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Calculus 3 : Directional Derivatives

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

Example Questions

Example Question #1328 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ze^{(x)} + 2cos(z^{2})ln(y^{2})\\&\text{In the direction of }\overrightarrow{v}=(-11,10,0).\end{align*}$

Possible Answers:

${\frac{(2.68cos(z^{2}))}{y}- 0.74ze^{(x)}}$

${14.9e^{(x)} +\frac{ (59.5cos(z^{2}))}{y}+ 14.9ze^{(x)} - 59.5zln(y^{2})sin(z^{2})}$

${\frac{(40cos(z^{2}))}{y}- 11ze^{(x)}}$

Correct answer:

${\frac{(2.68cos(z^{2}))}{y}- 0.74ze^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11)^2+(10)^2+(0)^2}=14.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11}{14.87}=-0.74\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{14.87}=0.67\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.87}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.74)(ze^{(x)})+(0.67)(\frac{(4cos(z^{2}))}{y})+(0)(e^{(x)} - 4zln(y^{2})sin(z^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(2.68cos(z^{2}))}{y}- 0.74ze^{(x)}\end{align*}$

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Example Question #1329 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4x^{2}cos(z)sin(y) - 2^{(x^{2})}sin(z^{3})e^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(6,-1,0).\end{align*}$

Possible Answers:

${4x^{2}cos(y)cos(z) - 48xcos(z)sin(y) + 3\cdot 2^{(x^{2})}y^{2}sin(z^{3})e^{(y^{3})} - 8.32\cdot 2^{(x^{2})}xsin(z^{3})e^{(y^{3})}}$

${0.64x^{2}cos(y)cos(z) - 7.92xcos(z)sin(y) + 0.48\cdot 2^{(x^{2})}y^{2}sin(z^{3})e^{(y^{3})} - 1.37\cdot 2^{(x^{2})}xsin(z^{3})e^{(y^{3})}}$

${24.3x^{2}sin(y)sin(z) - 48.6xcos(z)sin(y) - 24.3x^{2}cos(y)cos(z) - 18.2\cdot 2^{(x^{2})}z^{2}cos(z^{3})e^{(y^{3})} - 18.2\cdot 2^{(x^{2})}y^{2}sin(z^{3})e^{(y^{3})} - 8.43\cdot 2^{(x^{2})}xsin(z^{3})e^{(y^{3})}}$

${48.6xcos(y)sin(z) - 75.9\cdot 2^{(x^{2})}xy^{2}z^{2}cos(z^{3})e^{(y^{3})}}$

Correct answer:

${0.64x^{2}cos(y)cos(z) - 7.92xcos(z)sin(y) + 0.48\cdot 2^{(x^{2})}y^{2}sin(z^{3})e^{(y^{3})} - 1.37\cdot 2^{(x^{2})}xsin(z^{3})e^{(y^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-1)^2+(0)^2}=6.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{6.08}=0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-1}{6.08}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{6.08}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.99)(- 8xcos(z)sin(y) - 1.39\cdot 2^{(x^{2})}xsin(z^{3})e^{(y^{3})})+(-0.16)(- 4x^{2}cos(y)cos(z) - 3\cdot 2^{(x^{2})}y^{2}sin(z^{3})e^{(y^{3})})+(0)(4x^{2}sin(y)sin(z) - 3\cdot 2^{(x^{2})}z^{2}cos(z^{3})e^{(y^{3})})\\&D_{\overrightarrow{u}}(x,y,z)=0.64x^{2}cos(y)cos(z) - 7.92xcos(z)sin(y) + 0.48\cdot 2^{(x^{2})}y^{2}sin(z^{3})e^{(y^{3})} - 1.37\cdot 2^{(x^{2})}xsin(z^{3})e^{(y^{3})}\end{align*}$

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Example Question #1330 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos(y^{3})e^{(x^{3})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(5,-18,0).\end{align*}$

Possible Answers:

${56y^{2}sin(y^{3})e^{(x^{3})}e^{(z^{2})} - 56x^{2}cos(y^{3})e^{(x^{3})}e^{(z^{2})} - 37.4zcos(y^{3})e^{(x^{3})}e^{(z^{2})}}$

