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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1228 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^ye^{(x)}cos{(z)}\\&\text{In the direction of }\overrightarrow{v}=(0,20,20).\end{align*}$

Possible Answers:

${0.78\cdot3^ye^{(x)}cos{(z)} - 0.71\cdot3^ye^{(x)}sin{(z)}}$

${59.3\cdot3^ye^{(x)}cos{(z)} - 28.3\cdot3^ye^{(x)}sin{(z)}}$

${-31.1\cdot3^ye^{(x)}sin{(z)}}$

${22\cdot3^ye^{(x)}cos{(z)} - 20\cdot3^ye^{(x)}sin{(z)}}$

Correct answer:

${0.78\cdot3^ye^{(x)}cos{(z)} - 0.71\cdot3^ye^{(x)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(20)^2+(20)^2}=28.28\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{28.28}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{28.28}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{28.28}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3^ye^{(x)}cos{(z)})+(0.71)(1.1\cdot3^ye^{(x)}cos{(z)})+(0.71)(-1\cdot3^ye^{(x)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.78\cdot3^ye^{(x)}cos{(z)} - 0.71\cdot3^ye^{(x)}sin{(z)}\end{align*}$

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Example Question #1229 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(z^4)}cos{(x)}ln{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-1,15,0).\end{align*}$

Possible Answers:

${15cos{(z^4)}ln{(y)}sin{(x)} -\frac{ {(15cos{(z^4)}cos{(x)})}}{y} + 60.1z^3sin{(z^4)}cos{(x)}ln{(y)}}$

${-\frac{ {(cos{(z^4)}cos{(x)})}}{y} - 0.07cos{(z^4)}ln{(y)}sin{(x)}}$

${0.0049cos{(z^4)}ln{(y)}sin{(x)} -\frac{ {(cos{(z^4)}cos{(x)})}}{y}}$

${\frac{-{(60.1z^3sin{(z^4)}sin{(x)})}}{y}}$

Correct answer:

${-\frac{ {(cos{(z^4)}cos{(x)})}}{y} - 0.07cos{(z^4)}ln{(y)}sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(15)^2+(0)^2}=15.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{15.03}=-0.07\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{15.03}=1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.07)(cos{(z^4)}ln{(y)}sin{(x)})+(1)(\frac{-{(cos{(z^4)}cos{(x)})}}{y})+(0)(4z^3sin{(z^4)}cos{(x)}ln{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(1cos{(z^4)}cos{(x)})}}{y} - 0.07cos{(z^4)}ln{(y)}sin{(x)}\end{align*}$

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Example Question #1230 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-sin{(x^2)}sin{(y^4)}sin{(z^4)}\\&\text{In the direction of }\overrightarrow{v}=(-7,11,0).\end{align*}$

Possible Answers:

${- 52.2y^3cos{(y^4)}sin{(x^2)}sin{(z^4)} - 52.2z^3cos{(z^4)}sin{(x^2)}sin{(y^4)} - 26.1xcos{(x^2)}sin{(y^4)}sin{(z^4)}}$

${-417xy^3z^3cos{(x^2)}cos{(y^4)}cos{(z^4)}}$

${1.08xcos{(x^2)}sin{(y^4)}sin{(z^4)} - 3.36y^3cos{(y^4)}sin{(x^2)}sin{(z^4)}}$

${- 2.82y^3cos{(y^4)}sin{(x^2)}sin{(z^4)} - 0.583xcos{(x^2)}sin{(y^4)}sin{(z^4)}}$

Correct answer:

${1.08xcos{(x^2)}sin{(y^4)}sin{(z^4)} - 3.36y^3cos{(y^4)}sin{(x^2)}sin{(z^4)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(11)^2+(0)^2}=13.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{13.04}=-0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{13.04}=0.84\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{13.04}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.54)(-2xcos{(x^2)}sin{(y^4)}sin{(z^4)})+(0.84)(-4y^3cos{(y^4)}sin{(x^2)}sin{(z^4)})+(0)(-4z^3cos{(z^4)}sin{(x^2)}sin{(y^4)})\\&D_{\overrightarrow{u}}(x,y,z)=1.08xcos{(x^2)}sin{(y^4)}sin{(z^4)} - 3.36y^3cos{(y^4)}sin{(x^2)}sin{(z^4)}\end{align*}$

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Example Question #161 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(y^4)}sin{(x^2)}sin{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-11,0,-5).\end{align*}$

Possible Answers:

${1.66xcos{(x^2)}cos{(y^4)}sin{(z^2)} + 0.336zcos{(y^4)}cos{(z^2)}sin{(x^2)}}$

${- 1.82xcos{(x^2)}cos{(y^4)}sin{(z^2)} - 0.82zcos{(y^4)}cos{(z^2)}sin{(x^2)}}$

${24.2xcos{(x^2)}cos{(y^4)}sin{(z^2)} - 48.3y^3sin{(x^2)}sin{(y^4)}sin{(z^2)} + 24.2zcos{(y^4)}cos{(z^2)}sin{(x^2)}}$

${-193xy^3zcos{(x^2)}cos{(z^2)}sin{(y^4)}}$

Correct answer:

${- 1.82xcos{(x^2)}cos{(y^4)}sin{(z^2)} - 0.82zcos{(y^4)}cos{(z^2)}sin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11)^2+(0)^2+(-5)^2}=12.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11}{12.08}=-0.91\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{12.08}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{12.08}=-0.41\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.91)(2xcos{(x^2)}cos{(y^4)}sin{(z^2)})+(0)(-4y^3sin{(x^2)}sin{(y^4)}sin{(z^2)})+(-0.41)(2zcos{(y^4)}cos{(z^2)}sin{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.82xcos{(x^2)}cos{(y^4)}sin{(z^2)} - 0.82zcos{(y^4)}cos{(z^2)}sin{(x^2)}\end{align*}$

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Example Question #162 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(x^4)}e^{(z^2)})}}{y}\\&\text{In the direction of }\overrightarrow{v}=(0,2,-12).\end{align*}$

Possible Answers:

${\frac{{(97.4x^3zsin{(x^4)}e^{(z^2)})}}{y^2}}$

${-\frac{ {(2cos{(x^4)}e^{(z^2)})}}{y^2} -\frac{ {(24zcos{(x^4)}e^{(z^2)})}}{y}}$

${-\frac{ {(0.16cos{(x^4)}e^{(z^2)})}}{y^2} -\frac{ {(1.98zcos{(x^4)}e^{(z^2)})}}{y}}$

${\frac{{(24.3zcos{(x^4)}e^{(z^2)})}}{y} -\frac{ {(12.2cos{(x^4)}e^{(z^2)})}}{y^2} -\frac{ {(48.7x^3sin{(x^4)}e^{(z^2)})}}{y}}$

Correct answer:

${-\frac{ {(0.16cos{(x^4)}e^{(z^2)})}}{y^2} -\frac{ {(1.98zcos{(x^4)}e^{(z^2)})}}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(2)^2+(-12)^2}=12.17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{2}{12.17}=0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{12.17}=-0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(4x^3sin{(x^4)}e^{(z^2)})}}{y})+(0.16)(\frac{-{(cos{(x^4)}e^{(z^2)})}}{y^2})+(-0.99)(\frac{{(2zcos{(x^4)}e^{(z^2)})}}{y})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(0.16cos{(x^4)}e^{(z^2)})}}{y^2} -\frac{ {(1.98zcos{(x^4)}e^{(z^2)})}}{y}\end{align*}$

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Example Question #1231 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(x^4)}sin{(y^3)}e^{(z^4)}\\&\text{In the direction of }\overrightarrow{v}=(-6,5,0).\end{align*}$

Possible Answers:

${23.4y^2cos{(x^4)}cos{(y^3)}e^{(z^4)} - 31.2x^3sin{(x^4)}sin{(y^3)}e^{(z^4)} + 31.2z^3cos{(x^4)}sin{(y^3)}e^{(z^4)}}$

${-375x^3y^2z^3cos{(y^3)}sin{(x^4)}e^{(z^4)}}$

${24x^3sin{(x^4)}sin{(y^3)}e^{(z^4)} + 15y^2cos{(x^4)}cos{(y^3)}e^{(z^4)}}$

${3.08x^3sin{(x^4)}sin{(y^3)}e^{(z^4)} + 1.92y^2cos{(x^4)}cos{(y^3)}e^{(z^4)}}$

Correct answer:

${3.08x^3sin{(x^4)}sin{(y^3)}e^{(z^4)} + 1.92y^2cos{(x^4)}cos{(y^3)}e^{(z^4)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(5)^2+(0)^2}=7.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{7.81}=-0.77\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{7.81}=0.64\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{7.81}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.77)(-4x^3sin{(x^4)}sin{(y^3)}e^{(z^4)})+(0.64)(3y^2cos{(x^4)}cos{(y^3)}e^{(z^4)})+(0)(4z^3cos{(x^4)}sin{(y^3)}e^{(z^4)})\\&D_{\overrightarrow{u}}(x,y,z)=3.08x^3sin{(x^4)}sin{(y^3)}e^{(z^4)} + 1.92y^2cos{(x^4)}cos{(y^3)}e^{(z^4)}\end{align*}$

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Example Question #164 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(z^3)}e^{(y^3)}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,-9,18).\end{align*}$

Possible Answers:

${2.38z^2cos{(z^3)}e^{(y^3)}e^{(x)} + 0.607y^2sin{(z^3)}e^{(y^3)}e^{(x)}}$

${2.67z^2cos{(z^3)}e^{(y^3)}e^{(x)} - 1.35y^2sin{(z^3)}e^{(y^3)}e^{(x)}}$

${20.1sin{(z^3)}e^{(y^3)}e^{(x)} + 60.4z^2cos{(z^3)}e^{(y^3)}e^{(x)} + 60.4y^2sin{(z^3)}e^{(y^3)}e^{(x)}}$

${181y^2z^2cos{(z^3)}e^{(y^3)}e^{(x)}}$

Correct answer:

