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Calculus 3 : Calculus 3

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Example Questions

Example Question #12 : Multi Variable Chain Rule

Determine $\frac{\mathrm{dg} }{\mathrm{d} t}$ , where g=x^2+yz , x=t , y=e^t , z=4t

Possible Answers:

2t+4e^t+1

2t+4te^t(t+1)

2t+4e^t(t+1)

2x+ze^t+4y

Correct answer:

2t+4e^t(t+1)

Explanation:

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that, for x, $\frac{\mathrm{d} g}{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}$ . (The same rule applies for the other two variables, and we add the results of the three variables to get the total derivative.)

So, the derivatives are

$\frac{\mathrm{d} g}{\mathrm{d} x}=2x$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=1$ , $\frac{\mathrm{d} g}{\mathrm{d} y}=z$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=e^t$ , $\frac{\mathrm{d} g}{\mathrm{d} z}=y$ , $\frac{\mathrm{d} z}{\mathrm{d} t}=4$

Using the above rule, we get

2x(1)+z(e^t)+y(4)

which rewritten in terms of t, and simplified, becomes

2t+4e^t(t+1)

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Example Question #1331 : Calculus 3

Find $\frac{\mathrm{dk} }{\mathrm{d} t}$ , where k=x^2+xy , $x=e^{t^2}$ , $y=4t+\cos(t)$

Possible Answers:

$4te^{t^3}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

$2te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

$4te^{t^2+t}+8t^2e^t+2t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

Correct answer:

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

Explanation:

To find the given derivative of the function, we must use the multivariable chain rule, which states that

$\frac{\mathrm{d}k }{\mathrm{d} t}=\frac{\mathrm{d} k}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} k}{\mathrm{d} y}\cdot \frac{\mathrm{d} y}{\mathrm{d} t}$

So, we find all of these derivatives:

$\frac{\mathrm{d} k}{\mathrm{d} x}=2x+y$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=2te^t$ , $\frac{\mathrm{d} k}{\mathrm{d} y}=x$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=4-\sin(t)$

Plugging this into the formula above, and rewriting x and y in terms of t, we get

$(2e^{t^2}+4t+\cos(t))(2te^t)+e^{t^2}(4-\sin(t))$

which simplified becomes

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

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Example Question #16 : Multi Variable Chain Rule

Find $\frac{\mathrm{d} g}{\mathrm{d} t}$ of the following function:

g(x,y,z)=x+y^2+3z , where x=2t , y=t^2 , z=e^t

Possible Answers:

4t^2+4t^3+3e^t

2t+4yt+3e^t

2t+4t^3+3e^t

2t+4t^4+3e^t

Correct answer:

2t+4t^3+3e^t

Explanation:

To find the derivative of the function g(x,y,z) with respect to t we must use the multivariable chain rule, which states that

$\frac{\mathrm{d} g}{\mathrm{d} t}= \frac{\mathrm{d} g}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} g}{\mathrm{d} y}\cdot \frac{\mathrm{d} y}{\mathrm{d} t}+\frac{\mathrm{d} g}{\mathrm{d} z}\cdot \frac{\mathrm{d} z}{\mathrm{d} t}$

Our partial derivatives are:

$\frac{\mathrm{d} g}{\mathrm{d} x}= 1$ , $\frac{\mathrm{d} g}{\mathrm{d} y}= 2y$ , $\frac{\mathrm{d} g}{\mathrm{d} z}= 3$ , $\frac{\mathrm{d} x}{\mathrm{d} t}= 2t$ , $\frac{\mathrm{d} y}{\mathrm{d} t}= 2t$ , $\frac{\mathrm{d} z}{\mathrm{d} t}= e^t$

Plugging all of this in - and rewriting any remaining variables in terms of t - we get

2t+4t^3+3e^t

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Example Question #21 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x+y , x=t^2 and y=2t .

Possible Answers:

2t^2+2t

2t+2

t^2

Correct answer:

2t+2

Explanation:

Find $\frac{dz}{dt}$ if z=x+y , x=t^2 and y=2t .

We use the chain rule to find the total derivative of with respect to .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=1(2t)+1(2)=2t+2$

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Example Question #373 : Partial Derivatives

Find $\frac{dz}{dt}$ if $z=e^{x+y}$ , x=lnt and y=1 .

Possible Answers:

$\frac{e^{lnt+1}}{t}$

2t^2+2t

$\frac{e^{lnt-1}}{7}$

$\frac{e^{lnt^2}}{4}$

Correct answer:

$\frac{e^{lnt+1}}{t}$

Explanation:

Find $\frac{dz}{dt}$ if $z=e^{x+y}$ , x=lnt and y=1 .

We use the chain rule to find the total derivative of with respect to .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=\left (e^{x+y}(1) \right )(1/t)+\left (e^{x+y}(1) \right )(0)=\frac{e^{x+y}}{t}$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=lnt and y=1 .

$\frac{dz}{dt}=\frac{e^{lnt+1}}{t}$

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Example Question #1331 : Calculus 3

Find $\frac{dz}{dt}$ if z=x^3y , x=t^2+2t and y=5t-3 .

Possible Answers:

t^6-312t^5+224t^4-46t^3+117t^2

-5t^6+112t^5-424t^4-26t^3+72t^2

t^6+12t^5+24t^4-6t^3+7t^2

35t^6+162t^5+210t^4+16t^3-72t^2

Correct answer:

35t^6+162t^5+210t^4+16t^3-72t^2

Explanation:

Find $\frac{dz}{dt}$ if z=x^3y , x=t^2+2t and y=5t-3 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(y3x^2)(2t+2)+(x^3)(5)=(3x^2y)(2t+2)+5x^3$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=t^2+2t and y=5t-3 .

