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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #231 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2x^{2}y^{3}tan(z) - xy^{2}z^{3}\\&\text{In the direction of }\overrightarrow{v}=(0,-13,-12).\end{align*}$

Possible Answers:

${0.925x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 1.39xy^{2}z^{2} - 1.07xyz^{3} + 3.2x^{2}y^{2}tan(z)}$

${212xy^{2}\cdot (tan(z)^{2} + 1) - 106yz^{2}}$

${70.8xy^{3}tan(z) - 35.4xyz^{3} - 53.1xy^{2}z^{2} - 17.7y^{2}z^{3} + 35.4x^{2}y^{3}\cdot (tan(z)^{2} + 1) + 106x^{2}y^{2}tan(z)}$

${1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)}$

Correct answer:

${1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(-12)^2}=17.69\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.69}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{17.69}=-0.73\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{17.69}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4xy^{3}tan(z) - y^{2}z^{3})+(-0.73)(6x^{2}y^{2}tan(z) - 2xyz^{3})+(-0.68)(2x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 3xy^{2}z^{2})\\&D_{\overrightarrow{u}}(x,y,z)=1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)\end{align*}$

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Example Question #232 : Directional Derivatives

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

Possible Answers:

${\frac{(56\cdot 2^{(y^{2})}x^{2})}{z}+ 8\cdot 2^{(y^{2})}xln(z^{4}) - 56zcos(z^{2})ln(y)}$

${\frac{(29.1sin(z^{2}))}{y}-\frac{ (58.2\cdot 2^{(y^{2})}x^{2})}{z}- 29.1\cdot 2^{(y^{2})}xln(z^{4}) + 58.2zcos(z^{2})ln(y) - 20.2\cdot 2^{(y^{2})}x^{2}yln(z^{4})}$

${\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)}$

${-\frac{(161\cdot 2^{(y^{2})}xy)}{z}}$

Correct answer:

${\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(-14)^2}=14.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{14.56}=-0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{14.56}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.27)(-2\cdot 2^{(y^{2})}xln(z^{4}))+(0)(\frac{(2sin(z^{2}))}{y}- 1.39\cdot 2^{(y^{2})}x^{2}yln(z^{4}))+(-0.96)(4zcos(z^{2})ln(y) -\frac{ (4\cdot 2^{(y^{2})}x^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)\end{align*}$

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Example Question #233 : Directional Derivatives

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

Possible Answers:

${-63.5\cdot 3^{(x^{4})}x^{3}z^{3}sin(z^{4})cos(y)}$

${12\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 8.79\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${3.61\cdot 3^{(x^{4})}cos(z^{4})cos(y) - 14.4\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 15.9\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Correct answer:

${3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(0)^2+(-3)^2}=3.61\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{3.61}=0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{3.61}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{3.61}=-0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(4.39\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y))+(0)(3^{(x^{4})}cos(z^{4})cos(y))+(-0.83)(-4\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)\end{align*}$

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Example Question #1301 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2^{(x^{3})}ln(y^{3})ln(z^{3}) - 4ln(x^{2})ln(y^{2})ln(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-2,-12).\end{align*}$

Possible Answers:

${-\frac{ (0.205ln(x^{2})ln(z^{2}))}{y}-\frac{ (7.84ln(x^{2})ln(y^{2}))}{z}-\frac{ (0.0768\cdot 2^{(x^{3})}ln(z^{3}))}{y}-\frac{ (2.94\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$

${\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$

${-\frac{ (97.4ln(y^{2})ln(z^{2}))}{x}-\frac{ (97.4ln(x^{2})ln(z^{2}))}{y}-\frac{ (97.4ln(x^{2})ln(y^{2}))}{z}-\frac{ (36.5\cdot 2^{(x^{3})}ln(z^{3}))}{y}-\frac{ (36.5\cdot 2^{(x^{3})}ln(y^{3}))}{z}- 25.3\cdot 2^{(x^{3})}x^{2}ln(y^{3})ln(z^{3})}$

${-\frac{ 389.0}{(xyz)}-\frac{ (228\cdot 2^{(x^{3})}x^{2})}{(yz)}}$

Correct answer:

${\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{12.17}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{12.17}=-0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{ (8ln(y^{2})ln(z^{2}))}{x}- 2.08\cdot 2^{(x^{3})}x^{2}ln(y^{3})ln(z^{3}))+(-0.16)(-\frac{ (8ln(x^{2})ln(z^{2}))}{y}-\frac{ (3\cdot 2^{(x^{3})}ln(z^{3}))}{y})+(-0.99)(-\frac{ (8ln(x^{2})ln(y^{2}))}{z}-\frac{ (3\cdot 2^{(x^{3})}ln(y^{3}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}\end{align*}$

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Example Question #1302 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(5cos(y))}{x}+ x^{4}z^{4}cos(y^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-18,18).\end{align*}$

Possible Answers:

${\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})}$

${-815x^{3}yz^{3}sin(y^{2})}$

${102x^{3}z^{4}cos(y^{2}) -\frac{ (127sin(y))}{x}-\frac{ (127cos(y))}{x^{2}}+ 102x^{4}z^{3}cos(y^{2}) - 50.9x^{4}yz^{4}sin(y^{2})}$

${2.02x^{4}z^{3}cos(y^{2}) -\frac{ (2.52sin(y))}{x}- 1.01x^{4}yz^{4}sin(y^{2})}$

Correct answer:

${\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(18)^2}=25.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{25.46}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{25.46}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{25.46}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}z^{4}cos(y^{2}) -\frac{ (5cos(y))}{x^{2}})+(-0.71)(-\frac{ (5sin(y))}{x}- 2x^{4}yz^{4}sin(y^{2}))+(0.71)(4x^{4}z^{3}cos(y^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})\end{align*}$

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Example Question #1303 : Partial Derivatives

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

Possible Answers:

${- 20.4\cdot 2^{(y^{2})}ln(x^{2}) -\frac{ (40.8\cdot 2^{(y^{2})}z)}{x}- 28.3\cdot 2^{(y^{2})}yzln(x^{2})}$

${-\frac{(56.6\cdot 2^{(y^{2})}y)}{x}}$

${\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})}$

${-\frac{ (0.08\cdot 2^{(y^{2})}z)}{x}- 1.33\cdot 2^{(y^{2})}yzln(x^{2})}$

Correct answer:

${\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(20)^2+(0)^2}=20.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{20.4}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{20.4}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.4}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(-\frac{(2\cdot 2^{(y^{2})}z)}{x})+(0.98)(-1.39\cdot 2^{(y^{2})}yzln(x^{2}))+(0)(-1\cdot 2^{(y^{2})}ln(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})\end{align*}$

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Example Question #1304 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(z^{4})}x^{4}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(0,-17,9).\end{align*}$

Possible Answers:

${0.774\cdot 3^{(z^{4})}x^{4}cos(y) + 0.971\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${338\cdot 3^{(z^{4})}x^{3}z^{3}cos(y)}$

${19.2\cdot 3^{(z^{4})}x^{4}cos(y) + 77\cdot 3^{(z^{4})}x^{3}sin(y) + 84.5\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)}$

Correct answer:

${2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-17)^2+(9)^2}=19.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.24}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{19.24}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{19.24}=0.47\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4\cdot 3^{(z^{4})}x^{3}sin(y))+(-0.88)(3^{(z^{4})}x^{4}cos(y))+(0.47)(4.39\cdot 3^{(z^{4})}x^{4}z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)\end{align*}$

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Example Question #1305 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}ln(y^{2})e^{(z)} - cos(y^{2})cos(z)ln(x)\\&\text{In the direction of }\overrightarrow{v}=(-20,14,0).\end{align*}$

Possible Answers:

${\frac{(195x^{3}e^{(z)})}{y}-\frac{ (48.8ysin(y^{2})sin(z))}{x}}$

${\frac{(20cos(y^{2})cos(z))}{x}- 80x^{3}ln(y^{2})e^{(z)} +\frac{ (28x^{4}e^{(z)})}{y}+ 28ysin(y^{2})cos(z)ln(x)}$

${97.6x^{3}ln(y^{2})e^{(z)} -\frac{ (24.4cos(y^{2})cos(z))}{x}+ 24.4x^{4}ln(y^{2})e^{(z)} + 24.4cos(y^{2})ln(x)sin(z) +\frac{ (48.8x^{4}e^{(z)})}{y}+ 48.8ysin(y^{2})cos(z)ln(x)}$

