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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #111 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-xsin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-13,19,-10).\end{align*}$

Possible Answers:

{- 0.27sin{(z)} - 0.16xcos{(z)}}

{0}

{0.52sin{(z)} + 0.4xcos{(z)}}

{- 25.1sin{(z)} - 25.1xcos{(z)}}

Correct answer:

{0.52sin{(z)} + 0.4xcos{(z)}}

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(19)^2+(-10)^2}=25.1\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{25.1}=-0.52\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{25.1}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{25.1}=-0.4\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.52)(-1sin{(z)})+(0.76)(0.0)+(-0.4)(-1xcos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.52sin{(z)} + 0.4xcos{(z)}\end{align*}$

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Example Question #111 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ycos{(x^2)} - 2cos{(z^2)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(17,12,10).\end{align*}$

Possible Answers:

${0.52cos{(x^2)} - 1.04cos{(z^2)}cos{(y)} + 1.72zsin{(z^2)}sin{(y)} - 1.48xysin{(x^2)}}$

${0.27cos{(x^2)} - 0.541cos{(z^2)}cos{(y)} + 0.74zsin{(z^2)}sin{(y)} - 1.1xysin{(x^2)}}$

${23.1cos{(x^2)} - 46.2cos{(z^2)}cos{(y)} + 92.4zsin{(z^2)}sin{(y)} - 46.2xysin{(x^2)}}$

Correct answer:

${0.52cos{(x^2)} - 1.04cos{(z^2)}cos{(y)} + 1.72zsin{(z^2)}sin{(y)} - 1.48xysin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17)^2+(12)^2+(10)^2}=23.09\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17}{23.09}=0.74\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{23.09}=0.52\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{23.09}=0.43\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.74)(-2xysin{(x^2)})+(0.52)(cos{(x^2)} - 2cos{(z^2)}cos{(y)})+(0.43)(4zsin{(z^2)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=0.52cos{(x^2)} - 1.04cos{(z^2)}cos{(y)} + 1.72zsin{(z^2)}sin{(y)} - 1.48xysin{(x^2)}\end{align*}$

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Example Question #112 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^3z^2sin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-17,12,20).\end{align*}$

Possible Answers:

${51x^2y^3z^2cos{(x^3)} - 40y^3zsin{(x^3)} - 36y^2z^2sin{(x^3)}}$

${1.77x^2y^3z^2cos{(x^3)} - 1.38y^3zsin{(x^3)} - 1.26y^2z^2sin{(x^3)}}$

${-519x^2y^2zcos{(x^3)}}$

${- 86.6y^2z^2sin{(x^3)} - 57.7y^3zsin{(x^3)} - 86.6x^2y^3z^2cos{(x^3)}}$

Correct answer:

${1.77x^2y^3z^2cos{(x^3)} - 1.38y^3zsin{(x^3)} - 1.26y^2z^2sin{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-17)^2+(12)^2+(20)^2}=28.86\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-17}{28.86}=-0.59\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{28.86}=0.42\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{28.86}=0.69\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.59)(-3x^2y^3z^2cos{(x^3)})+(0.42)(-3y^2z^2sin{(x^3)})+(0.69)(-2y^3zsin{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.77x^2y^3z^2cos{(x^3)} - 1.38y^3zsin{(x^3)} - 1.26y^2z^2sin{(x^3)}\end{align*}$

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Example Question #113 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^xsin{(y^3)}e^{(z)} -\frac{ {(x^2y)}}{z}\\&\text{In the direction of }\overrightarrow{v}=(5,18,-6).\end{align*}$

Possible Answers:

${\frac{{(19.6x^2y)}}{z^2} -\frac{ {(39.2xy)}}{z} -\frac{ {(19.6x^2)}}{z} + 41.2\cdot3^xsin{(y^3)}e^{(z)} + 58.9\cdot3^xy^2cos{(y^3)}e^{(z)}}$

${\frac{{(39.2x)}}{z^2} + 64.7\cdot3^xy^2cos{(y^3)}e^{(z)}}$

${54\cdot3^xy^2cos{(y^3)}e^{(z)} -\frac{ {(10xy)}}{z} -\frac{ {(6x^2y)}}{z^2} - 0.507\cdot3^xsin{(y^3)}e^{(z)} -\frac{ {(18x^2)}}{z}}$

