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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #114 : 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:

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

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

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

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

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 #115 : 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:

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

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

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

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

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 #116 : 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:

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

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

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

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

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 #1181 : Partial 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:

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

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

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

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 #1182 : Partial 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:

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

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

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

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

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 #1183 : Partial 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:

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

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

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

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

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 #1184 : Partial 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:

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

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

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

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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Example Question #121 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(y^3)}sin{(x^2)}e^{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(7,-9,-11).\end{align*}$

Possible Answers:

${0.975y^2sin{(x^2)}sin{(y^3)}e^{(z^2)} - 0.387xcos{(x^2)}cos{(y^3)}e^{(z^2)} - 0.952zcos{(y^3)}sin{(x^2)}e^{(z^2)}}$

${1.38zcos{(y^3)}sin{(x^2)}e^{(z^2)} - 0.88xcos{(x^2)}cos{(y^3)}e^{(z^2)} - 1.71y^2sin{(x^2)}sin{(y^3)}e^{(z^2)}}$

${190xy^2zcos{(x^2)}sin{(y^3)}e^{(z^2)}}$

${47.5y^2sin{(x^2)}sin{(y^3)}e^{(z^2)} - 31.7xcos{(x^2)}cos{(y^3)}e^{(z^2)} - 31.7zcos{(y^3)}sin{(x^2)}e^{(z^2)}}$

Correct answer:

${1.38zcos{(y^3)}sin{(x^2)}e^{(z^2)} - 0.88xcos{(x^2)}cos{(y^3)}e^{(z^2)} - 1.71y^2sin{(x^2)}sin{(y^3)}e^{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(-9)^2+(-11)^2}=15.84\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{15.84}=0.44\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{15.84}=-0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{15.84}=-0.69\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.44)(-2xcos{(x^2)}cos{(y^3)}e^{(z^2)})+(-0.57)(3y^2sin{(x^2)}sin{(y^3)}e^{(z^2)})+(-0.69)(-2zcos{(y^3)}sin{(x^2)}e^{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.38zcos{(y^3)}sin{(x^2)}e^{(z^2)} - 0.88xcos{(x^2)}cos{(y^3)}e^{(z^2)} - 1.71y^2sin{(x^2)}sin{(y^3)}e^{(z^2)}\end{align*}$

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Example Question #122 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- y^4sin{(z^2)} - 4y^3cos{(z^3)}ln{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(14,-15,4).\end{align*}$

Possible Answers:

${\frac{{(1500y^2z^2sin{(z^3)})}}{x}}$

${60y^3sin{(z^2)} -\frac{ {(112y^3cos{(z^3)})}}{x} + 180y^2cos{(z^3)}ln{(x^2)} - 8y^4zcos{(z^2)} + 48y^3z^2ln{(x^2)}sin{(z^3)}}$

${251y^3z^2ln{(x^2)}sin{(z^3)} -\frac{ {(167y^3cos{(z^3)})}}{x} - 251y^2cos{(z^3)}ln{(x^2)} - 41.8y^4zcos{(z^2)} - 83.6y^3sin{(z^2)}}$

${2.88y^3sin{(z^2)} -\frac{ {(5.36y^3cos{(z^3)})}}{x} + 8.64y^2cos{(z^3)}ln{(x^2)} - 0.38y^4zcos{(z^2)} + 2.28y^3z^2ln{(x^2)}sin{(z^3)}}$

Correct answer:

${2.88y^3sin{(z^2)} -\frac{ {(5.36y^3cos{(z^3)})}}{x} + 8.64y^2cos{(z^3)}ln{(x^2)} - 0.38y^4zcos{(z^2)} + 2.28y^3z^2ln{(x^2)}sin{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(-15)^2+(4)^2}=20.9\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{20.9}=0.67\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{20.9}=-0.72\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{20.9}=0.19\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.67)(\frac{-{(8y^3cos{(z^3)})}}{x})+(-0.72)(- 4y^3sin{(z^2)} - 12y^2cos{(z^3)}ln{(x^2)})+(0.19)(12y^3z^2ln{(x^2)}sin{(z^3)} - 2y^4zcos{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=2.88y^3sin{(z^2)} -\frac{ {(5.36y^3cos{(z^3)})}}{x} + 8.64y^2cos{(z^3)}ln{(x^2)} - 0.38y^4zcos{(z^2)} + 2.28y^3z^2ln{(x^2)}sin{(z^3)}\end{align*}$

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Example Question #121 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^4sin{(z^2)}e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(7,-6,5).\end{align*}$

Possible Answers:

${1.3y^3sin{(z^2)}e^{(x)} + 0.449y^4sin{(z^2)}e^{(x)} + 0.461y^4zcos{(z^2)}e^{(x)}}$

${0.67y^4sin{(z^2)}e^{(x)} - 2.28y^3sin{(z^2)}e^{(x)} + 0.96y^4zcos{(z^2)}e^{(x)}}$

${42y^3sin{(z^2)}e^{(x)} + 10.5y^4sin{(z^2)}e^{(x)} + 21y^4zcos{(z^2)}e^{(x)}}$

${83.9y^3zcos{(z^2)}e^{(x)}}$

Correct answer:

${0.67y^4sin{(z^2)}e^{(x)} - 2.28y^3sin{(z^2)}e^{(x)} + 0.96y^4zcos{(z^2)}e^{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(-6)^2+(5)^2}=10.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{10.49}=0.67\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-6}{10.49}=-0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{10.49}=0.48\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.67)(y^4sin{(z^2)}e^{(x)})+(-0.57)(4y^3sin{(z^2)}e^{(x)})+(0.48)(2y^4zcos{(z^2)}e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.67y^4sin{(z^2)}e^{(x)} - 2.28y^3sin{(z^2)}e^{(x)} + 0.96y^4zcos{(z^2)}e^{(x)}\end{align*}$

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