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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1141 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3y^4\\&\text{In the direction of }\overrightarrow{v}=(2,9,0).\end{align*}$

Possible Answers:

${27.7x^2y^4 + 36.9x^3y^3}$

${6x^2y^4 + 36x^3y^3}$

${0.66x^2y^4 + 3.92x^3y^3}$

{0}

Correct answer:

${0.66x^2y^4 + 3.92x^3y^3}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(9)^2+(0)^2}=9.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{9.22}=0.22\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{9.22}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.22}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.22)(3x^2y^4)+(0.98)(4x^3y^3)+(0)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.66x^2y^4 + 3.92x^3y^3\end{align*}$

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Example Question #82 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4e^{(z)}ln{(y)} - sin{(z^2)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,-15,8).\end{align*}$

Possible Answers:

${1.88e^{(z)}ln{(y)} -\frac{ {(3.52e^{(z)})}}{y} + 0.88sin{(z^2)}cos{(y)} - 0.94zcos{(z^2)}sin{(y)}}$

${32e^{(z)}ln{(y)} -\frac{ {(60e^{(z)})}}{y} + 15sin{(z^2)}cos{(y)} - 16zcos{(z^2)}sin{(y)}}$

${\frac{{(68e^{(z)})}}{y} + 68e^{(z)}ln{(y)} - 17sin{(z^2)}cos{(y)} - 34zcos{(z^2)}sin{(y)}}$

Correct answer:

${1.88e^{(z)}ln{(y)} -\frac{ {(3.52e^{(z)})}}{y} + 0.88sin{(z^2)}cos{(y)} - 0.94zcos{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(8)^2}=17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{17}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8}{17}=0.47\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(-0.88)(\frac{{(4e^{(z)})}}{y} - sin{(z^2)}cos{(y)})+(0.47)(4e^{(z)}ln{(y)} - 2zcos{(z^2)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=1.88e^{(z)}ln{(y)} -\frac{ {(3.52e^{(z)})}}{y} + 0.88sin{(z^2)}cos{(y)} - 0.94zcos{(z^2)}sin{(y)}\end{align*}$

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Example Question #83 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(z^2)}sin{(x^2)}sin{(y)} - 3x^4ycos{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-19,0,-11).\end{align*}$

Possible Answers:

${65.8x^4ysin{(z)} - 263x^3ycos{(z)} - 65.8x^4cos{(z)} + 22cos{(z^2)}sin{(x^2)}cos{(y)} + 43.9xcos{(x^2)}cos{(z^2)}sin{(y)} - 43.9zsin{(x^2)}sin{(z^2)}sin{(y)}}$

${228x^3ycos{(z)} - 33x^4ysin{(z)} - 38xcos{(x^2)}cos{(z^2)}sin{(y)} + 22zsin{(x^2)}sin{(z^2)}sin{(y)}}$

${10.4x^3ycos{(z)} - 1.5x^4ysin{(z)} - 1.74xcos{(x^2)}cos{(z^2)}sin{(y)} + zsin{(x^2)}sin{(z^2)}sin{(y)}}$

${263x^3sin{(z)} - 87.8xzcos{(x^2)}sin{(z^2)}cos{(y)}}$

Correct answer:

${10.4x^3ycos{(z)} - 1.5x^4ysin{(z)} - 1.74xcos{(x^2)}cos{(z^2)}sin{(y)} + zsin{(x^2)}sin{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(0)^2+(-11)^2}=21.95\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{21.95}=-0.87\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{21.95}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{21.95}=-0.5\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.87)(2xcos{(x^2)}cos{(z^2)}sin{(y)} - 12x^3ycos{(z)})+(0)(cos{(z^2)}sin{(x^2)}cos{(y)} - 3x^4cos{(z)})+(-0.5)(3x^4ysin{(z)} - 2zsin{(x^2)}sin{(z^2)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=10.4x^3ycos{(z)} - 1.5x^4ysin{(z)} - 1.74xcos{(x^2)}cos{(z^2)}sin{(y)} + zsin{(x^2)}sin{(z^2)}sin{(y)}\end{align*}$

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Example Question #81 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- cos{(y^2)}cos{(z^3)}sin{(x)} - x^4z^3sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(13,0,16).\end{align*}$

Possible Answers:

${41.2ycos{(z^3)}sin{(y^2)}sin{(x)} - 20.6x^4z^3cos{(y)} - 82.5x^3z^3sin{(y)} - 61.9x^4z^2sin{(y)} - 20.6cos{(y^2)}cos{(z^3)}cos{(x)} + 61.9z^2cos{(y^2)}sin{(z^3)}sin{(x)}}$

${2.34z^2cos{(y^2)}sin{(z^3)}sin{(x)} - 2.52x^3z^3sin{(y)} - 2.34x^4z^2sin{(y)} - 0.63cos{(y^2)}cos{(z^3)}cos{(x)}}$

