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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1118 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3ze^{(x^2)}e^{(y^2)} - cos{(y^2)}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,-1,12).\end{align*}$

Possible Answers:

$36e^{(x^2)}e^{(y^2)} - 2ysin{(y^2)}e^{(x)} - 6yze^{(x^2)}e^{(y^2)}$

$36.1e^{(x^2)}e^{(y^2)} - 12cos{(y^2)}e^{(x)} + 24.1ysin{(y^2)}e^{(x)} + 72.2xze^{(x^2)}e^{(y^2)} + 72.2yze^{(x^2)}e^{(y^2)}$

$144xye^{(x^2)}e^{(y^2)}$

$3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}$

Correct answer:

$3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-1)^2+(12)^2}=12.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.04}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-1}{12.04}=-0.08\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{12.04}=1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(6xze^{(x^2)}e^{(y^2)} -cos{(y^2)}e^{(x)})+(-0.08)(2ysin{(y^2)}e^{(x)} + 6yze^{(x^2)}e^{(y^2)})+(1)(3e^{(x^2)}e^{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=3e^{(x^2)}e^{(y^2)} - 0.16ysin{(y^2)}e^{(x)} - 0.48yze^{(x^2)}e^{(y^2)}\end{align*}$

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Example Question #1119 : Partial Derivatives

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

Possible Answers:

$-\frac{ {(19cos{(x)})}}{{(yz)}} - 105\cdot2^{(x^2)}xy^2z^3sin{(y^3)}$

$\frac{{(6ln{(z)}sin{(x)})}}{y} -\frac{ {(18ln{(y)}sin{(x)})}}{z} + 24\cdot2^{(x^2)}z^3cos{(y^3)} + 6\cdot2^{(x^2)}y^2z^4sin{(y^3)}$

${25.3\cdot2^{(x^2)}z^3cos{(y^3)} -\frac{ {(19ln{(z)}sin{(x)})}}{y} -\frac{ {(19ln{(y)}sin{(x)})}}{z} - 19cos{(x)}ln{(y)}ln{(z)} + 8.76\cdot2^{(x^2)}xz^4cos{(y^3)} - 19\cdot2^{(x^2)}y^2z^4sin{(y^3)}}$

$\frac{{(0.96ln{(z)}sin{(x)})}}{y} -\frac{ {(2.85ln{(y)}sin{(x)})}}{z} + 3.8\cdot2^{(x^2)}z^3cos{(y^3)} + 0.96\cdot2^{(x^2)}y^2z^4sin{(y^3)}$

Correct answer:

$\frac{{(0.96ln{(z)}sin{(x)})}}{y} -\frac{ {(2.85ln{(y)}sin{(x)})}}{z} + 3.8\cdot2^{(x^2)}z^3cos{(y^3)} + 0.96\cdot2^{(x^2)}y^2z^4sin{(y^3)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-2)^2+(6)^2}=6.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{6.32}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{6.32}=-0.32\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{6.32}=0.95\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(1.39\cdot2^{(x^2)}xz^4cos{(y^3)} - 3cos{(x)}ln{(y)}ln{(z)})+(-0.32)(-\frac{ {(3ln{(z)}sin{(x)})}}{y} - 3\cdot2^{(x^2)}y^2z^4sin{(y^3)})+(0.95)(4\cdot2^{(x^2)}z^3cos{(y^3)} -\frac{ {(3ln{(y)}sin{(x)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.96ln{(z)}sin{(x)})}}{y} -\frac{ {(2.85ln{(y)}sin{(x)})}}{z} + 3.8\cdot2^{(x^2)}z^3cos{(y^3)} + 0.96\cdot2^{(x^2)}y^2z^4sin{(y^3)}\end{align*}$

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Example Question #1120 : Partial Derivatives

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

Possible Answers:

${120y^3e^{(x^2)}cos{(z)} - 60xy^4e^{(x^2)}cos{(z)} + 20xzcos{(y^2)}sin{(x^2)} - 20yzcos{(x^2)}sin{(y^2)}}$