${2.76\cdot3^xy^2cos{(y^3)}e^{(z)} -\frac{ {(0.5xy)}}{z} -\frac{ {(0.31x^2y)}}{z^2} - 0.0353\cdot3^xsin{(y^3)}e^{(z)} -\frac{ {(0.92x^2)}}{z}}$

Correct answer:

${2.76\cdot3^xy^2cos{(y^3)}e^{(z)} -\frac{ {(0.5xy)}}{z} -\frac{ {(0.31x^2y)}}{z^2} - 0.0353\cdot3^xsin{(y^3)}e^{(z)} -\frac{ {(0.92x^2)}}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(18)^2+(-6)^2}=19.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{19.62}=0.25\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{19.62}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{19.62}=-0.31\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.25)(1.1\cdot3^xsin{(y^3)}e^{(z)} -\frac{ {(2xy)}}{z})+(0.92)(3\cdot3^xy^2cos{(y^3)}e^{(z)} -\frac{ {(1x^2)}}{z})+(-0.31)(\frac{{(x^2y)}}{z^2} + 3^xsin{(y^3)}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=2.76\cdot3^xy^2cos{(y^3)}e^{(z)} -\frac{ {(0.5xy)}}{z} -\frac{ {(0.31x^2y)}}{z^2} - 0.0353\cdot3^xsin{(y^3)}e^{(z)} -\frac{ {(0.92x^2)}}{z}\end{align*}$

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Example Question #114 : Directional Derivatives

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

Possible Answers:

${42.5\cdot2^yz^3e^{(x^2)} + 7.37\cdot2^yz^4e^{(x^2)} + 21.3\cdot2^yxz^4e^{(x^2)}}$

${0.144\cdot2^yz^3e^{(x^2)} + 0.0543\cdot2^yz^4e^{(x^2)} + 1.77\cdot2^yxz^4e^{(x^2)}}$

${0.76\cdot2^yz^3e^{(x^2)} + 0.194\cdot2^yz^4e^{(x^2)} + 1.88\cdot2^yxz^4e^{(x^2)}}$

${58.9\cdot2^yxz^3e^{(x^2)}}$

Correct answer:

${0.76\cdot2^yz^3e^{(x^2)} + 0.194\cdot2^yz^4e^{(x^2)} + 1.88\cdot2^yxz^4e^{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(3)^2+(2)^2}=10.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{10.63}=0.94\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{3}{10.63}=0.28\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{10.63}=0.19\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.94)(2\cdot2^yxz^4e^{(x^2)})+(0.28)(0.693\cdot2^yz^4e^{(x^2)})+(0.19)(4\cdot2^yz^3e^{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)=0.76\cdot2^yz^3e^{(x^2)} + 0.194\cdot2^yz^4e^{(x^2)} + 1.88\cdot2^yxz^4e^{(x^2)}\end{align*}$

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Example Question #115 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 5z^2sin{(x^2)}sin{(y)} - y^4e^{(x^2)}e^{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-8,9,12).\end{align*}$

Possible Answers:

${- 170zsin{(x^2)}sin{(y)} - 85z^2sin{(x^2)}cos{(y)} - 68y^3e^{(x^2)}e^{(z^2)} - 34xy^4e^{(x^2)}e^{(z^2)} - 34y^4ze^{(x^2)}e^{(z^2)} - 170xz^2cos{(x^2)}sin{(y)}}$

${0.94xy^4e^{(x^2)}e^{(z^2)} - 2.65z^2sin{(x^2)}cos{(y)} - 2.12y^3e^{(x^2)}e^{(z^2)} - 7.1zsin{(x^2)}sin{(y)} - 1.42y^4ze^{(x^2)}e^{(z^2)} + 4.7xz^2cos{(x^2)}sin{(y)}}$

${- 340xzcos{(x^2)}cos{(y)} - 272xy^3ze^{(x^2)}e^{(z^2)}}$

${- 5.04zsin{(x^2)}sin{(y)} - 1.4z^2sin{(x^2)}cos{(y)} - 1.12y^3e^{(x^2)}e^{(z^2)} - 0.442xy^4e^{(x^2)}e^{(z^2)} - 1.01y^4ze^{(x^2)}e^{(z^2)} - 2.21xz^2cos{(x^2)}sin{(y)}}$

