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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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All Calculus 3 Resources

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

Example Questions

Example Question #1273 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-x^{3}ze^{(y^{3})}\\&\text{In the direction of }\overrightarrow{v}=(0,10,8).\end{align*}$

Possible Answers:

${- 12.8x^{3}e^{(y^{3})} - 38.4x^{2}ze^{(y^{3})} - 38.4x^{3}y^{2}ze^{(y^{3})}}$

${-115x^{2}y^{2}e^{(y^{3})}}$

${- 8x^{3}e^{(y^{3})} - 30x^{3}y^{2}ze^{(y^{3})}}$

${- 0.62x^{3}e^{(y^{3})} - 2.34x^{3}y^{2}ze^{(y^{3})}}$

Correct answer:

${- 0.62x^{3}e^{(y^{3})} - 2.34x^{3}y^{2}ze^{(y^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(8)^2}=12.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.81}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{12.81}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8}{12.81}=0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-3x^{2}ze^{(y^{3})})+(0.78)(-3x^{3}y^{2}ze^{(y^{3})})+(0.62)(-1x^{3}e^{(y^{3})})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.62x^{3}e^{(y^{3})} - 2.34x^{3}y^{2}ze^{(y^{3})}\end{align*}$

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Example Question #1274 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-e^{(x^{2})}e^{(z)}ln(y)\\&\text{In the direction of }\overrightarrow{v}=(-8,0,18).\end{align*}$

Possible Answers:

${16xe^{(x^{2})}e^{(z)}ln(y) - 18e^{(x^{2})}e^{(z)}ln(y)}$

${0.82xe^{(x^{2})}e^{(z)}ln(y) - 0.91e^{(x^{2})}e^{(z)}ln(y)}$

${-\frac{ (19.7e^{(x^{2})}e^{(z)})}{y}- 19.7e^{(x^{2})}e^{(z)}ln(y) - 39.4xe^{(x^{2})}e^{(z)}ln(y)}$

${-\frac{(39.4xe^{(x^{2})}e^{(z)})}{y}}$

Correct answer:

${0.82xe^{(x^{2})}e^{(z)}ln(y) - 0.91e^{(x^{2})}e^{(z)}ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(0)^2+(18)^2}=19.7\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{19.7}=-0.41\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{19.7}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{19.7}=0.91\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.41)(-2xe^{(x^{2})}e^{(z)}ln(y))+(0)(-\frac{(e^{(x^{2})}e^{(z)})}{y})+(0.91)(-1e^{(x^{2})}e^{(z)}ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.82xe^{(x^{2})}e^{(z)}ln(y) - 0.91e^{(x^{2})}e^{(z)}ln(y)\end{align*}$

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Example Question #1274 : Partial Derivatives

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

Possible Answers:

${\frac{(28.6\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (47.2\cdot 3^{(z^{3})}z^{2}cos(y^{4}))}{x^{2}}+\frac{ (57.3\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

${-\frac{(378\cdot 3^{(z^{3})}y^{3}z^{2}sin(y^{4}))}{x^{3}}}$

${\frac{(28\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (12\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

${\frac{(1.96\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (0.84\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

Correct answer:

${\frac{(1.96\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (0.84\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(-3)^2+(0)^2}=14.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{14.32}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{14.32}=-0.21\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(\frac{(2\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}})+(-0.21)(\frac{(4\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}})+(0)(-\frac{(3.3\cdot 3^{(z^{3})}z^{2}cos(y^{4}))}{x^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.96\cdot 3^{(z^{3})}cos(y^{4}))}{x^{3}}-\frac{ (0.84\cdot 3^{(z^{3})}y^{3}sin(y^{4}))}{x^{2}}\end{align*}$

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Example Question #1276 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{e^{(y^{2})}}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-5,16,0).\end{align*}$

Possible Answers:

${\frac{(33.5e^{(y^{2})})}{z^{3}}-\frac{ (33.5ye^{(y^{2})})}{z^{2}}}$

${-\frac{(1.9ye^{(y^{2})})}{z^{2}}}$

${-\frac{(1.8ye^{(y^{2})})}{z^{2}}}$

Correct answer:

${-\frac{(1.9ye^{(y^{2})})}{z^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(16)^2+(0)^2}=16.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{16.76}=-0.3\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{16.76}=0.95\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.76}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.3)(0.0)+(0.95)(-\frac{(2ye^{(y^{2})})}{z^{2}})+(0)(\frac{(2e^{(y^{2})})}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{(1.9ye^{(y^{2})})}{z^{2}}\end{align*}$

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Example Question #1277 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2x^{3}z^{4}cos(y^{3}) +\frac{ (y^{3}ln(x^{2}))}{z^{2}}\\&\text{In the direction of }\overrightarrow{v}=(16,18,0).\end{align*}$

Possible Answers:

${144x^{2}z^{4}cos(y^{3}) + 193x^{3}z^{3}cos(y^{3}) +\frac{ (72.2y^{2}ln(x^{2}))}{z^{2}}-\frac{ (48.2y^{3}ln(x^{2}))}{z^{3}}+\frac{ (48.2y^{3})}{(xz^{2})}- 144x^{3}y^{2}z^{4}sin(y^{3})}$

${-\frac{ (289y^{2})}{(xz^{3})}- 1733x^{2}y^{2}z^{3}sin(y^{3})}$

${96x^{2}z^{4}cos(y^{3}) +\frac{ (54y^{2}ln(x^{2}))}{z^{2}}+\frac{ (32y^{3})}{(xz^{2})}- 108x^{3}y^{2}z^{4}sin(y^{3})}$

${3.96x^{2}z^{4}cos(y^{3}) +\frac{ (2.25y^{2}ln(x^{2}))}{z^{2}}+\frac{ (1.32y^{3})}{(xz^{2})}- 4.5x^{3}y^{2}z^{4}sin(y^{3})}$

Correct answer:

${3.96x^{2}z^{4}cos(y^{3}) +\frac{ (2.25y^{2}ln(x^{2}))}{z^{2}}+\frac{ (1.32y^{3})}{(xz^{2})}- 4.5x^{3}y^{2}z^{4}sin(y^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(16)^2+(18)^2+(0)^2}=24.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{16}{24.08}=0.66\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{24.08}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{24.08}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.66)(6x^{2}z^{4}cos(y^{3}) +\frac{ (2y^{3})}{(xz^{2})})+(0.75)(\frac{(3y^{2}ln(x^{2}))}{z^{2}}- 6x^{3}y^{2}z^{4}sin(y^{3}))+(0)(8x^{3}z^{3}cos(y^{3}) -\frac{ (2y^{3}ln(x^{2}))}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=3.96x^{2}z^{4}cos(y^{3}) +\frac{ (2.25y^{2}ln(x^{2}))}{z^{2}}+\frac{ (1.32y^{3})}{(xz^{2})}- 4.5x^{3}y^{2}z^{4}sin(y^{3})\end{align*}$

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Example Question #1275 : Partial Derivatives

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

Possible Answers:

${\frac{(38x^{3}sin(x^{4})cos(z))}{y}}$

${38x^{3}sin(x^{4})ln(y)sin(z) -\frac{ (9.49cos(x^{4})sin(z))}{y}- 9.49cos(x^{4})cos(z)ln(y)}$

${\frac{(0.95cos(x^{4})sin(z))}{y}+ 1.28x^{3}sin(x^{4})ln(y)sin(z)}$

${\frac{(9cos(x^{4})sin(z))}{y}+ 12x^{3}sin(x^{4})ln(y)sin(z)}$

Correct answer:

${\frac{(0.95cos(x^{4})sin(z))}{y}+ 1.28x^{3}sin(x^{4})ln(y)sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(3)^2+(-9)^2+(0)^2}=9.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{3}{9.49}=0.32\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{9.49}=-0.95\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.49}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.32)(4x^{3}sin(x^{4})ln(y)sin(z))+(-0.95)(-\frac{(cos(x^{4})sin(z))}{y})+(0)(-1cos(x^{4})cos(z)ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.95cos(x^{4})sin(z))}{y}+ 1.28x^{3}sin(x^{4})ln(y)sin(z)\end{align*}$