${8.52y^3e^{(x^2)}cos{(z)} - 4.26xy^4e^{(x^2)}cos{(z)} + 1.42xzcos{(y^2)}sin{(x^2)} - 1.42yzcos{(x^2)}sin{(y^2)}}$

${42.4y^4e^{(x^2)}sin{(z)} - 170y^3e^{(x^2)}cos{(z)} - 14.1cos{(x^2)}cos{(y^2)} - 84.8xy^4e^{(x^2)}cos{(z)} + 28.3xzcos{(y^2)}sin{(x^2)} + 28.3yzcos{(x^2)}sin{(y^2)}}$

${339xy^3e^{(x^2)}sin{(z)} - 56.6xysin{(x^2)}sin{(y^2)}}$

Correct answer:

${8.52y^3e^{(x^2)}cos{(z)} - 4.26xy^4e^{(x^2)}cos{(z)} + 1.42xzcos{(y^2)}sin{(x^2)} - 1.42yzcos{(x^2)}sin{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(-10)^2+(0)^2}=14.14\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{14.14}=0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{14.14}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.14}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.71)(2xzcos{(y^2)}sin{(x^2)} - 6xy^4e^{(x^2)}cos{(z)})+(-0.71)(2yzcos{(x^2)}sin{(y^2)} - 12y^3e^{(x^2)}cos{(z)})+(0)(3y^4e^{(x^2)}sin{(z)} -cos{(x^2)}cos{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=8.52y^3e^{(x^2)}cos{(z)} - 4.26xy^4e^{(x^2)}cos{(z)} + 1.42xzcos{(y^2)}sin{(x^2)} - 1.42yzcos{(x^2)}sin{(y^2)}\end{align*}$

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Example Question #51 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5x^3z^2ln{(y^2)} - yzsin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-6,0,-10).\end{align*}$

Possible Answers:

${\frac{{(700x^2z)}}{y} - 35x^2cos{(x^3)}}$

${10ysin{(x^3)} - 90x^2z^2ln{(y^2)} - 100x^3zln{(y^2)} + 18x^2yzcos{(x^3)}}$

${175x^2z^2ln{(y^2)} - 11.7zsin{(x^3)} - 11.7ysin{(x^3)} +\frac{ {(117x^3z^2)}}{y} + 117x^3zln{(y^2)} - 35x^2yzcos{(x^3)}}$

${0.86ysin{(x^3)} - 7.65x^2z^2ln{(y^2)} - 8.6x^3zln{(y^2)} + 1.53x^2yzcos{(x^3)}}$

Correct answer:

${0.86ysin{(x^3)} - 7.65x^2z^2ln{(y^2)} - 8.6x^3zln{(y^2)} + 1.53x^2yzcos{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(0)^2+(-10)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{11.66}=-0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{11.66}=-0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.51)(15x^2z^2ln{(y^2)} - 3x^2yzcos{(x^3)})+(0)(\frac{{(10x^3z^2)}}{y} -zsin{(x^3)})+(-0.86)(10x^3zln{(y^2)} -ysin{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86ysin{(x^3)} - 7.65x^2z^2ln{(y^2)} - 8.6x^3zln{(y^2)} + 1.53x^2yzcos{(x^3)}\end{align*}$

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Example Question #52 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4cos{(z)}ln{(x)}sin{(y)} - sin{(y^3)}e^{(z)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-18,1,0).\end{align*}$

Possible Answers:

${\frac{{(72.1cos{(y)}sin{(z)})}}{x} + 54.1y^2cos{(y^3)}e^{(z)}sin{(x)}}$

${72.1ln{(x)}sin{(y)}sin{(z)} -\frac{ {(72.1cos{(z)}sin{(y)})}}{x} - 18sin{(y^3)}e^{(z)}cos{(x)} + 18sin{(y^3)}e^{(z)}sin{(x)} - 72.1cos{(y)}cos{(z)}ln{(x)} - 54.1y^2cos{(y^3)}e^{(z)}cos{(x)}}$