Correct answer:

${0.94xy^4e^{(x^2)}e^{(z^2)} - 2.65z^2sin{(x^2)}cos{(y)} - 2.12y^3e^{(x^2)}e^{(z^2)} - 7.1zsin{(x^2)}sin{(y)} - 1.42y^4ze^{(x^2)}e^{(z^2)} + 4.7xz^2cos{(x^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(9)^2+(12)^2}=17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{17}=-0.47\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{17}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{17}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.47)(- 2xy^4e^{(x^2)}e^{(z^2)} - 10xz^2cos{(x^2)}sin{(y)})+(0.53)(- 5z^2sin{(x^2)}cos{(y)} - 4y^3e^{(x^2)}e^{(z^2)})+(0.71)(- 10zsin{(x^2)}sin{(y)} - 2y^4ze^{(x^2)}e^{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=0.94xy^4e^{(x^2)}e^{(z^2)} - 2.65z^2sin{(x^2)}cos{(y)} - 2.12y^3e^{(x^2)}e^{(z^2)} - 7.1zsin{(x^2)}sin{(y)} - 1.42y^4ze^{(x^2)}e^{(z^2)} + 4.7xz^2cos{(x^2)}sin{(y)}\end{align*}$

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Example Question #116 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^xsin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-11,-13,14).\end{align*}$

Possible Answers:

${22\cdot4^xcos{(z)} + 30.6\cdot4^xsin{(z)}}$

${0.41\cdot4^xcos{(z)} + 0.347\cdot4^xsin{(z)}}$

${0.64\cdot4^xcos{(z)} - 0.693\cdot4^xsin{(z)}}$

Correct answer:

${0.64\cdot4^xcos{(z)} - 0.693\cdot4^xsin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11)^2+(-13)^2+(14)^2}=22.05\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11}{22.05}=-0.5\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{22.05}=-0.59\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{14}{22.05}=0.64\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.5)(1.39\cdot4^xsin{(z)})+(-0.59)(0.0)+(0.64)(4^xcos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.64\cdot4^xcos{(z)} - 0.693\cdot4^xsin{(z)}\end{align*}$

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Example Question #117 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^2sin{(y^2)} -\frac{ {(sin{(y^3)}ln{(z)})}}{x}\\&\text{In the direction of }\overrightarrow{v}=(-13,-7,-12).\end{align*}$

Possible Answers:

${\frac{{(0.63sin{(y^3)})}}{{(xz)}} - 1.26zsin{(y^2)} -\frac{ {(0.68sin{(y^3)}ln{(z)})}}{x^2} - 0.74yz^2cos{(y^2)} +\frac{ {(1.11y^2cos{(y^3)}ln{(z)})}}{x}}$

${0.794zsin{(y^2)} -\frac{ {(0.397sin{(y^3)})}}{{(xz)}} +\frac{ {(0.462sin{(y^3)}ln{(z)})}}{x^2} + 0.274yz^2cos{(y^2)} -\frac{ {(0.411y^2cos{(y^3)}ln{(z)})}}{x}}$

${38.1zsin{(y^2)} -\frac{ {(19sin{(y^3)})}}{{(xz)}} +\frac{ {(19sin{(y^3)}ln{(z)})}}{x^2} + 38.1yz^2cos{(y^2)} -\frac{ {(57.1y^2cos{(y^3)}ln{(z)})}}{x}}$

${\frac{{(57.1y^2cos{(y^3)})}}{{(x^2z)}}}$

Correct answer:

${\frac{{(0.63sin{(y^3)})}}{{(xz)}} - 1.26zsin{(y^2)} -\frac{ {(0.68sin{(y^3)}ln{(z)})}}{x^2} - 0.74yz^2cos{(y^2)} +\frac{ {(1.11y^2cos{(y^3)}ln{(z)})}}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-7)^2+(-12)^2}=19.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{19.03}=-0.68\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{19.03}=-0.37\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{19.03}=-0.63\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.68)(\frac{{(sin{(y^3)}ln{(z)})}}{x^2})+(-0.37)(2yz^2cos{(y^2)} -\frac{ {(3y^2cos{(y^3)}ln{(z)})}}{x})+(-0.63)(2zsin{(y^2)} -\frac{ {(1sin{(y^3)})}}{{(xz)}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.63sin{(y^3)})}}{{(xz)}} - 1.26zsin{(y^2)} -\frac{ {(0.68sin{(y^3)}ln{(z)})}}{x^2} - 0.74yz^2cos{(y^2)} +\frac{ {(1.11y^2cos{(y^3)}ln{(z)})}}{x}\end{align*}$

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Example Question #118 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- sin{(z^3)}sin{(x)} - 5sin{(y^2)}e^{(z)}ln{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-5,18,17).\end{align*}$

Possible Answers:

${- 25.3sin{(z^3)}cos{(x)} - 75.8z^2cos{(z^3)}sin{(x)} -\frac{ {(126sin{(y^2)}e^{(z)})}}{x} - 126sin{(y^2)}e^{(z)}ln{(x)} - 253ycos{(y^2)}e^{(z)}ln{(x)}}$

${0.2sin{(z^3)}cos{(x)} - 2.01z^2cos{(z^3)}sin{(x)} +\frac{ {(sin{(y^2)}e^{(z)})}}{x} - 3.35sin{(y^2)}e^{(z)}ln{(x)} - 7.1ycos{(y^2)}e^{(z)}ln{(x)}}$

${\frac{-{(253ycos{(y^2)}e^{(z)})}}{x}}$

${- 0.04sin{(z^3)}cos{(x)} - 1.35z^2cos{(z^3)}sin{(x)} -\frac{ {(0.2sin{(y^2)}e^{(z)})}}{x} - 2.24sin{(y^2)}e^{(z)}ln{(x)} - 5.04ycos{(y^2)}e^{(z)}ln{(x)}}$

Correct answer:

${0.2sin{(z^3)}cos{(x)} - 2.01z^2cos{(z^3)}sin{(x)} +\frac{ {(sin{(y^2)}e^{(z)})}}{x} - 3.35sin{(y^2)}e^{(z)}ln{(x)} - 7.1ycos{(y^2)}e^{(z)}ln{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(18)^2+(17)^2}=25.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{25.26}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{25.26}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{17}{25.26}=0.67\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(- sin{(z^3)}cos{(x)} -\frac{ {(5sin{(y^2)}e^{(z)})}}{x})+(0.71)(-10ycos{(y^2)}e^{(z)}ln{(x)})+(0.67)(- 3z^2cos{(z^3)}sin{(x)} - 5sin{(y^2)}e^{(z)}ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.2sin{(z^3)}cos{(x)} - 2.01z^2cos{(z^3)}sin{(x)} +\frac{ {(sin{(y^2)}e^{(z)})}}{x} - 3.35sin{(y^2)}e^{(z)}ln{(x)} - 7.1ycos{(y^2)}e^{(z)}ln{(x)}\end{align*}$

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Example Question #119 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(y^3)}cos{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(20,2,-7).\end{align*}$

Possible Answers:

${- 42.6zcos{(y^3)}sin{(z^2)} - 63.8y^2cos{(z^2)}sin{(y^3)}}$

${- 0.218zcos{(y^3)}sin{(z^2)} - 0.0243y^2cos{(z^2)}sin{(y^3)}}$

${0.66zcos{(y^3)}sin{(z^2)} - 0.27y^2cos{(z^2)}sin{(y^3)}}$

Correct answer:

${0.66zcos{(y^3)}sin{(z^2)} - 0.27y^2cos{(z^2)}sin{(y^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(2)^2+(-7)^2}=21.28\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{21.28}=0.94\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{2}{21.28}=0.09\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{21.28}=-0.33\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.94)(0.0)+(0.09)(-3y^2cos{(z^2)}sin{(y^3)})+(-0.33)(-2zcos{(y^3)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=0.66zcos{(y^3)}sin{(z^2)} - 0.27y^2cos{(z^2)}sin{(y^3)}\end{align*}$

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Calculus 3 : Directional Derivatives

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Example Question #111 : Directional Derivatives

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