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Calculus 3 : Partial Derivatives

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

Example Questions

Example Question #1321 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2e^{(z^{2})}e^{(y)} +\frac{ (cos(z)ln(x))}{y}\\&\text{In the direction of }\overrightarrow{v}=(-13,0,16).\end{align*}$

Possible Answers:

${\frac{(0.397cos(z))}{(xy)}-\frac{ (0.608ln(x)sin(z))}{y}+ 2.43ze^{(z^{2})}e^{(y)}}$

${3.12ze^{(z^{2})}e^{(y)} -\frac{ (0.63cos(z))}{(xy)}-\frac{ (0.78ln(x)sin(z))}{y}}$

${41.2e^{(z^{2})}e^{(y)} -\frac{ (20.6ln(x)sin(z))}{y}+\frac{ (20.6cos(z))}{(xy)}+ 82.5ze^{(z^{2})}e^{(y)} -\frac{ (20.6cos(z)ln(x))}{y^{2}}}$

${\frac{(20.6sin(z))}{(xy^{2})}}$

Correct answer:

${3.12ze^{(z^{2})}e^{(y)} -\frac{ (0.63cos(z))}{(xy)}-\frac{ (0.78ln(x)sin(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(0)^2+(16)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{20.62}=-0.63\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{20.62}=0.78\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.63)(\frac{cos(z)}{(xy)})+(0)(2e^{(z^{2})}e^{(y)} -\frac{ (cos(z)ln(x))}{y^{2}})+(0.78)(4ze^{(z^{2})}e^{(y)} -\frac{ (ln(x)sin(z))}{y})\\&D_{\overrightarrow{u}}(x,y,z)=3.12ze^{(z^{2})}e^{(y)} -\frac{ (0.63cos(z))}{(xy)}-\frac{ (0.78ln(x)sin(z))}{y}\end{align*}$

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Example Question #251 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(e^{(x)}ln(z))}{y}- 5e^{(x^{2})}e^{(y)}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(-4,-15,0).\end{align*}$

Possible Answers:

${75e^{(x^{2})}e^{(y)}cos(z) -\frac{ (4e^{(x)}ln(z))}{y}+\frac{ (15e^{(x)}ln(z))}{y^{2}}+ 40xe^{(x^{2})}e^{(y)}cos(z)}$

${77.6e^{(x^{2})}e^{(y)}sin(z) - 77.6e^{(x^{2})}e^{(y)}cos(z) +\frac{ (15.5e^{(x)})}{(yz)}+\frac{ (15.5e^{(x)}ln(z))}{y}-\frac{ (15.5e^{(x)}ln(z))}{y^{2}}- 155xe^{(x^{2})}e^{(y)}cos(z)}$

${155xe^{(x^{2})}e^{(y)}sin(z) -\frac{ (15.5e^{(x)})}{(y^{2}z)}}$

${4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)}$

Correct answer:

${4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-15)^2+(0)^2}=15.52\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{15.52}=-0.26\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{15.52}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15.52}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.26)(\frac{(e^{(x)}ln(z))}{y}- 10xe^{(x^{2})}e^{(y)}cos(z))+(-0.97)(- 5e^{(x^{2})}e^{(y)}cos(z) -\frac{ (e^{(x)}ln(z))}{y^{2}})+(0)(5e^{(x^{2})}e^{(y)}sin(z) +\frac{ e^{(x)}}{(yz)})\\&D_{\overrightarrow{u}}(x,y,z)=4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)\end{align*}$

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Example Question #1323 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(4^{(x^{3})}e^{(z^{4})})}{y^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-7,16,0).\end{align*}$

Possible Answers:

${\frac{(72.6\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}-\frac{ (34.9\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}+\frac{ (69.8\cdot 4^{(x^{3})}z^{3}e^{(z^{4})})}{y^{2}}}$

${-\frac{ (1.84\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}-\frac{ (1.66\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}}$

${\frac{(0.665\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}-\frac{ (1.69\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}}$

${-\frac{(581\cdot 4^{(x^{3})}x^{2}z^{3}e^{(z^{4})})}{y^{3}}}$

Correct answer:

${-\frac{ (1.84\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}-\frac{ (1.66\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(16)^2+(0)^2}=17.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{17.46}=-0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{17.46}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.46}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.4)(\frac{(4.16\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}})+(0.92)(-\frac{(2\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}})+(0)(\frac{(4\cdot 4^{(x^{3})}z^{3}e^{(z^{4})})}{y^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ (1.84\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}-\frac{ (1.66\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}\end{align*}$

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Example Question #1324 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos(y^{2})e^{(z^{2})}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(-7,-13,0).\end{align*}$

Possible Answers:

${- 0.221cos(y^{2})e^{(z^{2})}sin(x) - 1.55ysin(y^{2})e^{(z^{2})}cos(x)}$

${0.47cos(y^{2})e^{(z^{2})}sin(x) + 1.76ysin(y^{2})e^{(z^{2})}cos(x)}$

${29.5zcos(y^{2})e^{(z^{2})}cos(x) - 14.8cos(y^{2})e^{(z^{2})}sin(x) - 29.5ysin(y^{2})e^{(z^{2})}cos(x)}$

${59yzsin(y^{2})e^{(z^{2})}sin(x)}$

Correct answer:

${0.47cos(y^{2})e^{(z^{2})}sin(x) + 1.76ysin(y^{2})e^{(z^{2})}cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(-13)^2+(0)^2}=14.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{14.76}=-0.47\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{14.76}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.76}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.47)(-1cos(y^{2})e^{(z^{2})}sin(x))+(-0.88)(-2ysin(y^{2})e^{(z^{2})}cos(x))+(0)(2zcos(y^{2})e^{(z^{2})}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.47cos(y^{2})e^{(z^{2})}sin(x) + 1.76ysin(y^{2})e^{(z^{2})}cos(x)\end{align*}$

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Example Question #1325 : Partial Derivatives

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

Possible Answers:

${- 72.4y\cdot (tan(z)^{2} + 1) - 39.8\cdot 3^zxe^{(x^{2})}e^{(y)}}$

${- 36.2y^{2}tan(z) - 72.4xytan(z) - 36.2xy^{2}\cdot (tan(z)^{2} + 1) - 38\cdot 3^ze^{(x^{2})}e^{(y)} - 36.2\cdot 3^zxe^{(x^{2})}e^{(y)}}$

${- 3.92xytan(z) - 0.0242xy^{2}\cdot (tan(z)^{2} + 1) - 0.993\cdot 3^ze^{(x^{2})}e^{(y)}}$

${3.96xytan(z) + 0.22xy^{2}\cdot (tan(z)^{2} + 1) + 1.11\cdot 3^ze^{(x^{2})}e^{(y)}}$

Correct answer:

${3.96xytan(z) + 0.22xy^{2}\cdot (tan(z)^{2} + 1) + 1.11\cdot 3^ze^{(x^{2})}e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(-2)^2}=18.11\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.11}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.11}=-0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{18.11}=-0.11\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(- 2y^{2}tan(z) - 2\cdot 3^zxe^{(x^{2})}e^{(y)})+(-0.99)(- 4xytan(z) - 1\cdot 3^ze^{(x^{2})}e^{(y)})+(-0.11)(- 2xy^{2}\cdot (tan(z)^{2} + 1) - 1.1\cdot 3^ze^{(x^{2})}e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=3.96xytan(z) + 0.22xy^{2}\cdot (tan(z)^{2} + 1) + 1.11\cdot 3^ze^{(x^{2})}e^{(y)}\end{align*}$

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Example Question #1321 : Partial Derivatives

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

Possible Answers:

${\frac{(3.84cos(x^{2}))}{z}}$

${-\frac{(3.69cos(x^{2}))}{z}}$

${35.4xln(z^{4})sin(x^{2}) -\frac{ (70.9cos(x^{2}))}{z}}$

Correct answer:

${\frac{(3.84cos(x^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(-17)^2}=17.72\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.72}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{17.72}=0.28\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{17.72}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xln(z^{4})sin(x^{2}))+(0.28)(0.0)+(-0.96)(-\frac{(4cos(x^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.84cos(x^{2}))}{z}\end{align*}$

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Example Question #1327 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin(y^{3})e^{(x^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(0,8,12).\end{align*}$

Possible Answers:

${14.4sin(y^{3})e^{(x^{4})}cos(z) + 43.3y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 57.7x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

${173x^{3}y^{2}cos(y^{3})e^{(x^{4})}cos(z)}$

${0.689sin(y^{3})e^{(x^{4})}cos(z) + 0.908y^{2}cos(y^{3})e^{(x^{4})}sin(z)}$

${0.83sin(y^{3})e^{(x^{4})}cos(z) + 1.65y^{2}cos(y^{3})e^{(x^{4})}sin(z)}$

Correct answer:

${0.83sin(y^{3})e^{(x^{4})}cos(z) + 1.65y^{2}cos(y^{3})e^{(x^{4})}sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(8)^2+(12)^2}=14.42\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.42}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{8}{14.42}=0.55\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{14.42}=0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}sin(y^{3})e^{(x^{4})}sin(z))+(0.55)(3y^{2}cos(y^{3})e^{(x^{4})}sin(z))+(0.83)(sin(y^{3})e^{(x^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.83sin(y^{3})e^{(x^{4})}cos(z) + 1.65y^{2}cos(y^{3})e^{(x^{4})}sin(z)\end{align*}$

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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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If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

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Calculus 3 : Partial Derivatives

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