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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #351 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0+,0+)}-\frac{(2xy^{3} - 5x^{4} + 3y^{4})}{(x^{4} + 2y^{4})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:-\frac{3}{10}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{51}{2}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:-24\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{3}{2}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }-\frac{3}{2}\\&y=x:0\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&-\frac{(2(0)y^{3} - 5(0)^{4} + 3y^{4})}{((0)^{4} + 2y^{4})}=\frac{-3y^{4}}{2y^{4}}\\&\lim_{(0,y)\rightarrow(0+,0+)}\frac{-3y^{4}}{2y^{4}}=-\frac{3}{2}\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0+,0+)}-\frac{(2x(x)^{3} - 5x^{4} + 3(x)^{4})}{(x^{4} + 2(x)^{4})}=0\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #351 : Limits

$\begin{align*}&\text{Calculate the limits of the function}\\&\lim_{(x,y)\rightarrow(0-,0+)}\frac{(26x^{3}y^{2} - 14x^{5})}{(18xy^{4} - 5x^{4}y + 3x^{5} + 3y^{5})}\\&\text{along the y-axis and the line, }y=x\end{align*}$

Possible Answers:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{6}{133}\end{align*}$

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{12}{19}\end{align*}$

$\begin{align*}&\text{y-axis: }\frac{36}{19}\\&y=x:0\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{120}{19}\\&y=x:0\end{align*}$

Correct answer:

$\begin{align*}&\text{y-axis: }0\\&y=x:\frac{12}{19}\end{align*}$

Explanation:

$\begin{align*}&\text{To approach this problem, it is important to understand what}\\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\\&\frac{(26(0)^{3}y^{2} - 14(0)^{5})}{(18(0)y^{4} - 5(0)^{4}y + 3(0)^{5} + 3y^{5})}=\frac{0}{3y^{5}}\\&\lim_{(0,y)\rightarrow(0-,0+)}\frac{0}{3y^{5}}=0\\&\text{We can make a similar substitution for }y=x:\\&\lim_{(x,x)\rightarrow(0-,0-)}\frac{(26x^{3}(x)^{2} - 14x^{5})}{(18x(x)^{4} - 5x^{4}(x) + 3x^{5} + 3(x)^{5})}=\frac{12}{19}\\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

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Example Question #351 : Partial Derivatives

Evaluate:

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}$

Possible Answers:

$-\infty$

$\infty$

Does not exist

Correct answer:

Explanation:

Since we won't have any divided by zero problems, we can just evaluate at the point.

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0$

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Example Question #352 : Partial Derivatives

Evaluate:

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}$

Possible Answers:

$-\infty$

Does not exist

$\infty$

Correct answer:

Explanation:

Since we won't have any divided by zero problems, we can just evaluate at the point.

$\lim_{(x,y)\rightarrow (0,1)} \frac{x^2+xy}{y}=\frac{0^2+0\cdot1}{1}=0$

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Example Question #1 : Multi Variable Chain Rule

Compute $\frac{dz}{dt}$ for $z=2xe^{xy}$ , x=t^3 , $y=t^{-2}$ .

Possible Answers:

$\frac{dz}{dt}=6t^2e^{xy}-4t^{-3}x^2e^{xy}$

$\frac{dz}{dt}=(6t^2e^{xy}-2xye^{xy})-(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(4t^{-3}x^2e^{xy})$

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})$

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

Correct answer:

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

Explanation:

All we need to do is use the formula for multivariable chain rule.

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

$\frac{\partial f}{\partial x}=2e^{xy}+2xye^{xy}$

$\frac{dx}{dt}=3t^2$

$\frac{\partial f}{\partial y}=2x^2e^{xy}$

$\frac{dy}{dt}=-2t^{-3}$

When we put this all together, we get.

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

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Example Question #2 : Multi Variable Chain Rule

Evaluate $\frac{\partial a}{\partial x}$ in terms of and/or if a= x-3y+z^2 , x=r^3 , y= s+2r , and z=6r .

Possible Answers:

3r^2-3s-6r+12r^2

3r^2+72r-6

3r^2+72rs-6

3r^6 +72r-6

Correct answer:

3r^2+72r-6

Explanation:

Expand the equation for chain rule.

$\frac{\partial a}{\partial x} = \frac{\partial a}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial a}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial a}{\partial z}\frac{\partial z}{\partial r}$

Evaluate each partial derivative.

$\frac{\partial a}{\partial x} = 1$ , $\frac{\partial x}{\partial r} =3r^2$

$\frac{\partial a}{\partial y}= -3$ , $\frac{\partial y}{\partial r} = 2$

$\frac{\partial a}{\partial z} =2z$ , $\frac{\partial z}{\partial r} = 6$

Substitute the terms in the equation.

