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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1233 : 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:

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

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

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

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

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 #1234 : Partial 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:

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

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

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

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

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 #1235 : Partial 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:

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

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

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

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

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 #1236 : Partial 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:

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

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

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

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

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 #1237 : Partial Derivatives

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

Possible Answers:

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

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

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

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

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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Example Question #1238 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(y^{4})sin(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,5,-17).\end{align*}$

Possible Answers:

${-\frac{ (0.314sin(z^{2}))}{y}- 1.84zcos(z^{2})ln(y^{4})}$

${1.92zcos(z^{2})ln(y^{4}) -\frac{ (1.12sin(z^{2}))}{y}}$

{0}

${-\frac{ (70.9sin(z^{2}))}{y}- 35.4zcos(z^{2})ln(y^{4})}$

Correct answer:

${1.92zcos(z^{2})ln(y^{4}) -\frac{ (1.12sin(z^{2}))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(-17)^2}=17.72\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.72}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{17.72}=0.28\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{17.72}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(0.28)(-\frac{(4sin(z^{2}))}{y})+(-0.96)(-2zcos(z^{2})ln(y^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.92zcos(z^{2})ln(y^{4}) -\frac{ (1.12sin(z^{2}))}{y}\end{align*}$

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Example Question #1239 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin(y^{3})e^{(x^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(1,9,0).\end{align*}$

Possible Answers:

${2.97y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 0.44x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

${109x^{3}y^{2}cos(y^{3})e^{(x^{4})}cos(z)}$

${27y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 4x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

${9.06sin(y^{3})e^{(x^{4})}cos(z) + 27.2y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 36.2x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

Correct answer:

${2.97y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 0.44x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(9)^2+(0)^2}=9.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{9.06}=0.11\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{9.06}=0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.06}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.11)(4x^{3}sin(y^{3})e^{(x^{4})}sin(z))+(0.99)(3y^{2}cos(y^{3})e^{(x^{4})}sin(z))+(0)(sin(y^{3})e^{(x^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=2.97y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 0.44x^{3}sin(y^{3})e^{(x^{4})}sin(z)\end{align*}$

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Example Question #1240 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(sin(z^{3})e^{(y^{3})})}{x}\\&\text{In the direction of }\overrightarrow{v}=(-1,16,0).\end{align*}$

Possible Answers:

${\frac{(sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (48y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

${-\frac{(144y^{2}z^{2}cos(z^{3})e^{(y^{3})})}{x^{2}}}$

${\frac{(48.1z^{2}cos(z^{3})e^{(y^{3})})}{x}-\frac{ (16sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (48.1y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

${\frac{(0.06sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (3y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

Correct answer:

${\frac{(0.06sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (3y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(16)^2+(0)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{16.03}=-0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{16.03}=1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.06)(-\frac{(sin(z^{3})e^{(y^{3})})}{x^{2}})+(1)(\frac{(3y^{2}sin(z^{3})e^{(y^{3})})}{x})+(0)(\frac{(3z^{2}cos(z^{3})e^{(y^{3})})}{x})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.06sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (3y^{2}sin(z^{3})e^{(y^{3})})}{x}\end{align*}$

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Example Question #1241 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(y^{4})}zln(x^{3})\\&\text{In the direction of }\overrightarrow{v}=(0,3,10).\end{align*}$

Possible Answers:

${0.96\cdot2^{(y^{4})}ln(x^{3}) + 0.804\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

${\frac{(86.8\cdot2^{(y^{4})}y^{3})}{x}}$

${0.922\cdot2^{(y^{4})}ln(x^{3}) + 0.233\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

${10.4\cdot2^{(y^{4})}ln(x^{3}) +\frac{ (31.3\cdot2^{(y^{4})}z)}{x}+ 28.9\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

Correct answer:

${0.96\cdot2^{(y^{4})}ln(x^{3}) + 0.804\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(3)^2+(10)^2}=10.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.44}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{3}{10.44}=0.29\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{10.44}=0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(3\cdot2^{(y^{4})}z)}{x})+(0.29)(2.77\cdot2^{(y^{4})}y^{3}zln(x^{3}))+(0.96)(2^{(y^{4})}ln(x^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=0.96\cdot2^{(y^{4})}ln(x^{3}) + 0.804\cdot2^{(y^{4})}y^{3}zln(x^{3})\end{align*}$

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Example Question #1242 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^{4})}sin(x^{3})sin(y^{3})\\&\text{In the direction of }\overrightarrow{v}=(0,-5,2).\end{align*}$

Possible Answers:

${2.05\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}) - 2.79\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3})}$

${2.59\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3}) + 0.759\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3})}$

${269\cdot4^{(z^{4})}x^{2}y^{2}z^{3}cos(x^{3})cos(y^{3})}$

${16.2\cdot4^{(z^{4})}x^{2}cos(x^{3})sin(y^{3}) + 16.2\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3}) + 29.9\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3})}$

Correct answer:

${2.05\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}) - 2.79\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-5)^2+(2)^2}=5.39\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{5.39}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{5.39}=-0.93\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{5.39}=0.37\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3\cdot4^{(z^{4})}x^{2}cos(x^{3})sin(y^{3}))+(-0.93)(3\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3}))+(0.37)(5.55\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=2.05\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}) - 2.79\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3})\end{align*}$

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Calculus 3 : Calculus 3

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