${\frac{{(4cos{(z)}sin{(y)})}}{x} -sin{(y^3)}e^{(z)}sin{(x)} - 0.24cos{(y)}cos{(z)}ln{(x)} - 0.18y^2cos{(y^3)}e^{(z)}cos{(x)}}$

${\frac{{(72cos{(z)}sin{(y)})}}{x} - 18sin{(y^3)}e^{(z)}sin{(x)} - 4cos{(y)}cos{(z)}ln{(x)} - 3y^2cos{(y^3)}e^{(z)}cos{(x)}}$

Correct answer:

${\frac{{(4cos{(z)}sin{(y)})}}{x} -sin{(y^3)}e^{(z)}sin{(x)} - 0.24cos{(y)}cos{(z)}ln{(x)} - 0.18y^2cos{(y^3)}e^{(z)}cos{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(1)^2+(0)^2}=18.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{18.03}=-1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{18.03}=0.06\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-1)(sin{(y^3)}e^{(z)}sin{(x)} -\frac{ {(4cos{(z)}sin{(y)})}}{x})+(0.06)(- 4cos{(y)}cos{(z)}ln{(x)} - 3y^2cos{(y^3)}e^{(z)}cos{(x)})+(0)(4ln{(x)}sin{(y)}sin{(z)} -sin{(y^3)}e^{(z)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(4cos{(z)}sin{(y)})}}{x} -sin{(y^3)}e^{(z)}sin{(x)} - 0.24cos{(y)}cos{(z)}ln{(x)} - 0.18y^2cos{(y^3)}e^{(z)}cos{(x)}\end{align*}$

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Example Question #1121 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^4z - 2ysin{(x^3)}cos{(z)}\\&\text{In the direction of }\overrightarrow{v}=(6,0,10).\end{align*}$

Possible Answers:

${10y^4 + 20ysin{(x^3)}sin{(z)} - 36x^2ycos{(x^3)}cos{(z)}}$

${70x^2cos{(x^3)}sin{(z)}}$

${0.86y^4 + 1.72ysin{(x^3)}sin{(z)} - 3.06x^2ycos{(x^3)}cos{(z)}}$

${46.6y^3z - 23.3sin{(x^3)}cos{(z)} + 11.7y^4 + 23.3ysin{(x^3)}sin{(z)} - 70x^2ycos{(x^3)}cos{(z)}}$

Correct answer:

${0.86y^4 + 1.72ysin{(x^3)}sin{(z)} - 3.06x^2ycos{(x^3)}cos{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(0)^2+(10)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{11.66}=0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{11.66}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.51)(-6x^2ycos{(x^3)}cos{(z)})+(0)(4y^3z - 2sin{(x^3)}cos{(z)})+(0.86)(y^4 + 2ysin{(x^3)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86y^4 + 1.72ysin{(x^3)}sin{(z)} - 3.06x^2ycos{(x^3)}cos{(z)}\end{align*}$

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Example Question #53 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(3e^{(y^2)})}}{z} + 2^{(y^2)}xsin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(13,0,-17).\end{align*}$

Possible Answers:

${13\cdot2^{(y^2)}sin{(z^3)} +\frac{ {(51e^{(y^2)})}}{z^2} - 51\cdot2^{(y^2)}xz^2cos{(z^3)}}$

${89\cdot2^{(y^2)}yz^2cos{(z^3)}}$

${21.4\cdot2^{(y^2)}sin{(z^3)} -\frac{ {(64.2e^{(y^2)})}}{z^2} +\frac{ {(128ye^{(y^2)})}}{z} + 29.7\cdot2^{(y^2)}xysin{(z^3)} + 64.2\cdot2^{(y^2)}xz^2cos{(z^3)}}$

${0.61\cdot2^{(y^2)}sin{(z^3)} +\frac{ {(2.37e^{(y^2)})}}{z^2} - 2.37\cdot2^{(y^2)}xz^2cos{(z^3)}}$

Correct answer:

${0.61\cdot2^{(y^2)}sin{(z^3)} +\frac{ {(2.37e^{(y^2)})}}{z^2} - 2.37\cdot2^{(y^2)}xz^2cos{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(-17)^2}=21.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{21.4}=0.61\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{21.4}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{21.4}=-0.79\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.61)(2^{(y^2)}sin{(z^3)})+(0)(\frac{{(6ye^{(y^2)})}}{z} + 1.39\cdot2^{(y^2)}xysin{(z^3)})+(-0.79)(3\cdot2^{(y^2)}xz^2cos{(z^3)} -\frac{ {(3e^{(y^2)})}}{z^2})\\&D_{\overrightarrow{u}}(x,y,z)=0.61\cdot2^{(y^2)}sin{(z^3)} +\frac{ {(2.37e^{(y^2)})}}{z^2} - 2.37\cdot2^{(y^2)}xz^2cos{(z^3)}\end{align*}$

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Example Question #54 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- yz^3e^{(x)} - ln{(x^2)}sin{(z^3)}cos{(y)}\\&\text{In the direction of }\overrightarrow{v}=(14,0,-6).\end{align*}$

Possible Answers:

${\frac{{(91.4z^2cos{(z^3)}sin{(y)})}}{x} - 45.7z^2e^{(x)}}$

${18yz^2e^{(x)} -\frac{ {(28sin{(z^3)}cos{(y)})}}{x} - 14yz^3e^{(x)} + 18z^2cos{(z^3)}ln{(x^2)}cos{(y)}}$

${15.2ln{(x^2)}sin{(z^3)}sin{(y)} -\frac{ {(30.5sin{(z^3)}cos{(y)})}}{x} - 45.7yz^2e^{(x)} - 15.2yz^3e^{(x)} - 15.2z^3e^{(x)} - 45.7z^2cos{(z^3)}ln{(x^2)}cos{(y)}}$

${1.17yz^2e^{(x)} -\frac{ {(1.84sin{(z^3)}cos{(y)})}}{x} - 0.92yz^3e^{(x)} + 1.17z^2cos{(z^3)}ln{(x^2)}cos{(y)}}$

Correct answer:

${1.17yz^2e^{(x)} -\frac{ {(1.84sin{(z^3)}cos{(y)})}}{x} - 0.92yz^3e^{(x)} + 1.17z^2cos{(z^3)}ln{(x^2)}cos{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(0)^2+(-6)^2}=15.23\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{15.23}=0.92\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.23}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{15.23}=-0.39\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.92)(-\frac{ {(2sin{(z^3)}cos{(y)})}}{x} - yz^3e^{(x)})+(0)(ln{(x^2)}sin{(z^3)}sin{(y)} - z^3e^{(x)})+(-0.39)(- 3yz^2e^{(x)} - 3z^2cos{(z^3)}ln{(x^2)}cos{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=1.17yz^2e^{(x)} -\frac{ {(1.84sin{(z^3)}cos{(y)})}}{x} - 0.92yz^3e^{(x)} + 1.17z^2cos{(z^3)}ln{(x^2)}cos{(y)}\end{align*}$

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Example Question #1122 : Partial Derivatives

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

Possible Answers:

${10yz - 16z^2}$

${0.6yz - 0.95z^2}$

${- 33.5yz - 16.8z^2}$

Correct answer:

${0.6yz - 0.95z^2}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(16)^2+(-5)^2}=16.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{16.76}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{16.76}=0.95\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{16.76}=-0.3\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(0.95)(-1z^2)+(-0.3)(-2yz)\\&D_{\overrightarrow{u}}(x,y,z)=0.6yz - 0.95z^2\end{align*}$

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Example Question #1123 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=zsin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(0,-20,-19).\end{align*}$

Possible Answers:

{-0.69sin{(x)}}

{27.6sin{(x)} + 27.6zcos{(x)}}

{0}

{-19sin{(x)}}

Correct answer:

{-0.69sin{(x)}}

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-20)^2+(-19)^2}=27.59\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{27.59}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{27.59}=-0.72\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{27.59}=-0.69\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(zcos{(x)})+(-0.72)(0.0)+(-0.69)(sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=-0.69sin{(x)}\end{align*}$

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Calculus 3 : Directional Derivatives

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