${\frac{(0.82cos(y^{2})cos(z))}{x}- 3.28x^{3}ln(y^{2})e^{(z)} +\frac{ (1.14x^{4}e^{(z)})}{y}+ 1.14ysin(y^{2})cos(z)ln(x)}$

Correct answer:

${\frac{(0.82cos(y^{2})cos(z))}{x}- 3.28x^{3}ln(y^{2})e^{(z)} +\frac{ (1.14x^{4}e^{(z)})}{y}+ 1.14ysin(y^{2})cos(z)ln(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(14)^2+(0)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{24.41}=-0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{24.41}=0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.82)(4x^{3}ln(y^{2})e^{(z)} -\frac{ (cos(y^{2})cos(z))}{x})+(0.57)(\frac{(2x^{4}e^{(z)})}{y}+ 2ysin(y^{2})cos(z)ln(x))+(0)(x^{4}ln(y^{2})e^{(z)} + cos(y^{2})ln(x)sin(z))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.82cos(y^{2})cos(z))}{x}- 3.28x^{3}ln(y^{2})e^{(z)} +\frac{ (1.14x^{4}e^{(z)})}{y}+ 1.14ysin(y^{2})cos(z)ln(x)\end{align*}$

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Example Question #1306 : Partial Derivatives

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

Possible Answers:

${45.3x^{2}ln(z^{2})e^{(y)} - 22.6\cdot 3^{(z^{4})}x^{4}cos(y) - 90.5\cdot 3^{(z^{4})}x^{3}sin(y) +\frac{ (90.5x^{2}e^{(y)})}{z}+ 90.5xln(z^{2})e^{(y)} - 99.4\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${2.84\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (2.84x^{2}e^{(y)})}{z}- 2.84xln(z^{2})e^{(y)} + 3.12\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${64\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (64x^{2}e^{(y)})}{z}- 64xln(z^{2})e^{(y)} + 70.3\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${\frac{(181xe^{(y)})}{z}- 398\cdot 3^{(z^{4})}x^{3}z^{3}cos(y)}$

Correct answer:

${2.84\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (2.84x^{2}e^{(y)})}{z}- 2.84xln(z^{2})e^{(y)} + 3.12\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(0)^2+(-16)^2}=22.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{22.63}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{22.63}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{22.63}=-0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(4xln(z^{2})e^{(y)} - 4\cdot 3^{(z^{4})}x^{3}sin(y))+(0)(2x^{2}ln(z^{2})e^{(y)} - 1\cdot 3^{(z^{4})}x^{4}cos(y))+(-0.71)(\frac{(4x^{2}e^{(y)})}{z}- 4.39\cdot 3^{(z^{4})}x^{4}z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=2.84\cdot 3^{(z^{4})}x^{3}sin(y) -\frac{ (2.84x^{2}e^{(y)})}{z}- 2.84xln(z^{2})e^{(y)} + 3.12\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)\end{align*}$

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Example Question #1307 : Partial Derivatives

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

Possible Answers:

${111z^{2}sin(z^{3})tan(y) - 20.3\cdot 3^xsin(z^{2})e^{(y^{2})} - 36.9cos(z^{3})\cdot (tan(y)^{2} + 1) - 36.9\cdot 3^xzcos(z^{2})e^{(y^{2})} - 36.9\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}}$

${-81\cdot 3^xyzcos(z^{2})e^{(y^{2})}}$

${36cos(z^{3})\cdot (tan(y)^{2} + 1) + 4.39\cdot 3^xsin(z^{2})e^{(y^{2})} + 36\cdot 3^xysin(z^{2})e^{(y^{2})}}$

Correct answer:

${1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-18)^2+(0)^2}=18.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{18.44}=-0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.44}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.22)(-1.1\cdot 3^xsin(z^{2})e^{(y^{2})})+(-0.98)(- 2cos(z^{3})\cdot (tan(y)^{2} + 1) - 2\cdot 3^xysin(z^{2})e^{(y^{2})})+(0)(6z^{2}sin(z^{3})tan(y) - 2\cdot 3^xzcos(z^{2})e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.96cos(z^{3})\cdot (tan(y)^{2} + 1) + 0.242\cdot 3^xsin(z^{2})e^{(y^{2})} + 1.96\cdot 3^xysin(z^{2})e^{(y^{2})}\end{align*}$

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Calculus 3 : Directional Derivatives

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