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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(4,2,-1)\text{ where }f(x,y,z)=x^3sin{(y^2)}sin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(-16,4,-11).\end{align*}$

Possible Answers:

-79.36

47.3

6.12

2.5

Correct answer:

47.3

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(4,2,-1)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(4)^2+(-11)^2}=19.82\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{19.82}=-0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{19.82}=0.2\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{19.82}=-0.55\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(4,2,-1)=(-0.81)(3x^2sin{(y^2)}sin{(z^3)})+(0.2)(2x^3ycos{(y^2)}sin{(z^3)})+(-0.55)(3x^3z^2cos{(z^3)}sin{(y^2)})=47.3\end{align*}$

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Example Question #3 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-1,1,-3)\text{ where }f(x,y,z)=x^4z^4cos{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(13,20,9).\end{align*}$

Possible Answers:

2.51

138.98

1.37

-216.8

Correct answer:

-216.8

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-1,1,-3)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(20)^2+(9)^2}=25.5\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{25.5}=0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{25.5}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{25.5}=0.35\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-1,1,-3)=(0.51)(4x^3z^4cos{(y^2)})+(0.78)(-2x^4yz^4sin{(y^2)})+(0.35)(4x^4z^3cos{(y^2)})=-216.8\end{align*}$

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Example Question #1073 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-2,-1,-3)\text{ where }f(x,y,z)=cos{(x^2)}cos{(z^3)}sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(8,6,-19).\end{align*}$

Possible Answers:

-12.76

-2.27

5.69

-0.04

Correct answer:

-12.76

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-1,-3)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(6)^2+(-19)^2}=21.47\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{21.47}=0.37\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{6}{21.47}=0.28\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{21.47}=-0.88\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-2,-1,-3)=(0.37)(-2xcos{(z^3)}sin{(x^2)}sin{(y^3)})+(0.28)(3y^2cos{(x^2)}cos{(y^3)}cos{(z^3)})+(-0.88)(-3z^2cos{(x^2)}sin{(y^3)}sin{(z^3)})=-12.76\end{align*}$

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Example Question #2 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-3,1,2)\text{ where }f(x,y,z)=y^2z^4sin{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-14,-9,-2).\end{align*}$

Possible Answers:

-395.41

20.47

-81.7

-19.93

Correct answer:

-81.7

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-3,1,2)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(-9)^2+(-2)^2}=16.76\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{16.76}=-0.84\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{16.76}=-0.54\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{16.76}=-0.12\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-3,1,2)=(-0.84)(2xy^2z^4cos{(x^2)})+(-0.54)(2yz^4sin{(x^2)})+(-0.12)(4y^2z^3sin{(x^2)})=-81.7\end{align*}$

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Example Question #1075 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-3,-3,-2)\text{ where }f(x,y,z)=2zcos{(x^2)}cos{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(9,10,-18).\end{align*}$

Possible Answers:

6.29

-4.43

-3.27

-2.43

Correct answer:

6.29

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-3,-3,-2)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(9)^2+(10)^2+(-18)^2}=22.47\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{9}{22.47}=0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{22.47}=0.44\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{22.47}=-0.8\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-3,-3,-2)=(0.4)(-4xzcos{(y^2)}sin{(x^2)})+(0.44)(-4yzcos{(x^2)}sin{(y^2)})+(-0.8)(2cos{(x^2)}cos{(y^2)})=6.29\end{align*}$

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Example Question #1076 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(1,3,2)\text{ where }f(x,y,z)=2^zsin{(x^2)}sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(-18,1,-7).\end{align*}$

Possible Answers:

9.79

17.15

-10.24

-6.03

Correct answer:

-6.03

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(1,3,2)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(1)^2+(-7)^2}=19.34\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{19.34}=-0.93\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{19.34}=0.05\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{19.34}=-0.36\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(1,3,2)=(-0.93)((2)2^zxcos{(x^2)}sin{(y^3)})+(0.05)((3)2^zy^2cos{(y^3)}sin{(x^2)})+(-0.36)(2^zsin{(x^2)}sin{(y^3)}ln{(2)})=-6.03\end{align*}$

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Example Question #3441 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-2,-1,3)\text{ where }f(x,y,z)=2^{(1/y^2)}e^{(1/x^2)}cos{(z)} + 2^ye^{(x)}cos{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-8,-8,-7).\end{align*}$

Possible Answers:

-1.78

-0.13

-5.62

2.77

Correct answer:

2.77

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-1,3)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(-8)^2+(-7)^2}=13.3\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{13.3}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{13.3}=-0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{13.3}=-0.53\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-2,-1,3)=(-0.6)(2^ye^{(x)}cos{(z)} - {((2)2^{(1/y^2)}e^{(1/x^2)}cos{(z)})}/x^3)+(-0.6)(2^ye^{(x)}ln{(2)}cos{(z)} - {((2)2^{(1/y^2)}e^{(1/x^2)}ln{(2)}cos{(z)})}/y^3)+(-0.53)(- 2^ye^{(x)}sin{(z)} - 2^{(1/y^2)}e^{(1/x^2)}sin{(z)})=2.77\end{align*}$

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Example Question #11 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-2,-3,-1)\text{ where }f(x,y,z)=y^2sin{(x^3)}e^{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-18,-8,-19).\end{align*}$

Possible Answers:

107.71

-28775.6

-10.23

Correct answer:

-10.23

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-3,-1)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(-8)^2+(-19)^2}=27.37\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{27.37}=-0.66\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{27.37}=-0.29\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{27.37}=-0.69\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-2,-3,-1)=(-0.66)(3*x^2*y^2*cos{(x^3)}*e^{(z^2)})+(-0.29)(2*y*sin{(x^3)}*e^{(z^2)})+(-0.69)(2*y^2*z*sin{(x^3)}*e^{(z^2)})=-10.23\end{align*}$

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Example Question #11 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-3,2,-2)\text{ where }f(x,y,z)=2^zy^2cos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(6,10,3).\end{align*}$

Possible Answers:

214.96

-2.91

12.57

208.94

Correct answer:

12.57

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-3,2,-2)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(10)^2+(3)^2}=12.04\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{12.04}=0.5\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{12.04}=0.83\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{12.04}=0.25\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-3,2,-2)=(0.5)(-3*2^z*x^2*y^2*sin{(x^3)})+(0.83)(2*2^z*y*cos{(x^3)})+(0.25)(2^z*y^2*cos{(x^3)}*ln{(2)})=12.57\end{align*}$

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Example Question #12 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-1,-1,2)\text{ where }f(x,y,z)=z^2e^{(x^2)}e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(4,12,-5).\end{align*}$

Possible Answers:

20.67

-0.29

56.12

Correct answer:

-0.29

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-1,-1,2)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(4)^2+(12)^2+(-5)^2}=13.6\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{4}{13.6}=0.29\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{13.6}=0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{13.6}=-0.37\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-1,-1,2)=(0.29)(2*x*z^2*e^{(x^2)}*e^{(y)})+(0.88)(z^2*e^{(x^2)}*e^{(y)})+(-0.37)(2*z*e^{(x^2)}*e^{(y)})=-0.29\end{align*}$

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Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1 : Directional Derivatives

Example Question #3 : Directional Derivatives

Example Question #1073 : Partial Derivatives

Example Question #2 : Directional Derivatives

Example Question #1075 : Partial Derivatives

Example Question #1076 : Partial Derivatives

Example Question #3441 : Calculus 3

Example Question #11 : Directional Derivatives

Example Question #11 : Directional Derivatives

Example Question #12 : Directional Derivatives

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