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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #131 : Directional Derivatives

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

Possible Answers:

${\frac{-{(102sin{(y)})}}{{(xz)}}}$

${\frac{{(50.9ln{(z^2)}cos{(y)})}}{x} +\frac{ {(50.9ln{(x^2)}cos{(y)})}}{z} - 25.5ln{(x^2)}ln{(z^2)}sin{(y)}}$

${\frac{{(1.01ln{(z^2)}cos{(y)})}}{x} +\frac{ {(1.01ln{(x^2)}cos{(y)})}}{z}}$

${-\frac{ {(1.42ln{(z^2)}cos{(y)})}}{x} -\frac{ {(1.42ln{(x^2)}cos{(y)})}}{z}}$

Correct answer:

${-\frac{ {(1.42ln{(z^2)}cos{(y)})}}{x} -\frac{ {(1.42ln{(x^2)}cos{(y)})}}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(0)^2+(-18)^2}=25.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{25.46}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{25.46}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{25.46}=-0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(\frac{{(2ln{(z^2)}cos{(y)})}}{x})+(0)(-1ln{(x^2)}ln{(z^2)}sin{(y)})+(-0.71)(\frac{{(2ln{(x^2)}cos{(y)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(1.42ln{(z^2)}cos{(y)})}}{x} -\frac{ {(1.42ln{(x^2)}cos{(y)})}}{z}\end{align*}$

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Example Question #132 : Directional Derivatives

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

Possible Answers:

${- 17.7e^{(x^2)}ln{(y)} -\frac{ {(17.7ze^{(x^2)})}}{y} - 35.4xze^{(x^2)}ln{(y)}}$

${0.68e^{(x^2)}ln{(y)} - 1.46xze^{(x^2)}ln{(y)}}$

${\frac{-{(35.4xe^{(x^2)})}}{y}}$

${12e^{(x^2)}ln{(y)} - 26xze^{(x^2)}ln{(y)}}$

Correct answer:

${0.68e^{(x^2)}ln{(y)} - 1.46xze^{(x^2)}ln{(y)}}$

Explanation:

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

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Example Question #133 : Directional Derivatives

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

Possible Answers:

${0.666\cdot2^xy^4sin{(z^3)} + 0.145\cdot2^xy^4z^2cos{(z^3)}}$

${0.679\cdot2^xy^4sin{(z^3)} - 0.66\cdot2^xy^4z^2cos{(z^3)}}$

${36.9\cdot2^xy^3sin{(z^3)} + 6.39\cdot2^xy^4sin{(z^3)} + 27.7\cdot2^xy^4z^2cos{(z^3)}}$

${76.7\cdot2^xy^3z^2cos{(z^3)}}$

Correct answer:

${0.679\cdot2^xy^4sin{(z^3)} - 0.66\cdot2^xy^4z^2cos{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(9)^2+(0)^2+(-2)^2}=9.22\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{9}{9.22}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{9.22}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{9.22}=-0.22\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(0.693\cdot2^xy^4sin{(z^3)})+(0)(4\cdot2^xy^3sin{(z^3)})+(-0.22)(3\cdot2^xy^4z^2cos{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=0.679\cdot2^xy^4sin{(z^3)} - 0.66\cdot2^xy^4z^2cos{(z^3)}\end{align*}$

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Example Question #134 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(y^2)}\cdot3^zln{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-20,15,0).\end{align*}$

Possible Answers:

${-\frac{ {(0.64\cdot3^z\cdot3^{(y^2)})}}{x} - 0.791\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

${-\frac{ {(25\cdot3^z\cdot3^{(y^2)})}}{x} - 27.5\cdot3^z\cdot3^{(y^2)}ln{(x)} - 54.9\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

${\frac{-{(60.3\cdot3^z\cdot3^{(y^2)}y)}}{x}}$

${\frac{{(0.8\cdot3^z\cdot3^{(y^2)})}}{x} - 1.32\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

Correct answer:

${\frac{{(0.8\cdot3^z\cdot3^{(y^2)})}}{x} - 1.32\cdot3^z\cdot3^{(y^2)}yln{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(15)^2+(0)^2}=25\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{25}=-0.8\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{25}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{25}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.8)(\frac{-{(1\cdot3^z\cdot3^{(y^2)})}}{x})+(0.6)(-2.2\cdot3^z\cdot3^{(y^2)}yln{(x)})+(0)(-1.1\cdot3^z\cdot3^{(y^2)}ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.8\cdot3^z\cdot3^{(y^2)})}}{x} - 1.32\cdot3^z\cdot3^{(y^2)}yln{(x)}\end{align*}$

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Example Question #135 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{-{(cos{(x^3)}cos{(y^2)})}}{z}\\&\text{In the direction of }\overrightarrow{v}=(0,1,15).\end{align*}$

Possible Answers:

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.0098ycos{(x^3)}sin{(y^2)})}}{z}}$

${\frac{{(90.2x^2ysin{(x^3)}sin{(y^2)})}}{z^2}}$

${\frac{{(15cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(30.1ycos{(x^3)}sin{(y^2)})}}{z} +\frac{ {(45.1x^2cos{(y^2)}sin{(x^3)})}}{z}}$

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}}$

Correct answer:

${\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(1)^2+(15)^2}=15.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.03}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{15.03}=0.07\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{15.03}=1\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(3x^2cos{(y^2)}sin{(x^3)})}}{z})+(0.07)(\frac{{(2ycos{(x^3)}sin{(y^2)})}}{z})+(1)(\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(cos{(x^3)}cos{(y^2)})}}{z^2} +\frac{ {(0.14ycos{(x^3)}sin{(y^2)})}}{z}\end{align*}$

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Example Question #136 : Directional Derivatives

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

Possible Answers:

${\frac{{(30.5\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(8\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

${\frac{{(64.9\cdot4^{(z^2)}zsin{(x)})}}{y^3}}$

${\frac{{(2.61\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(0.68\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

${\frac{{(32.4\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(11.7\cdot4^{(z^2)}sin{(x)})}}{y^2} -\frac{ {(23.4\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

Correct answer:

${\frac{{(2.61\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(0.68\cdot4^{(z^2)}cos{(x)})}}{y^3}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(4)^2+(11)^2}=11.7\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{11.7}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{11.7}=0.34\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{11.7}=0.94\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(1\cdot4^{(z^2)}sin{(x)})}}{y^2})+(0.34)(\frac{-{(2\cdot4^{(z^2)}cos{(x)})}}{y^3})+(0.94)(\frac{{(2.77\cdot4^{(z^2)}zcos{(x)})}}{y^2})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(2.61\cdot4^{(z^2)}zcos{(x)})}}{y^2} -\frac{ {(0.68\cdot4^{(z^2)}cos{(x)})}}{y^3}\end{align*}$

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Example Question #137 : Directional Derivatives

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

Possible Answers:

${\frac{-{(34.4yzcos{(z^2)}sin{(y^2)})}}{x}}$

${0.673zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.31ysin{(y^2)}sin{(z^2)}ln{(x)}}$

${\frac{{(8.6cos{(y^2)}sin{(z^2)})}}{x} + 17.2zcos{(y^2)}cos{(z^2)}ln{(x)} - 17.2ysin{(y^2)}sin{(z^2)}ln{(x)}}$

${- 1.16zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.62ysin{(y^2)}sin{(z^2)}ln{(x)}}$

Correct answer:

${- 1.16zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.62ysin{(y^2)}sin{(z^2)}ln{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(7)^2+(-5)^2}=8.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{8.6}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{8.6}=0.81\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{8.6}=-0.58\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{{(cos{(y^2)}sin{(z^2)})}}{x})+(0.81)(-2ysin{(y^2)}sin{(z^2)}ln{(x)})+(-0.58)(2zcos{(y^2)}cos{(z^2)}ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.16zcos{(y^2)}cos{(z^2)}ln{(x)} - 1.62ysin{(y^2)}sin{(z^2)}ln{(x)}\end{align*}$

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Example Question #138 : Directional Derivatives

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

Possible Answers:

${104xy^2cos{(x^2)}cos{(z)}}$

${- 0.17y^3sin{(x^2)}cos{(z)} - 2.94y^2sin{(x^2)}sin{(z)}}$

${17.3y^3sin{(x^2)}cos{(z)} + 51.8y^2sin{(x^2)}sin{(z)} + 34.5xy^3cos{(x^2)}sin{(z)}}$

${- 3y^3sin{(x^2)}cos{(z)} - 51y^2sin{(x^2)}sin{(z)}}$

Correct answer:

${- 0.17y^3sin{(x^2)}cos{(z)} - 2.94y^2sin{(x^2)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-17)^2+(-3)^2}=17.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.26}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{17.26}=-0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{17.26}=-0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xy^3cos{(x^2)}sin{(z)})+(-0.98)(3y^2sin{(x^2)}sin{(z)})+(-0.17)(y^3sin{(x^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.17y^3sin{(x^2)}cos{(z)} - 2.94y^2sin{(x^2)}sin{(z)}\end{align*}$

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Example Question #139 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(z^2)}\cdot3^xy^2\\&\text{In the direction of }\overrightarrow{v}=(-10,0,-15).\end{align*}$

Possible Answers:

${- 0.332\cdot2^{(z^2)}\cdot3^xy^2 - 0.955\cdot2^{(z^2)}\cdot3^xy^2z}$

${- 36.1\cdot2^{(z^2)}\cdot3^xy - 19.8\cdot2^{(z^2)}\cdot3^xy^2 - 25\cdot2^{(z^2)}\cdot3^xy^2z}$

${-54.9\cdot2^{(z^2)}\cdot3^xyz}$

${0.604\cdot2^{(z^2)}\cdot3^xy^2 + 1.15\cdot2^{(z^2)}\cdot3^xy^2z}$

Correct answer:

${0.604\cdot2^{(z^2)}\cdot3^xy^2 + 1.15\cdot2^{(z^2)}\cdot3^xy^2z}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-10)^2+(0)^2+(-15)^2}=18.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-10}{18.03}=-0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{18.03}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-15}{18.03}=-0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.55)(-1.1\cdot2^{(z^2)}\cdot3^xy^2)+(0)(-2\cdot2^{(z^2)}\cdot3^xy)+(-0.83)(-1.39\cdot2^{(z^2)}\cdot3^xy^2z)\\&D_{\overrightarrow{u}}(x,y,z)=0.604\cdot2^{(z^2)}\cdot3^xy^2 + 1.15\cdot2^{(z^2)}\cdot3^xy^2z\end{align*}$

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Example Question #140 : Directional Derivatives

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

Possible Answers:

${- 0.75z^2sin{(y^3)}sin{(x)} - 2.27y^2z^3cos{(y^3)}sin{(x)}}$

${-198y^2z^2cos{(y^3)}cos{(x)}}$

${- 22z^3sin{(y^3)}cos{(x)} - 65.8z^2sin{(y^3)}sin{(x)} - 65.8y^2z^3cos{(y^3)}sin{(x)}}$

${- 1.5z^2sin{(y^3)}sin{(x)} - 2.61y^2z^3cos{(y^3)}sin{(x)}}$

Correct answer:

${- 1.5z^2sin{(y^3)}sin{(x)} - 2.61y^2z^3cos{(y^3)}sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(19)^2+(11)^2}=21.95\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.95}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{21.95}=0.87\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{21.95}=0.5\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1z^3sin{(y^3)}cos{(x)})+(0.87)(-3y^2z^3cos{(y^3)}sin{(x)})+(0.5)(-3z^2sin{(y^3)}sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.5z^2sin{(y^3)}sin{(x)} - 2.61y^2z^3cos{(y^3)}sin{(x)}\end{align*}$

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Calculus 3 : Directional Derivatives

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Example Question #131 : Directional Derivatives

Example Question #132 : Directional Derivatives

Example Question #133 : Directional Derivatives

Example Question #134 : Directional Derivatives

Example Question #135 : Directional Derivatives

Example Question #136 : Directional Derivatives

Example Question #137 : Directional Derivatives

Example Question #138 : Directional Derivatives

Example Question #139 : Directional Derivatives

Example Question #140 : Directional Derivatives

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