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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #361 : Partial Derivatives

Use the chain rule to find $\frac{dr}{dt}$ when $r=e^{xy}$ , x=-2t^3+1 , y=t^2-t-1 .

Possible Answers:

$\frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

$\frac{dr}{dt}=e^{t^2-2t+7}$

$\frac{dr}{dt}=(-9t^4+23t^3+t^2-5t-1)e^{(-2t^3+2t^2-t-1)}$

$\frac{dr}{dt}=(t^2+7t-6)e^{(3t^2-9t-11)}$

Correct answer:

$\frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=ye^{xy}=(t^2-t-1)e^{xy}$

$\frac{dx}{dt}=-6t^2$

$\frac{\partial r}{\partial y}=xe^{xy}=(-2t^3+1)e^{xy}$

$\frac{dy}{dt}=2t-1$

$\frac{dr}{dt}=(t^2-t-1)e^{xy}(-6t^2)+(-2t^3+1)e^{xy}(2t-1)$

$=(-6t^4+12t^3+6t^2)e^{xy}+(-4t^4+2t^3+2t-1)e^{xy}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{xy}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^3+1)(t^2-t-1)}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

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Example Question #4 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=ln(ab) , a=-2t^2-3t , b=4t+9 .

Possible Answers:

$\frac{dr}{dt}=\frac{3t^3-11t^2+7t-14}{t^4-t^3-7t^2+7t+1}$

$\frac{dr}{dt}=\frac{2t^2-6t+7}{-t^3+3t^2+7t}$

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

$\frac{dr}{dt}=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

Correct answer:

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$\frac{\partial r}{\partial a}=\frac{b}{ab}=\frac{1}{a} =\frac{1}{-2t^2-3t}$

$\frac{da}{dt}=-4t-3$

$\frac{\partial r}{\partial b}=\frac{a}{ab}=\frac{1}{b}=\frac{1}{4t+9}$

$\frac{db}{dt}=4$

$\frac{dr}{dt}=\left ( \frac{1}{-2t^2-3t} \right )(-4t-3)+\left ( \frac{1}{4t+9} \right )(4)$

$=\frac{-4t-3}{-2t^2-3t}+\frac{4}{4t+9}$

$=\frac{(-4t-3)(4t+9)+4(-2t^2-3t)}{(-2t^2-3t)(4t+9)}$

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

Report an Error

Example Question #5 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=ln(5a+3b) , a=8t-3 , b=t^2-7t+1 .

Possible Answers:

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

$\frac{dr}{dt}=\frac{-t^2-9t+11}{t^2-9t+4}$

$\frac{dr}{dt}=\frac{t-4}{t^2-9t+12}$

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

Correct answer:

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$\frac{\partial r}{\partial a}=\frac{5}{5a+3b}=\frac{5}{5(8t-3)+3(t^2-7t+1)}=\frac{5}{3t^2+19t-12}$

$\frac{da}{dt}=8$

$\frac{\partial r}{\partial b}=\frac{3}{5a+3b}=\frac{3}{5(8t-3)+3(t^2-7t+1)}=\frac{3}{3t^2+19t-12}$

$\frac{db}{dt}=2t-7$

$\frac{dr}{dt}=\left ( \frac{5}{3t^2+19t-12} \right )(8)+\left ( \frac{3}{3t^2+19t-12} \right )(2t-7)$

$=\left ( \frac{40}{3t^2+19t-12} \right )+\left ( \frac{6t-21}{3t^2+19t-12} \right )$

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

Report an Error

Example Question #11 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when $r=\frac{x^2-7y^2}{4}$ , x=sin(t) , y=cos(t) .

Possible Answers:

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

$\frac{dr}{dt}=4cos(t)$

$\frac{dr}{dt}=4sin(t)$

$\frac{dr}{dt}=4sin(t)cos(t)$

Correct answer:

$\frac{dr}{dt}=4sin(t)cos(t)$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=\frac{2x}{4}=\frac{x}{2}=\frac{sin(t)}{2}$

$\frac{dx}{dt}=cos(t)$

$\frac{\partial r}{\partial y}=\frac{-14y}{4}=\frac{-7y}{2}=\frac{-7cos(t)}{2}$

$\frac{dy}{dt}=-sin(t)$

$\frac{dr}{dt}=\left (\frac{sin(t)}{2} \right )(cos(t))+\left ( \frac{-7cos(t)}{2} \right )(-sin(t))$

$=\left (\frac{sin(t)cos(t)}{2} \right )+\left ( \frac{7sin(t)cos(t)}{2} \right )$

$\frac{dr}{dt}=4sin(t)cos(t)$

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Example Question #1321 : Calculus 3

Use the chain rule to find $\frac{dr}{dt}$ when $r=\frac{4x^2+8y^2}{x^3}$ , x=ln(t) , y=t^2-1 .