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Example Question #1276 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- xln(y^{2})sin(z) - 4yln(x)ln(z)\\&\text{In the direction of }\overrightarrow{v}=(18,16,0).\end{align*}$

Possible Answers:

${- 96.3ln(x)ln(z) - 24.1ln(y^{2})sin(z) -\frac{ (96.3yln(x))}{z}-\frac{ (96.3yln(z))}{x}-\frac{ (48.2xsin(z))}{y}- 24.1xln(y^{2})cos(z)}$

${- 1.74ln(x)ln(z) - 0.562ln(y^{2})sin(z) -\frac{ (2.25yln(z))}{x}-\frac{ (0.871xsin(z))}{y}}$

${- 2.64ln(x)ln(z) - 0.75ln(y^{2})sin(z) -\frac{ (3yln(z))}{x}-\frac{ (1.32xsin(z))}{y}}$

${-\frac{ 96.3}{(xz)}-\frac{ (48.2cos(z))}{y}}$

Correct answer:

${- 2.64ln(x)ln(z) - 0.75ln(y^{2})sin(z) -\frac{ (3yln(z))}{x}-\frac{ (1.32xsin(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(16)^2+(0)^2}=24.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{24.08}=0.75\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{24.08}=0.66\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{24.08}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.75)(- ln(y^{2})sin(z) -\frac{ (4yln(z))}{x})+(0.66)(- 4ln(x)ln(z) -\frac{ (2xsin(z))}{y})+(0)(-\frac{ (4yln(x))}{z}- xln(y^{2})cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=- 2.64ln(x)ln(z) - 0.75ln(y^{2})sin(z) -\frac{ (3yln(z))}{x}-\frac{ (1.32xsin(z))}{y}\end{align*}$

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Example Question #1277 : Partial Derivatives

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

Possible Answers:

${1.92xy^{2}cos(y^{3})cos(z^{2}) - 0.96y^{2}e^{(x^{3})}cos(z) - 1.9cos(z^{2})sin(y^{3}) + 2.85x^{2}y^{3}e^{(x^{3})}cos(z)}$

${28.4x^{2}y^{2}e^{(x^{3})}sin(z) - 37.9y^{2}zcos(y^{3})sin(z^{2})}$

${1.8cos(z^{2})sin(y^{3}) - 0.307y^{2}e^{(x^{3})}cos(z) + 0.614xy^{2}cos(y^{3})cos(z^{2}) - 2.71x^{2}y^{3}e^{(x^{3})}cos(z)}$

${6.32cos(z^{2})sin(y^{3}) - 9.48y^{2}e^{(x^{3})}cos(z) + 3.16y^{3}e^{(x^{3})}sin(z) + 19xy^{2}cos(y^{3})cos(z^{2}) - 9.48x^{2}y^{3}e^{(x^{3})}cos(z) - 12.6xzsin(y^{3})sin(z^{2})}$

Correct answer:

${1.92xy^{2}cos(y^{3})cos(z^{2}) - 0.96y^{2}e^{(x^{3})}cos(z) - 1.9cos(z^{2})sin(y^{3}) + 2.85x^{2}y^{3}e^{(x^{3})}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-3)^2+(1)^2+(0)^2}=3.16\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-3}{3.16}=-0.95\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{3.16}=0.32\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{3.16}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.95)(2cos(z^{2})sin(y^{3}) - 3x^{2}y^{3}e^{(x^{3})}cos(z))+(0.32)(6xy^{2}cos(y^{3})cos(z^{2}) - 3y^{2}e^{(x^{3})}cos(z))+(0)(y^{3}e^{(x^{3})}sin(z) - 4xzsin(y^{3})sin(z^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=1.92xy^{2}cos(y^{3})cos(z^{2}) - 0.96y^{2}e^{(x^{3})}cos(z) - 1.9cos(z^{2})sin(y^{3}) + 2.85x^{2}y^{3}e^{(x^{3})}cos(z)\end{align*}$