$\frac{\partial a}{\partial x} =(1)(3r^2)+(-3)(2)+(2z)(6) = 3r^2-6+12z$

Substitute z=6r .

3r^2-6+12(6r) = 3r^2+72r-6

The answer is: 3r^2+72r-6

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Example Question #2 : Multi Variable Chain Rule

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

Possible Answers:

$\frac{dr}{dt}=9-t$

$\frac{dr}{dt}=-t^3+5t^2+2t-3$

$\frac{dr}{dt}=12t^2+3t-7$

$\frac{dr}{dt}=4t^3-6t^2+4$

Correct answer:

$\frac{dr}{dt}=4t^3-6t^2+4$

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

$\frac{dr}{dt}=\frac{\partial r}{\partial z}\frac{dz}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$=\left ( 2z\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

$=\left ( 2t^2\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

=4t^3-6t^2+4

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Example Question #4 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=3a^4+5b^2 , a=t^2-6t+12 , b=t-5 .

Possible Answers:

$\frac{dr}{dt}=t^7-4t^6-52t^5-25t^4+88t^3-186t^2+2281t-1266$

$\frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

$\frac{dr}{dt}=4t^3-6t^2+4$

$\frac{dr}{dt}=-54t^6+52t^5-250t^4+828t^3-1624t^2+2206t-1246$

Correct answer:

$\frac{dr}{dt}=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466$

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

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

$=\left ( 12a^3\right )\left ( 2t-6\right )+\left ( 10b\right )\left ( 1\right )$

$=\left ( 12\left ( t^2-6t+12\right )^3\right )\left ( 2t-6\right )+\left ( 10\left ( t-5\right )\right )\left ( 1\right )$

$=\left12 ( t^6-18t^5+144t^4-648t^3+1728t^2-2592t+1728\right )\left ( 2t-6\right )+10t-50$

=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228096t-124416 +10t-50

=24t^7-504t^6+4752t^5-25920t^4+88128t^3-186624t^2+228106t-124466

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Example Question #5 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=cos(ab) , a=t^3+7t , b=5t^2-12t+1 .

Possible Answers:

$\frac{dr}{dt}=4t^3-6t^2+4sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(t^4-3t^3-128t^2-49t)sin(t^5-131t^4+66t^3-98t^2+17t-121)$

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

$\frac{dr}{dt}=(5t^4+8t^3-18t^2-9t)cos(4t^5-11t^4+6t^3-8t^2+7t-11)$

Correct answer:

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

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

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

=b*cos(ab)(3t^2+7)+a*cos(ab)(10t-12)

$=-(5t^2-12t+1)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 3t^2+7\right )-(t^3+7t)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 10t-12\right )$

=-(15^4-36t^3+38t^2-84t+7)sin(5t^5-12t^4+36t^3-84t^2+7t)-(10t^4-12t^3+70t^2-84t)sin(5t^5-12t^4+36t^3-84t^2+7t)

=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)

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Example Question #6 : Multi Variable Chain Rule

Use the chain rule to find $\frac{dr}{dt}$ when r=sin(ab^2) , a=5t-1 , b=-9t^3 .

Possible Answers:

$\frac{dr}{dt}=(4t^3-6t^2+4)sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(-90t^4+18t^3)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$

Correct answer:

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

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

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

$\frac{\partial r}{\partial a}=cos(ab^2)(b^2)$

=cos(405t^7-81t^6)(81t^6)

$\frac{da}{dt}=5$

$\frac{\partial r}{\partial b}=cos(ab^2)(2ab)$

=cos(405t^7-81t^6)(-90t^4+18t^3)

$\frac{db}{dt}=-27t^2$

$\frac{dr}{dt}=\left ( cos(405t^7-81t^6)(81t^6) \right )(5)+(cos(405t^7-81t^6)(-90t^4+18t^3))(-27t^2)$

=405t^6*cos(405t^7-81t^6)+(2430t^6-486t^5)cos(405t^7-81t^6)

=(2745t^6-486t^5)cos(405t^7-81t^6)

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #351 : Limits

Example Question #351 : Limits

Example Question #351 : Partial Derivatives

Example Question #352 : Partial Derivatives

Example Question #1 : Multi Variable Chain Rule

Example Question #2 : Multi Variable Chain Rule

Example Question #2 : Multi Variable Chain Rule

Example Question #4 : Multi Variable Chain Rule

Example Question #5 : Multi Variable Chain Rule

Example Question #6 : Multi Variable Chain Rule

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