${- 0.81x^{2}cos(y^{3})e^{(x^{3})}e^{(z^{2})} - 2.88y^{2}sin(y^{3})e^{(x^{3})}e^{(z^{2})}}$

${336x^{2}y^{2}zsin(y^{3})e^{(x^{3})}e^{(z^{2})}}$

${2.76y^{2}sin(y^{3})e^{(x^{3})}e^{(z^{2})} - 0.219x^{2}cos(y^{3})e^{(x^{3})}e^{(z^{2})}}$

Correct answer:

${- 0.81x^{2}cos(y^{3})e^{(x^{3})}e^{(z^{2})} - 2.88y^{2}sin(y^{3})e^{(x^{3})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(-18)^2+(0)^2}=18.68\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{18.68}=0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.68}=-0.96\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.68}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.27)(-3x^{2}cos(y^{3})e^{(x^{3})}e^{(z^{2})})+(-0.96)(3y^{2}sin(y^{3})e^{(x^{3})}e^{(z^{2})})+(0)(-2zcos(y^{3})e^{(x^{3})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.81x^{2}cos(y^{3})e^{(x^{3})}e^{(z^{2})} - 2.88y^{2}sin(y^{3})e^{(x^{3})}e^{(z^{2})}\end{align*}$

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Example Question #1331 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(3^{(z^{3})}cos(y))}{x^{2}}- yz\\&\text{In the direction of }\overrightarrow{v}=(0,-14,-12).\end{align*}$

Possible Answers:

${\frac{(60.8\cdot 3^{(z^{3})}z^{2}cos(y))}{x^{2}}- 18.4z -\frac{ (36.9\cdot 3^{(z^{3})}cos(y))}{x^{3}}-\frac{ (18.4\cdot 3^{(z^{3})}sin(y))}{x^{2}}- 18.4y}$

${12y + 14z +\frac{ (14\cdot 3^{(z^{3})}sin(y))}{x^{2}}-\frac{ (39.6\cdot 3^{(z^{3})}z^{2}cos(y))}{x^{2}}}$

${0.65y + 0.76z +\frac{ (0.76\cdot 3^{(z^{3})}sin(y))}{x^{2}}-\frac{ (2.14\cdot 3^{(z^{3})}z^{2}cos(y))}{x^{2}}}$

${\frac{(122\cdot 3^{(z^{3})}z^{2}sin(y))}{x^{3}}}$

Correct answer:

${0.65y + 0.76z +\frac{ (0.76\cdot 3^{(z^{3})}sin(y))}{x^{2}}-\frac{ (2.14\cdot 3^{(z^{3})}z^{2}cos(y))}{x^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-14)^2+(-12)^2}=18.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{18.44}=-0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{18.44}=-0.65\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(2\cdot 3^{(z^{3})}cos(y))}{x^{3}})+(-0.76)(- z -\frac{ (1\cdot 3^{(z^{3})}sin(y))}{x^{2}})+(-0.65)(\frac{(3.3\cdot 3^{(z^{3})}z^{2}cos(y))}{x^{2}}- y)\\&D_{\overrightarrow{u}}(x,y,z)=0.65y + 0.76z +\frac{ (0.76\cdot 3^{(z^{3})}sin(y))}{x^{2}}-\frac{ (2.14\cdot 3^{(z^{3})}z^{2}cos(y))}{x^{2}}\end{align*}$

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Example Question #1332 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(x^{4})}\cdot 4^yz^{2}\\&\text{In the direction of }\overrightarrow{v}=(20,0,-10).\end{align*}$

Possible Answers:

${20\cdot 2^{(x^{4})}\cdot 4^yz - 55.5\cdot 2^{(x^{4})}\cdot 4^yx^{3}z^{2}}$

${- 44.7\cdot 2^{(x^{4})}\cdot 4^yz - 31\cdot 2^{(x^{4})}\cdot 4^yz^{2} - 62\cdot 2^{(x^{4})}\cdot 4^yx^{3}z^{2}}$

${-172\cdot 2^{(x^{4})}\cdot 4^yx^{3}z}$

${0.9\cdot 2^{(x^{4})}\cdot 4^yz - 2.47\cdot 2^{(x^{4})}\cdot 4^yx^{3}z^{2}}$

Correct answer:

${0.9\cdot 2^{(x^{4})}\cdot 4^yz - 2.47\cdot 2^{(x^{4})}\cdot 4^yx^{3}z^{2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(0)^2+(-10)^2}=22.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{22.36}=0.89\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{22.36}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{22.36}=-0.45\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.89)(-2.77\cdot 2^{(x^{4})}\cdot 4^yx^{3}z^{2})+(0)(-1.39\cdot 2^{(x^{4})}\cdot 4^yz^{2})+(-0.45)(-2\cdot 2^{(x^{4})}\cdot 4^yz)\\&D_{\overrightarrow{u}}(x,y,z)=0.9\cdot 2^{(x^{4})}\cdot 4^yz - 2.47\cdot 2^{(x^{4})}\cdot 4^yx^{3}z^{2}\end{align*}$

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Example Question #1333 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}y^{4}cos(z^{3}) - 2^xe^{(y^{2})}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(18,-2,0).\end{align*}$

Possible Answers:

${3.92x^{3}y^{4}cos(z^{3}) + 0.0484x^{4}y^{3}cos(z^{3}) - 0.679\cdot 2^xe^{(y^{2})}cos(z) - 0.0242\cdot 2^xye^{(y^{2})}cos(z)}$

${3.96x^{3}y^{4}cos(z^{3}) - 0.44x^{4}y^{3}cos(z^{3}) - 0.686\cdot 2^xe^{(y^{2})}cos(z) + 0.22\cdot 2^xye^{(y^{2})}cos(z)}$

${72.4x^{3}y^{4}cos(z^{3}) + 72.4x^{4}y^{3}cos(z^{3}) - 12.6\cdot 2^xe^{(y^{2})}cos(z) + 18.1\cdot 2^xe^{(y^{2})}sin(z) - 36.2\cdot 2^xye^{(y^{2})}cos(z) - 54.3x^{4}y^{4}z^{2}sin(z^{3})}$

${25.1\cdot 2^xye^{(y^{2})}sin(z) - 869x^{3}y^{3}z^{2}sin(z^{3})}$

Correct answer:

${3.96x^{3}y^{4}cos(z^{3}) - 0.44x^{4}y^{3}cos(z^{3}) - 0.686\cdot 2^xe^{(y^{2})}cos(z) + 0.22\cdot 2^xye^{(y^{2})}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(-2)^2+(0)^2}=18.11\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{18.11}=0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{18.11}=-0.11\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.11}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.99)(4x^{3}y^{4}cos(z^{3}) - 0.693\cdot 2^xe^{(y^{2})}cos(z))+(-0.11)(4x^{4}y^{3}cos(z^{3}) - 2\cdot 2^xye^{(y^{2})}cos(z))+(0)(2^xe^{(y^{2})}sin(z) - 3x^{4}y^{4}z^{2}sin(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=3.96x^{3}y^{4}cos(z^{3}) - 0.44x^{4}y^{3}cos(z^{3}) - 0.686\cdot 2^xe^{(y^{2})}cos(z) + 0.22\cdot 2^xye^{(y^{2})}cos(z)\end{align*}$

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Example Question #1334 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(2^{(z^{4})}x)}{y}- 3zcos(y)\\&\text{In the direction of }\overrightarrow{v}=(-20,-3,0).\end{align*}$

Possible Answers:

${-\frac{(56.1\cdot 2^{(z^{4})}z^{3})}{y^{2}}}$

${\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}}$

${\frac{(20.2\cdot 2^{(z^{4})})}{y}- 60.7cos(y) + 60.7zsin(y) -\frac{ (20.2\cdot 2^{(z^{4})}x)}{y^{2}}+\frac{ (56.1\cdot 2^{(z^{4})}xz^{3})}{y}}$

${\frac{(3\cdot 2^{(z^{4})}x)}{y^{2}}- 9zsin(y) -\frac{ (20\cdot 2^{(z^{4})})}{y}}$

Correct answer:

${\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-3)^2+(0)^2}=20.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{20.22}=-0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{20.22}=-0.15\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.22}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{2^{(z^{4})}}{y})+(-0.15)(3zsin(y) -\frac{ (1\cdot 2^{(z^{4})}x)}{y^{2}})+(0)(\frac{(2.77\cdot 2^{(z^{4})}xz^{3})}{y}- 3cos(y))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.15\cdot 2^{(z^{4})}x)}{y^{2}}- 0.45zsin(y) -\frac{ (0.99\cdot 2^{(z^{4})})}{y}\end{align*}$