${2.67z^2cos{(z^3)}e^{(y^3)}e^{(x)} - 1.35y^2sin{(z^3)}e^{(y^3)}e^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-9)^2+(18)^2}=20.12\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{20.12}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{20.12}=-0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{20.12}=0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(sin{(z^3)}e^{(y^3)}e^{(x)})+(-0.45)(3y^2sin{(z^3)}e^{(y^3)}e^{(x)})+(0.89)(3z^2cos{(z^3)}e^{(y^3)}e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=2.67z^2cos{(z^3)}e^{(y^3)}e^{(x)} - 1.35y^2sin{(z^3)}e^{(y^3)}e^{(x)}\end{align*}$

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Example Question #165 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(x^3)}ln{(y^4)}sin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(18,0,19).\end{align*}$

Possible Answers:

${78.5x^2ln{(y^4)}sin{(x^3)}sin{(z^3)} -\frac{ {(105cos{(x^3)}sin{(z^3)})}}{y} - 78.5z^2cos{(x^3)}cos{(z^3)}ln{(y^4)}}$

${1.43x^2ln{(y^4)}sin{(x^3)}sin{(z^3)} - 1.6z^2cos{(x^3)}cos{(z^3)}ln{(y^4)}}$

${2.07x^2ln{(y^4)}sin{(x^3)}sin{(z^3)} - 2.19z^2cos{(x^3)}cos{(z^3)}ln{(y^4)}}$

${\frac{{(942x^2z^2cos{(z^3)}sin{(x^3)})}}{y}}$

Correct answer:

${2.07x^2ln{(y^4)}sin{(x^3)}sin{(z^3)} - 2.19z^2cos{(x^3)}cos{(z^3)}ln{(y^4)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(0)^2+(19)^2}=26.17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{26.17}=0.69\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{26.17}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{26.17}=0.73\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.69)(3x^2ln{(y^4)}sin{(x^3)}sin{(z^3)})+(0)(\frac{-{(4cos{(x^3)}sin{(z^3)})}}{y})+(0.73)(-3z^2cos{(x^3)}cos{(z^3)}ln{(y^4)})\\&D_{\overrightarrow{u}}(x,y,z)=2.07x^2ln{(y^4)}sin{(x^3)}sin{(z^3)} - 2.19z^2cos{(x^3)}cos{(z^3)}ln{(y^4)}\end{align*}$

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Example Question #166 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^4ln{(y^3)}e^{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(0,3,14).\end{align*}$

Possible Answers:

${\frac{{(0.63x^4e^{(z^2)})}}{y} + 1.96x^4zln{(y^3)}e^{(z^2)}}$

${\frac{{(43x^4e^{(z^2)})}}{y} + 57.3x^3ln{(y^3)}e^{(z^2)} + 28.6x^4zln{(y^3)}e^{(z^2)}}$

${\frac{{(344x^3ze^{(z^2)})}}{y}}$

${\frac{{(9x^4e^{(z^2)})}}{y} + 28x^4zln{(y^3)}e^{(z^2)}}$

Correct answer:

${\frac{{(0.63x^4e^{(z^2)})}}{y} + 1.96x^4zln{(y^3)}e^{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(3)^2+(14)^2}=14.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.32}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{3}{14.32}=0.21\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{14}{14.32}=0.98\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^3ln{(y^3)}e^{(z^2)})+(0.21)(\frac{{(3x^4e^{(z^2)})}}{y})+(0.98)(2x^4zln{(y^3)}e^{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.63x^4e^{(z^2)})}}{y} + 1.96x^4zln{(y^3)}e^{(z^2)}\end{align*}$

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Example Question #3601 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(z^4)}x^3e^{(y^4)}\\&\text{In the direction of }\overrightarrow{v}=(13,0,-6).\end{align*}$

Possible Answers:

${-755\cdot3^{(z^4)}x^2y^3z^3e^{(y^4)}}$

${- 43\cdot3^{(z^4)}x^2e^{(y^4)} - 57.3\cdot3^{(z^4)}x^3y^3e^{(y^4)} - 62.9\cdot3^{(z^4)}x^3z^3e^{(y^4)}}$

${- 2.48\cdot3^{(z^4)}x^2e^{(y^4)} - 0.775\cdot3^{(z^4)}x^3z^3e^{(y^4)}}$

${1.85\cdot3^{(z^4)}x^3z^3e^{(y^4)} - 2.73\cdot3^{(z^4)}x^2e^{(y^4)}}$

Correct answer:

${1.85\cdot3^{(z^4)}x^3z^3e^{(y^4)} - 2.73\cdot3^{(z^4)}x^2e^{(y^4)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(-6)^2}=14.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{14.32}=0.91\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.32}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{14.32}=-0.42\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.91)(-3\cdot3^{(z^4)}x^2e^{(y^4)})+(0)(-4\cdot3^{(z^4)}x^3y^3e^{(y^4)})+(-0.42)(-4.39\cdot3^{(z^4)}x^3z^3e^{(y^4)})\\&D_{\overrightarrow{u}}(x,y,z)=1.85\cdot3^{(z^4)}x^3z^3e^{(y^4)} - 2.73\cdot3^{(z^4)}x^2e^{(y^4)}\end{align*}$

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Calculus 3 : Directional Derivatives

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