${- 247x^3z^2cos{(y)} - 124yz^2sin{(y^2)}sin{(z^3)}cos{(x)}}$

${48z^2cos{(y^2)}sin{(z^3)}sin{(x)} - 52x^3z^3sin{(y)} - 48x^4z^2sin{(y)} - 13cos{(y^2)}cos{(z^3)}cos{(x)}}$

Correct answer:

${2.34z^2cos{(y^2)}sin{(z^3)}sin{(x)} - 2.52x^3z^3sin{(y)} - 2.34x^4z^2sin{(y)} - 0.63cos{(y^2)}cos{(z^3)}cos{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(13)^2+(0)^2+(16)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{13}{20.62}=0.63\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{20.62}=0.78\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.63)(- cos{(y^2)}cos{(z^3)}cos{(x)} - 4x^3z^3sin{(y)})+(0)(2ycos{(z^3)}sin{(y^2)}sin{(x)} - x^4z^3cos{(y)})+(0.78)(3z^2cos{(y^2)}sin{(z^3)}sin{(x)} - 3x^4z^2sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.34z^2cos{(y^2)}sin{(z^3)}sin{(x)} - 2.52x^3z^3sin{(y)} - 2.34x^4z^2sin{(y)} - 0.63cos{(y^2)}cos{(z^3)}cos{(x)}\end{align*}$

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Example Question #1151 : Partial Derivatives

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

Possible Answers:

${\frac{{(97.6sin{(y^2)})}}{z} + 24.4\cdot4^xz^2cos{(y)} + 33.8\cdot4^xz^2sin{(y)} + 97.6ycos{(y^2)}ln{(z^2)} + 48.8\cdot4^xzsin{(y)}}$

${67.7\cdot4^xzcos{(y)}}$

${\frac{{(3.28sin{(y^2)})}}{z} - 0.79\cdot4^xz^2sin{(y)} + 1.64\cdot4^xzsin{(y)}}$

${\frac{{(80sin{(y^2)})}}{z} - 19.4\cdot4^xz^2sin{(y)} + 40\cdot4^xzsin{(y)}}$

Correct answer:

${\frac{{(3.28sin{(y^2)})}}{z} - 0.79\cdot4^xz^2sin{(y)} + 1.64\cdot4^xzsin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(0)^2+(20)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{24.41}=-0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{24.41}=0.82\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.57)(1.39\cdot4^xz^2sin{(y)})+(0)(4^xz^2cos{(y)} + 4ycos{(y^2)}ln{(z^2)})+(0.82)(\frac{{(4sin{(y^2)})}}{z} + 2\cdot4^xzsin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(3.28sin{(y^2)})}}{z} - 0.79\cdot4^xz^2sin{(y)} + 1.64\cdot4^xzsin{(y)}\end{align*}$

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Example Question #85 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^2cos{(z^3)}sin{(y)} - 4x^4sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(-15,7,0).\end{align*}$

Possible Answers:

${240x^3sin{(y^3)} - 30xcos{(z^3)}sin{(y)} - 84x^4y^2cos{(y^3)} + 7x^2cos{(z^3)}cos{(y)}}$

${-99.3xz^2sin{(z^3)}cos{(y)}}$

${33.1xcos{(z^3)}sin{(y)} - 265x^3sin{(y^3)} - 199x^4y^2cos{(y^3)} + 16.5x^2cos{(z^3)}cos{(y)} - 49.7x^2z^2sin{(z^3)}sin{(y)}}$

${14.6x^3sin{(y^3)} - 1.82xcos{(z^3)}sin{(y)} - 5.04x^4y^2cos{(y^3)} + 0.42x^2cos{(z^3)}cos{(y)}}$

Correct answer:

${14.6x^3sin{(y^3)} - 1.82xcos{(z^3)}sin{(y)} - 5.04x^4y^2cos{(y^3)} + 0.42x^2cos{(z^3)}cos{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(7)^2+(0)^2}=16.55\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{16.55}=-0.91\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{16.55}=0.42\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.55}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.91)(2xcos{(z^3)}sin{(y)} - 16x^3sin{(y^3)})+(0.42)(x^2cos{(z^3)}cos{(y)} - 12x^4y^2cos{(y^3)})+(0)(-3x^2z^2sin{(z^3)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=14.6x^3sin{(y^3)} - 1.82xcos{(z^3)}sin{(y)} - 5.04x^4y^2cos{(y^3)} + 0.42x^2cos{(z^3)}cos{(y)}\end{align*}$

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Example Question #1152 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(y^2)}sin{(z^3)} - 3ln{(x^2)}e^{(z^2)}e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(15,0,20).\end{align*}$

Possible Answers:

${\frac{-{(300ze^{(z^2)}e^{(y)})}}{x}}$

${69.3\cdot4^{(y^2)}ysin{(z^3)} -\frac{ {(150e^{(z^2)}e^{(y)})}}{x} - 75ln{(x^2)}e^{(z^2)}e^{(y)} + 75\cdot4^{(y^2)}z^2cos{(z^3)} - 150zln{(x^2)}e^{(z^2)}e^{(y)}}$

${2.4\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(3.6e^{(z^2)}e^{(y)})}}{x} - 4.8zln{(x^2)}e^{(z^2)}e^{(y)}}$

${60\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(90e^{(z^2)}e^{(y)})}}{x} - 120zln{(x^2)}e^{(z^2)}e^{(y)}}$

Correct answer:

${2.4\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(3.6e^{(z^2)}e^{(y)})}}{x} - 4.8zln{(x^2)}e^{(z^2)}e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(15)^2+(0)^2+(20)^2}=25\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{15}{25}=0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{25}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{25}=0.8\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.6)(\frac{-{(6e^{(z^2)}e^{(y)})}}{x})+(0)(2.77\cdot4^{(y^2)}ysin{(z^3)} - 3ln{(x^2)}e^{(z^2)}e^{(y)})+(0.8)(3\cdot4^{(y^2)}z^2cos{(z^3)} - 6zln{(x^2)}e^{(z^2)}e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.4\cdot4^{(y^2)}z^2cos{(z^3)} -\frac{ {(3.6e^{(z^2)}e^{(y)})}}{x} - 4.8zln{(x^2)}e^{(z^2)}e^{(y)}\end{align*}$

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Example Question #1153 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(x^2)}e^{(y)}cos{(z)} - 4y^2cos{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(5,7,0).\end{align*}$

Possible Answers:

${0.81sin{(x^2)}e^{(y)}cos{(z)} - 6.48ycos{(z^2)} + 1.16xcos{(x^2)}e^{(y)}cos{(z)}}$

${7sin{(x^2)}e^{(y)}cos{(z)} - 56ycos{(z^2)} + 10xcos{(x^2)}e^{(y)}cos{(z)}}$

${8.6sin{(x^2)}e^{(y)}cos{(z)} - 68.8ycos{(z^2)} - 8.6sin{(x^2)}e^{(y)}sin{(z)} + 68.8y^2zsin{(z^2)} + 17.2xcos{(x^2)}e^{(y)}cos{(z)}}$

${-17.2xcos{(x^2)}e^{(y)}sin{(z)}}$

Correct answer:

${0.81sin{(x^2)}e^{(y)}cos{(z)} - 6.48ycos{(z^2)} + 1.16xcos{(x^2)}e^{(y)}cos{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(7)^2+(0)^2}=8.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{8.6}=0.58\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{8.6}=0.81\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{8.6}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.58)(2xcos{(x^2)}e^{(y)}cos{(z)})+(0.81)(sin{(x^2)}e^{(y)}cos{(z)} - 8ycos{(z^2)})+(0)(8y^2zsin{(z^2)} - sin{(x^2)}e^{(y)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.81sin{(x^2)}e^{(y)}cos{(z)} - 6.48ycos{(z^2)} + 1.16xcos{(x^2)}e^{(y)}cos{(z)}\end{align*}$

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Example Question #1154 : Partial Derivatives

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

Possible Answers:

${-5e^{(x)}cos{(y)}}$

{0}

${-0.71e^{(x)}cos{(y)}}$

${7.07e^{(x)}sin{(y)} - 7.07e^{(x)}cos{(y)}}$

Correct answer:

${-0.71e^{(x)}cos{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(0)^2+(5)^2}=7.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{7.07}=0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{7.07}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{7.07}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.71)(-1e^{(x)}cos{(y)})+(0)(e^{(x)}sin{(y)})+(0.71)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-0.71e^{(x)}cos{(y)}\end{align*}$

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Example Question #88 : Directional Derivatives

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

Possible Answers:

${1.35y^2sin{(z^3)} - 0.9yz + 0.89y^2 - 2.67y^3z^2cos{(z^3)}}$

{0}

${3y^2sin{(z^3)} - 2yz + 2y^2 - 6y^3z^2cos{(z^3)}}$

${6.72y^2sin{(z^3)} - 4.48yz - 2.24y^2 + 6.72y^3z^2cos{(z^3)}}$

Correct answer:

${1.35y^2sin{(z^3)} - 0.9yz + 0.89y^2 - 2.67y^3z^2cos{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(1)^2+(-2)^2}=2.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{2.24}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{2.24}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{2.24}=-0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(0.45)(3y^2sin{(z^3)} - 2yz)+(-0.89)(3y^3z^2cos{(z^3)} - y^2)\\&D_{\overrightarrow{u}}(x,y,z)=1.35y^2sin{(z^3)} - 0.9yz + 0.89y^2 - 2.67y^3z^2cos{(z^3)}\end{align*}$

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Calculus 3 : Directional Derivatives

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