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Example Question #1281 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(tan(y^{2})ln(z))}{x}- y^{4}sin(x^{3})e^{(z^{4})}\\&\text{In the direction of }\overrightarrow{v}=(0,7,17).\end{align*}$

Possible Answers:

${-\frac{ (36.8y\cdot (tan(y^{2})^{2} + 1))}{(x^{2}z)}- 882x^{2}y^{3}z^{3}cos(x^{3})e^{(z^{4})}}$

${\frac{(18.4\cdot tan(y^{2}))}{(xz)}-\frac{ (18.4\cdot tan(y^{2})ln(z))}{x^{2}}- 73.5y^{3}sin(x^{3})e^{(z^{4})} - 55.1x^{2}y^{4}cos(x^{3})e^{(z^{4})} - 73.5y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (36.8yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

${\frac{(0.92\cdot tan(y^{2}))}{(xz)}- 1.52y^{3}sin(x^{3})e^{(z^{4})} - 3.68y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.76yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

${\frac{(0.846\cdot tan(y^{2}))}{(xz)}- 0.578y^{3}sin(x^{3})e^{(z^{4})} - 3.39y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.289yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

Correct answer:

${\frac{(0.92\cdot tan(y^{2}))}{(xz)}- 1.52y^{3}sin(x^{3})e^{(z^{4})} - 3.68y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.76yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{18.38}=0.38\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{17}{18.38}=0.92\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{ (1\cdot tan(y^{2})ln(z))}{x^{2}}- 3x^{2}y^{4}cos(x^{3})e^{(z^{4})})+(0.38)(\frac{(2yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}- 4y^{3}sin(x^{3})e^{(z^{4})})+(0.92)(\frac{tan(y^{2})}{(xz)}- 4y^{4}z^{3}sin(x^{3})e^{(z^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.92\cdot tan(y^{2}))}{(xz)}- 1.52y^{3}sin(x^{3})e^{(z^{4})} - 3.68y^{4}z^{3}sin(x^{3})e^{(z^{4})} +\frac{ (0.76yln(z)\cdot (tan(y^{2})^{2} + 1))}{x}\end{align*}$

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Example Question #1282 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- e^{(x^{2})}e^{(z)} - 4x^{2}cos(z^{3})cos(y)\\&\text{In the direction of }\overrightarrow{v}=(-1,0,-16).\end{align*}$

Possible Answers:

${12x^{2}z^{2}sin(z^{3})cos(y) - 0.0288xcos(z^{3})cos(y) - 0.0072xe^{(x^{2})}e^{(z)} - e^{(x^{2})}e^{(z)}}$

${-385xz^{2}sin(z^{3})sin(y)}$

${e^{(x^{2})}e^{(z)} + 0.48xcos(z^{3})cos(y) + 0.12xe^{(x^{2})}e^{(z)} - 12x^{2}z^{2}sin(z^{3})cos(y)}$

${64.1x^{2}cos(z^{3})sin(y) - 16e^{(x^{2})}e^{(z)} - 128xcos(z^{3})cos(y) - 32.1xe^{(x^{2})}e^{(z)} + 192x^{2}z^{2}sin(z^{3})cos(y)}$

Correct answer:

${e^{(x^{2})}e^{(z)} + 0.48xcos(z^{3})cos(y) + 0.12xe^{(x^{2})}e^{(z)} - 12x^{2}z^{2}sin(z^{3})cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(0)^2+(-16)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{16.03}=-0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{16.03}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{16.03}=-1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.06)(- 8xcos(z^{3})cos(y) - 2xe^{(x^{2})}e^{(z)})+(0)(4x^{2}cos(z^{3})sin(y))+(-1)(12x^{2}z^{2}sin(z^{3})cos(y) - e^{(x^{2})}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=e^{(x^{2})}e^{(z)} + 0.48xcos(z^{3})cos(y) + 0.12xe^{(x^{2})}e^{(z)} - 12x^{2}z^{2}sin(z^{3})cos(y)\end{align*}$

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Calculus 3 : Calculus 3

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