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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1 : Directional Derivatives

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=x^2y+xz^{10}+xy^2z^5$ in the direction of $\vec{v}=(1,2,3)$ .

Possible Answers:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= 2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((2xy+z^{10}+y^2z^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+3(10xz^9+5xy^2z^4))$

Correct answer:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

Explanation:

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\neq1$

$\vec{u}=\frac{1}{\sqrt{14}}\vec{v}=\Big(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+0+2xyz^5)+\frac{3}{\sqrt{14}}(0+10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+2xyz^5)+\frac{3}{\sqrt{14}}(10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

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Example Question #1 : Directional Derivatives

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=z^2y\ln(x)+xy^2z^3$ in the direction of $\vec{v}=(0,1,1)$ .

Possible Answers:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ (2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ (z^2\ln(x)+2xyz^3)$

Correct answer:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

Explanation:

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{0^2+1^2+1^2}=\sqrt{0+1+1}=\sqrt{2}\neq1$

$\vec{u}=\frac{1}{\sqrt{2}}\vec{v}=\Big(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ 0(\frac{yz^2}{x}+y^2z^3)+\frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

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Example Question #3 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-5,1,4)\text{ where }f(x,y,z)=y^3cos{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-2,20,-14).\end{align*}$

Possible Answers:

23.13

-1.55

18.92

-66.53

Correct answer:

-1.55

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-5,1,4)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(20)^2+(-14)^2}=24.49\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{24.49}=-0.08\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{24.49}=0.82\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{24.49}=-0.57\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-5,1,4)=(-0.08)(-2xy^3sin{(x^2)}sin{(z)})+(0.82)(3y^2cos{(x^2)}sin{(z)})+(-0.57)(y^3cos{(x^2)}cos{(z)})=-1.55\end{align*}$

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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:

2.5

47.3

-79.36

6.12

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 #2 : 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

1.37

-216.8

138.98

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 #1074 : Partial 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:

-81.7

-395.41

-19.93

20.47

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 #3 : Directional Derivatives

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

Possible Answers:

2.77

-1.78

-0.13

-5.62

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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Calculus 3 : Directional Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #1 : Directional Derivatives

Example Question #1 : Directional Derivatives

Example Question #3 : Directional Derivatives

Example Question #1 : Directional Derivatives

Example Question #2 : Directional Derivatives

Example Question #1073 : Partial Derivatives

Example Question #1074 : Partial Derivatives

Example Question #1075 : Partial Derivatives

Example Question #1076 : Partial Derivatives

Example Question #3 : Directional Derivatives

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