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Example Question #1335 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(ln(x^{3})ln(y^{4}))}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-20,-5,0).\end{align*}$

Possible Answers:

${\frac{(20ln(x^{3}))}{(yz^{2})}+\frac{ (60ln(y^{4}))}{(xz^{2})}}$

${\frac{(41.2ln(x^{3})ln(y^{4}))}{z^{3}}-\frac{ (61.9ln(y^{4}))}{(xz^{2})}-\frac{ (82.5ln(x^{3}))}{(yz^{2})}}$

${\frac{(0.96ln(x^{3}))}{(yz^{2})}+\frac{ (2.91ln(y^{4}))}{(xz^{2})}}$

${\frac{495.0}{(xyz^{3})}}$

Correct answer:

${\frac{(0.96ln(x^{3}))}{(yz^{2})}+\frac{ (2.91ln(y^{4}))}{(xz^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-5)^2+(0)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{20.62}=-0.97\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{20.62}=-0.24\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.97)(-\frac{(3ln(y^{4}))}{(xz^{2})})+(-0.24)(-\frac{(4ln(x^{3}))}{(yz^{2})})+(0)(\frac{(2ln(x^{3})ln(y^{4}))}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.96ln(x^{3}))}{(yz^{2})}+\frac{ (2.91ln(y^{4}))}{(xz^{2})}\end{align*}$

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Example Question #1336 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2z^{3}tan(x^{2}) + ln(x^{4})e^{(y^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-17,1,0).\end{align*}$

Possible Answers:

${4y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 68xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (68e^{(y^{4})}sin(z))}{x}}$

${102z^{2}tan(x^{2}) +\frac{ (68.1e^{(y^{4})}sin(z))}{x}+ 17ln(x^{4})e^{(y^{4})}cos(z) + 68.1xz^{3}\cdot (tan(x^{2})^{2} + 1) + 68.1y^{3}ln(x^{4})e^{(y^{4})}sin(z)}$

${\frac{(272y^{3}e^{(y^{4})}cos(z))}{x}}$

${0.24y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 4xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (4e^{(y^{4})}sin(z))}{x}}$

Correct answer:

${0.24y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 4xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (4e^{(y^{4})}sin(z))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-17)^2+(1)^2+(0)^2}=17.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-17}{17.03}=-1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{17.03}=0.06\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-1)(\frac{(4e^{(y^{4})}sin(z))}{x}+ 4xz^{3}\cdot (tan(x^{2})^{2} + 1))+(0.06)(4y^{3}ln(x^{4})e^{(y^{4})}sin(z))+(0)(6z^{2}tan(x^{2}) + ln(x^{4})e^{(y^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.24y^{3}ln(x^{4})e^{(y^{4})}sin(z) - 4xz^{3}\cdot (tan(x^{2})^{2} + 1) -\frac{ (4e^{(y^{4})}sin(z))}{x}\end{align*}$

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Example Question #1337 : Partial Derivatives

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

Possible Answers:

${0.29\cdot 2^xln(z^{2}) -\frac{ (1.92\cdot 2^xy)}{z}}$

${10.4\cdot 2^xln(z^{2}) + 7.24\cdot 2^xyln(z^{2}) +\frac{ (20.9\cdot 2^xy)}{z}}$

${0.0841\cdot 2^xln(z^{2}) +\frac{ (1.84\cdot 2^xy)}{z}}$

${\frac{(14.5\cdot 2^x)}{z}}$

Correct answer:

${0.29\cdot 2^xln(z^{2}) -\frac{ (1.92\cdot 2^xy)}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(3)^2+(-10)^2}=10.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.44}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{3}{10.44}=0.29\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{10.44}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.693\cdot 2^xyln(z^{2}))+(0.29)(2^xln(z^{2}))+(-0.96)(\frac{(2\cdot 2^xy)}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.29\cdot 2^xln(z^{2}) -\frac{ (1.92\cdot 2^xy)}{z}\end{align*}$

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Calculus 3 : Directional Derivatives

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