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Calculus 3 : Directional Derivatives

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Example Questions

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

-10.23

-28775.6

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:

208.94

214.96

-2.91

12.57

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

20.67

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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Example Question #3441 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-2,-3,2)\text{ where }f(x,y,z)=x^3sin{(y^2)}cos{(z)}\\&\text{In the direction of }\overrightarrow{v}=(6,-1,-14).\end{align*}$

Possible Answers:

4.11

-11.77

5.67

-4.75

Correct answer:

-4.75

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-2,-3,2)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-1)^2+(-14)^2}=15.26\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{15.26}=0.39\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-1}{15.26}=-0.07\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{15.26}=-0.92\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-2,-3,2)=(0.39)(3*x^2*sin{(y^2)}*cos{(z)})+(-0.07)(2*x^3*y*cos{(y^2)}*cos{(z)})+(-0.92)(-x^3*sin{(y^2)}*sin{(z)})=-4.75\end{align*}$

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Example Question #3441 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^2cos{(y^2)}sin{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-14,11,15).\end{align*}$

Possible Answers:

${(23.28)(2*z*cos{(y^2)}*sin{(x^2)} + 2*x*z^2*cos{(x^2)}*cos{(y^2)} - 2*y*z^2*sin{(x^2)}*sin{(y^2)}) }$

$(23.28)(-8*x*y*z*cos{(x^2)}*sin{(y^2)})$

${(-0.6)(2*x*z^2*cos{(x^2)}*cos{(y^2)})+(0.47)(-2*y*z^2*sin{(x^2)}*sin{(y^2)})+(0.64)(2*z*cos{(y^2)}*sin{(x^2)})}$

${(-14)(2*x*z^2*cos{(x^2)}*cos{(y^2)})+(11)(-2*y*z^2*sin{(x^2)}*sin{(y^2)})+(15)(2*z*cos{(y^2)}*sin{(x^2)}) }$

Correct answer:

${(-0.6)(2*x*z^2*cos{(x^2)}*cos{(y^2)})+(0.47)(-2*y*z^2*sin{(x^2)}*sin{(y^2)})+(0.64)(2*z*cos{(y^2)}*sin{(x^2)})}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(11)^2+(15)^2}=23.28\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{23.28}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{23.28}=0.47\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{23.28}=0.64\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(2*x*z^2*cos{(x^2)}*cos{(y^2)})+(0.47)(-2*y*z^2*sin{(x^2)}*sin{(y^2)})+(0.64)(2*z*cos{(y^2)}*sin{(x^2)})\end{align*}$

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Example Question #16 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ycos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(7,9,6).\end{align*}$

Possible Answers:

${(12.88)(cos{(x^3)} - 3x^2ysin{(x^3)})}$

$(-21x^2ysin{(x^3)})+(9cos{(x^3)})$

$(-1.62x^2ysin{(x^3)})+(0.7cos{(x^3)})$

Correct answer:

$(-1.62x^2ysin{(x^3)})+(0.7cos{(x^3)})$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(9)^2+(6)^2}=12.88\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{12.88}=0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{12.88}=0.7\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{12.88}=0.47\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(-3x^2ysin{(x^3)})+(0.7)(cos{(x^3)})+(0.47)(0)\end{align*}$

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Example Question #17 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5cos{(y^2)}ln{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-12,-13,-19).\end{align*}$

Possible Answers:

$130ysin{(y^2)}ln{(z)}-{(95cos{(y^2)})}/z$

${(129.8cos{(y^2)})}/z - 259.6ysin{(y^2)}ln{(z)}$

$5ysin{(y^2)}ln{(z)}-{3.65cos{(y^2})}/z$

Correct answer:

$5ysin{(y^2)}ln{(z)}-{3.65cos{(y^2})}/z$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12)^2+(-13)^2+(-19)^2}=25.96\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12}{25.96}=-0.46\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{25.96}=-0.5\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{25.96}=-0.73\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.46)(0)+(-0.5)(-10ysin{(y^2)}ln{(z)})+(-0.73)({(5cos{(y^2)})}/z)\end{align*}$

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Example Question #1081 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3y^2zsin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(14,17,11).\end{align*}$

Possible Answers:

{147.72ycos{(x)}}

${1.35y^2sin{(x)} + 4.14yzsin{(x)} + 1.71y^2zcos{(x)}}$

${33y^2sin{(x)} + 102yzsin{(x)} + 42y^2zcos{(x)}}$

${73.86y^2sin{(x)} + 147.72yzsin{(x)} + 73.86y^2zcos{(x)}}$

Correct answer:

${1.35y^2sin{(x)} + 4.14yzsin{(x)} + 1.71y^2zcos{(x)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(17)^2+(11)^2}=24.62\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{24.62}=0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{24.62}=0.69\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{24.62}=0.45\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.57)(3y^2zcos{(x)})+(0.69)(6yzsin{(x)})+(0.45)(3y^2sin{(x)})\end{align*}$

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Example Question #3451 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=7y^2z^3cos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(14,11,18).\end{align*}$

Possible Answers:

${14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}$

${378y^2z^2cos{(x^3)} + 154yz^3cos{(x^3)} - 294x^2y^2z^3sin{(x^3)}}$

${531.72y^2z^2cos{(x^3)} + 354.48yz^3cos{(x^3)} - 531.72x^2y^2z^3sin{(x^3)}}$

${-3190.32x^2yz^2sin{(x^3)}}$

Correct answer:

${14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(11)^2+(18)^2}=25.32\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{25.32}=0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{25.32}=0.43\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{25.32}=0.71\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(-21x^2y^2z^3sin{(x^3)})+(0.43)(14yz^3cos{(x^3)})+(0.71)(21y^2z^2cos{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)={14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}\end{align*}$

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Example Question #20 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=9x^4e^{(z^2)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(7,-10,-8).\end{align*}$

Possible Answers:

${131.31x^4e^{(z^2)}cos{(y)} + 525.24x^3e^{(z^2)}sin{(y)} + 262.62x^4ze^{(z^2)}sin{(y)}}$

${17.28x^3e^{(z^2)}sin{(y)} - 6.21x^4e^{(z^2)}cos{(y)} - 9.9x^4ze^{(z^2)}sin{(y)}}$

${1050.48x^3ze^{(z^2)}cos{(y)}}$

${252x^3e^{(z^2)}sin{(y)} - 90x^4e^{(z^2)}cos{(y)} - 144x^4ze^{(z^2)}sin{(y)}}$

Correct answer:

${17.28x^3e^{(z^2)}sin{(y)} - 6.21x^4e^{(z^2)}cos{(y)} - 9.9x^4ze^{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(-10)^2+(-8)^2}=14.59\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{14.59}=0.48\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{14.59}=-0.69\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{14.59}=-0.55\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.48)(36x^3e^{(z^2)}sin{(y)})+(-0.69)(9x^4e^{(z^2)}cos{(y)})+(-0.55)(18x^4ze^{(z^2)}sin{(y)})\end{align*}$

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Calculus 3 : Directional Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Directional Derivatives

Example Question #11 : Directional Derivatives

Example Question #12 : Directional Derivatives

Example Question #3441 : Calculus 3

Example Question #3441 : Calculus 3

Example Question #16 : Directional Derivatives

Example Question #17 : Directional Derivatives

Example Question #1081 : Partial Derivatives

Example Question #3451 : Calculus 3

Example Question #20 : Directional Derivatives

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