Possible Answers:

$\frac{dr}{dt}=4sin(t)cos(t)$

$\frac{dr}{dt}=\frac{-5}{t(ln(t))}-\frac{4t^2-12t+24}{t(ln(t))^2} + \frac{2t^3-3t}{ln(t)^3}$

$\frac{dr}{dt}=\frac{-4}{t(ln(t))^2}-\frac{24t^4-48t^2+24}{t(ln(t))^4} + \frac{32t^3-32t}{ln(t)^3}$

$\frac{dr}{dt}=-\frac{7t^4-9t^2+4}{t(ln(t))^4}$

Correct answer:

$\frac{dr}{dt}=\frac{-4}{t(ln(t))^2}-\frac{24t^4-48t^2+24}{t(ln(t))^4} + \frac{32t^3-32t}{ln(t)^3}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$r=\frac{4x^2+8y^2}{x^3}=\frac{4x^2}{x^3}+\frac{8y^2}{x^3}=\frac{4}{x}+\frac{8y^2}{x^3}$

$\frac{\partial r}{\partial x}=\frac{-4}{x^2}+\frac{(-3)8y^2}{x^4}=\frac{-4}{x^2}-\frac{24y^2}{x^4}=\frac{-4}{(ln(t))^2}-\frac{24(t^2-1)^2}{(ln(t))^4}$

$\frac{\partial r}{\partial x}=\frac{-4}{(ln(t))^2}-\frac{24(t^4-2t^2+1)}{(ln(t))^4}=\frac{-4}{(ln(t))^2}-\frac{24t^4-48t^2+24}{(ln(t))^4}$

$\frac{dx}{dt}=\frac{1}{t}$

$\frac{\partial r}{\partial y}=\frac{16y}{x^3}=\frac{16(t^2-1)}{ln(t)^3}=\frac{16t^2-16}{ln(t)^3}$

$\frac{dy}{dt}=2t$

$\frac{dr}{dt}=\left (\frac{-4}{(ln(t))^2}-\frac{24t^4-48t^2+24}{(ln(t))^4} \right )\left ( \frac{1}{t} \right )+\left ( \frac{16t^2-16}{ln(t)^3}\right )(2t)$

$\frac{dr}{dt}=\frac{-4}{t(ln(t))^2}-\frac{24t^4-48t^2+24}{t(ln(t))^4} + \frac{32t^3-32t}{ln(t)^3}$

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Example Question #13 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when $r=e^{x+4y}$ , x=4t , y=3t^3-7t .

Possible Answers:

$\frac{dr}{dt}=4(9t^2-6)\left (e^{12t^3-24t} \right )$

$\frac{dr}{dt}=3(8t^2-11)\left (e^{t^3+37t+12} \right )$

$\frac{dr}{dt}=\frac{-4}{t(ln(t))^2}-\frac{24t^4-48t^2+24}{t(ln(t))^4} + \frac{32t^3-32t}{ln(t)^3}$

$\frac{dr}{dt}=(3t^2+7)\left (e^{2t^3-5t} \right )$

Correct answer:

$\frac{dr}{dt}=4(9t^2-6)\left (e^{12t^3-24t} \right )$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=e^{x+4y}=e^{4t+4(3t^3-7t)}=e^{12t^3-24t}$

$\frac{dx}{dt}=4$

$\frac{\partial r}{\partial y}=4e^{x+4y}=4e^{4t+4(3t^3-7t)}=4e^{12t^3-24t}$

$\frac{dy}{dt}=9t^2-7$

$\frac{dr}{dt}=\left ( e^{12t^3-24t} \right )(4)+4\left (e^{12t^3-24t} \right )(9t^2-7)$

$\frac{dr}{dt}=(36t^2-24)e^{12t^3-24t}$

$\frac{dr}{dt}=4(9t^2-6)\left (e^{12t^3-24t} \right )$

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Example Question #1321 : Calculus 3

Use the chain rule to find $\frac{dr}{dt}$ when $r=\sqrt{x^4-12y}$ , x=5t^2-12 , y=9-3t^2 .

Possible Answers:

$\frac{dr}{dt}=4(9t^2-6)\left (e^{12t^3-24t} \right )$

$\frac{dr}{dt}=\frac{250t^6-1800t^4+4320t^2-3456}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

$\frac{dr}{dt}=\frac{-6}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

$\frac{dr}{dt}= \frac{2500t^7-18000t^5+43200t^3-34596t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

Correct answer:

$\frac{dr}{dt}= \frac{2500t^7-18000t^5+43200t^3-34596t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

Explanation:

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$r=\sqrt{x^4-12y}=\left (x^4-12y \right )^{1/2}$

$\frac{\partial r}{\partial x}=\left ( 1/2\right )\left (x^4-12y \right )^{-1/2}\left ( 4x^3\right )=\left ( 2x^3\right )\left (x^4-12y \right )^{-1/2}$

