}{2} \right ) = \frac{8}{3} \pi + 4 \sqrt {3} \approx 15.3

\displaystyle f(\pi ) = 4\pi + 8 \sin \pi = 4 \pi + 8 \cdot 0 = 4 \pi \approx 12. 5

The minimum value is .

Example Question #1594 : Calculus Ii

Consider the equation

\displaystyle x^2 + y^2 = e^ { xy }.

Which of the following is equal to \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} ?

Possible Answers:

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x+ y e^ { xy } }{2y- x e^ { xy }}

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x+ 2y }{x e^ { xy }- y e^ { xy }}

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x+ 2y }{y e^ { xy }- x e^ { xy }}

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x- y e^ { xy } }{2y- x e^ { xy }}

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}

Correct answer:

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}

Explanation:

\displaystyle x^2 + y^2 = e^ { xy }

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^2 + y^2 \right )= \frac{\mathrm{d} }{\mathrm{d} x}e^ { xy }

\displaystyle 2x + 2y \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} xy \cdot e^ { xy }

\displaystyle 2x + 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} = \left ( x \cdot \frac{\mathrm{d}y }{\mathrm{d} x} + y \right ) \cdot e^ { xy }

\displaystyle 2x + 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} = x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} + y e^ { xy }

\displaystyle 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} - x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} =-2x+ y e^ { xy }

\displaystyle 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} - x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} =-2x+ y e^ { xy }

\displaystyle \left (2y- x e^ { xy } \right )\cdot \frac{\mathrm{d} y}{\mathrm{d} x} =-2x+ y e^ { xy }

\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}

Example Question #471 : Derivative Review

Define \displaystyle f(x) = \frac{x}{2}+ \sin x.

Give the maximum value of \displaystyle f(x) on the interval \displaystyle [0, \pi].

Possible Answers:

\displaystyle \frac{\pi + 2\sqrt{2}}{4}

\displaystyle \frac{ \pi + 3\sqrt{3} }{6}

\displaystyle \frac{3\pi + 2\sqrt{2}}{4}

\displaystyle \frac{\pi}{2}

\displaystyle \frac{ 2\pi + 3\sqrt{3} }{6}

Correct answer:

\displaystyle \frac{ 2\pi + 3\sqrt{3} }{6}

Explanation:

First, we determine if there are any points at which \displaystyle f'(x) = 0.

\displaystyle f(x) = \frac{x}{2}+ \sin x = \frac{1}{2}x+ \sin x

\displaystyle f'(x) = \frac{1}{2} + \cos x

 

\displaystyle f'(x) = 0

\displaystyle \frac{1}{2} + \cos x = 0

\displaystyle \cos x = -\frac{1}{2}

The only point on the interval on which this is true is \displaystyle x =\frac{ 2\pi }{3}.

We test this point as well as the two endpoints, \displaystyle x = 0 and \displaystyle x = \pi, by evaluating \displaystyle f for each of these values.

 

\displaystyle f(0) = \frac{0}{2}+ \sin 0 = 0 + 0 = 0

\displaystyle f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \left ( \frac{ 2\pi }{3} \right )}{2}+ \sin \frac{ 2\pi }{3} = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} \approx 2

\displaystyle f( \pi ) = \frac{ \pi}{2}+ \sin \pi = \frac{ \pi}{2} + 0 = \frac{ \pi}{2} \approx 1.6

Therefore, \displaystyle f assumes its maximum on this interval at the point \displaystyle x =\frac{ 2\pi }{3}, and \displaystyle f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} = \frac{ 2\pi + 3\sqrt{3} }{6}.

Example Question #1591 : Calculus Ii

Consider the equation

\displaystyle \sec y = x^2 + xy^2.

Which of the following is equal to \displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} ?

Possible Answers:

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y + 2xy}

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec ^2 y - 2xy}

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x - y^2}{ \sec y \tan y - 2xy}

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x - y^2}{ \sec ^2 y - 2xy}

Correct answer:

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}

Explanation:

\displaystyle \sec y = x^2 + xy^2

\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \sec y =\frac{\mathrm{d} }{\mathrm{d} x} \left (x^2 + xy^2 \right )

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y=2x + y^2 + x \cdot 2y \cdot \frac{\mathrm{d}y }{\mathrm{d} x}

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y=2x + y^2 + 2xy \frac{\mathrm{d}y }{\mathrm{d} x}

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y - \frac{\mathrm{d}y }{\mathrm{d} x} 2xy=2x + y^2

\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}

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