$\frac{dz}{dt}=\left [ 3(t^2+2t)^2(5t-3) \right ](2t+2)+5(t^2+2t)^3$

$\frac{dz}{dt}=\left [ 3(t^4+4t^3+4t^2)(5t-3) \right ](2t+2)+5(t^6+6t^5+12t^4+8t^3)$

$=\left [ 3(5t^5+20t^4+20t^3-3t^4-12t^3-12t^2) \right ](2t+2)+(5t^6+30t^5+60t^4+40t^3)$

$=\left [ 3(5t^5+17t^4+8t^3-12t^2) \right ](2t+2) +(5t^6+30t^5+60t^4+40t^3)$

=(15t^5+51t^4+24t^3-36t^2)(2t+2) +(5t^6+30t^5+60t^4+40t^3)

=(30t^6+102t^5+48t^4-72t^3 +30t^5+102t^4+48t^3-72t^2) +(5t^6+30t^5+60t^4+40t^3)

=35t^6+162t^5+210t^4+16t^3-72t^2

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Example Question #375 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=x^2cosy , x=sin(t) and y=t^2 .

Possible Answers:

2sin(t^2)cos(t)cos(t^2)+tsin^2(t^2)sin(t^2)

2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

2sin(t)cos(t)-2tsin^2(t)

cos(t)cos(t^2)-2sin(t^2)

Correct answer:

2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

Explanation:

Find $\frac{dz}{dt}$ if z=x^2cosy , x=sin(t) and y=t^2 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xcosy)(cost)+(x^2(-siny))(2t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=sin(t) and y=t^2 .

$\frac{dz}{dt}=(2sin(t)cos(t^2))(cost)+\left [ (sint)^2(-sin(t^2)) \right ](2t)$

=2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

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Example Question #1332 : Calculus 3

Find $\frac{dz}{dt}$ if z=ln(xy) , x=cos(t) and y=ln(t) .

Possible Answers:

$sin(t)+\frac{4}{tlnt}$

$-cos(t)+\frac{1}{t}$

$tan(2t)-\frac{1}{7lnt}$

$-tan(t)+\frac{1}{tlnt}$

Correct answer:

$-tan(t)+\frac{1}{tlnt}$

Explanation:

Find $\frac{dz}{dt}$ if z=ln(xy) , x=cos(t) and y=ln(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=\frac{y}{xy}(-sin(t))+\frac{x}{xy}(\frac{1}{t})$

$\frac{dz}{dt}=\frac{1}{x}(-sin(t))+\frac{1}{y}(\frac{1}{t})$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=cos(t) and y=ln(t) .

$\frac{dz}{dt}=\frac{1}{cos(t)}(-sin(t))+\frac{1}{lnt}(\frac{1}{t})$

$\frac{dz}{dt}=\frac{-sin(t)}{cos(t)}+\frac{1}{tlnt}=-tan(t)+\frac{1}{tlnt}$

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Example Question #24 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x^2y+cosy , x=-2t^2 and y=sin(t) .

Possible Answers:

$6t^3sin(t)+\left [ t^4-cos(sin(t)) \right ](sin(t))$

$t^3sin(t)-\left [ sin(cos(t)) \right ](cos(t))$

$16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

$-56t^3sin(t)+\left [ 3t^4-sin(sin(t)) \right ](sin(t))$

Correct answer:

$16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

Explanation:

Find $\frac{dz}{dt}$ if z=x^2y+cosy , x=-2t^2 and y=sin(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xy)(-4t)+(x^2-siny)(cos(t))$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=-2t^2 and y=sin(t) .

$\frac{dz}{dt}=(2(-2t^2)sin(t))(-4t)+\left [ (-2t^2)^2-sin(sin(t)) \right ](cos(t))$

$\frac{dz}{dt}=16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

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Example Question #1340 : Calculus 3

Find $\frac{dz}{dt}$ if z=sin(xy) , x=-t^5+4t and y=2t-3 .

Possible Answers:

(-12t^5+13t^4+6t-1)sin(-2t^4+4t^2+3t^5-12t)

(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

(4t^5+13t^4+3t+2)sin(-t^6+3t^2+9t^5-2t)

(-2t^5-5t^4+16t-42)cos(-t^6+8t^2+t^5-12t)

Correct answer:

(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

Explanation:

Find $\frac{dz}{dt}$ if z=sin(xy) , x=-t^5+4t and y=2t-3 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(ycos(xy))(-5t^4+4)+(xcos(xy))(2)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=-t^5+4t and y=2t-3 .

$\frac{dz}{dt}=\left [ (2t-3)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](-5t^4+4) +\left [ (-t^5+4t)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](2)$

$\frac{dz}{dt}=\left [ (2t-3)(-5t^4+4)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ] +\left [ 2(-t^5+4t)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ]$

$\frac{dz}{dt}=(-10t^5+8t+15t^4-12)cos(-2t^6+8t^2+3t^5-12t) +(-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$ $\frac{dz}{dt}=(-10t^5+8t+15t^4-12-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$

=(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #12 : Multi Variable Chain Rule

Example Question #1331 : Calculus 3

Example Question #16 : Multi Variable Chain Rule

Example Question #21 : Multi Variable Chain Rule

Example Question #373 : Partial Derivatives

Example Question #1331 : Calculus 3

Example Question #375 : Partial Derivatives

Example Question #1332 : Calculus 3

Example Question #24 : Multi Variable Chain Rule

Example Question #1340 : Calculus 3

All Calculus 3 Resources

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