$=\left ( 2(5t^2-12)^3\right )\left ((5t^2-12)^4-12(9-3t^2) \right )^{-1/2}$

$=\frac{250t^6-1800t^4+4320t^2-3456}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

$\frac{dx}{dt}=10t$

$\frac{\partial r}{\partial y}=\left ( 1/2\right )\left (x^4-12y \right )^{-1/2}(-12)=(-6)\left (x^4-12y \right )^{-1/2}$

$=-6\left ((5t^2-12)^4-12(9-3t^2) \right )^{-1/2}$

$=\frac{-6}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

$\frac{dy}{dt}=-6t$

$\frac{dr}{dt}=\left ( \frac{250t^6-1800t^4+4320t^2-3456}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}\right )(10t) +\left (\frac{-6}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}} \right )(-6t)$

$\frac{dr}{dt}=\left ( \frac{2500t^7-18000t^5+43200t^3-34560t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}\right ) +\left (\frac{36t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}} \right )$

$\frac{dr}{dt}= \frac{2500t^7-18000t^5+43200t^3-34596t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

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Example Question #1322 : Calculus 3

Find $\frac{\mathrm{dg} }{\mathrm{d} t}$ where g=4xy , x=t^3 , $y=\cos(t)$

Possible Answers:

$12t^2(\cos(t))+4t^3\sin(t)$

$3t^2(\cos(t))-4t^3\sin(t)$

$12yt^2-4x\sin(t)$

$12t^2(\cos(t))-4t^3\sin(t)$

Correct answer:

$12t^2(\cos(t))-4t^3\sin(t)$

Explanation:

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that

$\frac{\mathrm{dg} }{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x}\cdot \frac{\mathrm{dx} }{\mathrm{d} t}$ , where x=f(t)

Using this rule for both variables, we find that

$\frac{\mathrm{dg} }{\mathrm{d} x}=4y$ , $\frac{\mathrm{dg} }{\mathrm{d} y}=4x$ , $\frac{\mathrm{dx} }{\mathrm{d} t}=3t^2$ , $\frac{\mathrm{dy} }{\mathrm{d} t}=-\sin(t)$

Taking the products according to the formula above, and remembering to rewrite x and y in terms of t, we get

$\frac{\mathrm{dg} }{\mathrm{d} t}=12t^2(\cos(t))-4t^3\sin(t)$

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Example Question #11 : Multi Variable Chain Rule

Find $\frac{\mathrm{dh} }{\mathrm{d} t}$ , where h=x^2y and x=t , $y=e^{t^2}$

Possible Answers:

$2t^3e^{t^2}$

$4t^4e^{t^2}$

$e^{t^2}(2t+2t^3)$

Correct answer:

$e^{t^2}(2t+2t^3)$

Explanation:

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that $\frac{\mathrm{dg} }{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x} \cdot \frac{\mathrm{dx} }{\mathrm{d} t}$ for x. (We do the same for the rest of the variables, and add the products together.)

Using the above rule for both variables, we get

$\frac{\mathrm{dh} }{\mathrm{d} x}=2xy$ , $\frac{\mathrm{dx} }{\mathrm{d} t}=1$ , $\frac{\mathrm{dg} }{\mathrm{d} y}=x^2$ , $\frac{\mathrm{dy} }{\mathrm{d} t}=2te^{t^2}$

Plugging all of this into the above formula, and remembering to rewrite x and y in terms of t, we get

$\frac{\mathrm{d} g}{\mathrm{d} t}=e^{t^2}(2t+2t^3)$

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Example Question #1323 : Calculus 3

Find $\frac{\mathrm{dj} }{\mathrm{d} t}$ , where $j=xy, x=\cos(t), y=\sin(t)$

Possible Answers:

$-\cos(t)+\sin^2(t)$

$x\cos(t)-y\sin(t)$

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

Correct answer:

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

Explanation:

To find the derivative of the function, we must use the multivariable chain rule. For x, this states that $\frac{\mathrm{d}j }{\mathrm{d} t}=\frac{\mathrm{dj} }{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}$ . (We do the same for y and add the results for the total derivative.)

So, our derivatives are:

$\frac{\mathrm{d}j }{\mathrm{d} x}=y$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=-\sin(t)$ , $\frac{\mathrm{d}j }{\mathrm{d} y}=x$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=\cos(t)$

Now, using the above formula and remembering to rewrite x and y in terms of t, we get

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #361 : Partial Derivatives

Example Question #4 : Multi Variable Chain Rule

Example Question #5 : Multi Variable Chain Rule

Example Question #11 : Multi Variable Chain Rule

Example Question #1321 : Calculus 3

Example Question #13 : Multi Variable Chain Rule

Example Question #1321 : Calculus 3

Example Question #1322 : Calculus 3

Example Question #11 : Multi Variable Chain Rule

Example Question #1323 : Calculus 3

All Calculus 3 Resources

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