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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1191 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^2cos{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-16,16,-17).\end{align*}$

Possible Answers:

${56.6xy^2sin{(x^2)}sin{(z)} - 28.3y^2cos{(x^2)}cos{(z)} - 56.6ycos{(x^2)}sin{(z)}}$

${0.6y^2cos{(x^2)}cos{(z)} - 1.14ycos{(x^2)}sin{(z)} - 1.14xy^2sin{(x^2)}sin{(z)}}$

${0.65xy^2sin{(x^2)}sin{(z)} - 0.36y^2cos{(x^2)}cos{(z)} - 0.65ycos{(x^2)}sin{(z)}}$

${113xysin{(x^2)}cos{(z)}}$

Correct answer:

${0.6y^2cos{(x^2)}cos{(z)} - 1.14ycos{(x^2)}sin{(z)} - 1.14xy^2sin{(x^2)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(16)^2+(-17)^2}=28.3\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{28.3}=-0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{28.3}=0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{28.3}=-0.6\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.57)(2xy^2sin{(x^2)}sin{(z)})+(0.57)(-2ycos{(x^2)}sin{(z)})+(-0.6)(-1y^2cos{(x^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.6y^2cos{(x^2)}cos{(z)} - 1.14ycos{(x^2)}sin{(z)} - 1.14xy^2sin{(x^2)}sin{(z)}\end{align*}$

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Example Question #1192 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(z^3)}sin{(x^3)}e^{(y^2)} + 3z^3sin{(x^2)}sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(0,6,-8).\end{align*}$

Possible Answers:

${90z^2sin{(x^2)}sin{(y^3)} - 30z^2sin{(x^3)}sin{(z^3)}e^{(y^2)} + 60xz^3cos{(x^2)}sin{(y^3)} + 90y^2z^3cos{(y^3)}sin{(x^2)} + 20ycos{(z^3)}sin{(x^3)}e^{(y^2)} + 30x^2cos{(x^3)}cos{(z^3)}e^{(y^2)}}$

${540xy^2z^2cos{(x^2)}cos{(y^3)} - 180x^2yz^2cos{(x^3)}sin{(z^3)}e^{(y^2)}}$

${24z^2sin{(x^3)}sin{(z^3)}e^{(y^2)} - 72z^2sin{(x^2)}sin{(y^3)} + 54y^2z^3cos{(y^3)}sin{(x^2)} + 12ycos{(z^3)}sin{(x^3)}e^{(y^2)}}$

${2.4z^2sin{(x^3)}sin{(z^3)}e^{(y^2)} - 7.2z^2sin{(x^2)}sin{(y^3)} + 5.4y^2z^3cos{(y^3)}sin{(x^2)} + 1.2ycos{(z^3)}sin{(x^3)}e^{(y^2)}}$

Correct answer:

${2.4z^2sin{(x^3)}sin{(z^3)}e^{(y^2)} - 7.2z^2sin{(x^2)}sin{(y^3)} + 5.4y^2z^3cos{(y^3)}sin{(x^2)} + 1.2ycos{(z^3)}sin{(x^3)}e^{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(6)^2+(-8)^2}=10\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{6}{10}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{10}=-0.8\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(6xz^3cos{(x^2)}sin{(y^3)} + 3x^2cos{(x^3)}cos{(z^3)}e^{(y^2)})+(0.6)(9y^2z^3cos{(y^3)}sin{(x^2)} + 2ycos{(z^3)}sin{(x^3)}e^{(y^2)})+(-0.8)(9z^2sin{(x^2)}sin{(y^3)} - 3z^2sin{(x^3)}sin{(z^3)}e^{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=2.4z^2sin{(x^3)}sin{(z^3)}e^{(y^2)} - 7.2z^2sin{(x^2)}sin{(y^3)} + 5.4y^2z^3cos{(y^3)}sin{(x^2)} + 1.2ycos{(z^3)}sin{(x^3)}e^{(y^2)}\end{align*}$

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Example Question #1193 : Partial Derivatives

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

Possible Answers:

${138z^2e^{(x^2)}e^{(y^2)} - 25.2\cdot3^xzcos{(y^2)} - 23\cdot3^xcos{(y^2)} + 91.9xz^3e^{(x^2)}e^{(y^2)} + 91.9yz^3e^{(x^2)}e^{(y^2)} + 46\cdot3^xyzsin{(y^2)}}$

${4.2z^2e^{(x^2)}e^{(y^2)} - 0.769\cdot3^xzcos{(y^2)} - 0.7\cdot3^xcos{(y^2)} + 2.8xz^3e^{(x^2)}e^{(y^2)} + 0.68yz^3e^{(x^2)}e^{(y^2)} + 0.34\cdot3^xyzsin{(y^2)}}$

${2.94z^2e^{(x^2)}e^{(y^2)} - 0.538\cdot3^xzcos{(y^2)} - 0.49\cdot3^xcos{(y^2)} + 1.96xz^3e^{(x^2)}e^{(y^2)} + 0.116yz^3e^{(x^2)}e^{(y^2)} + 0.0578\cdot3^xyzsin{(y^2)}}$

${50.5\cdot3^xysin{(y^2)} + 552xyz^2e^{(x^2)}e^{(y^2)}}$

Correct answer:

${4.2z^2e^{(x^2)}e^{(y^2)} - 0.769\cdot3^xzcos{(y^2)} - 0.7\cdot3^xcos{(y^2)} + 2.8xz^3e^{(x^2)}e^{(y^2)} + 0.68yz^3e^{(x^2)}e^{(y^2)} + 0.34\cdot3^xyzsin{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(16)^2+(4)^2+(16)^2}=22.98\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{16}{22.98}=0.7\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{4}{22.98}=0.17\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{22.98}=0.7\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.7)(4xz^3e^{(x^2)}e^{(y^2)} - 1.1\cdot3^xzcos{(y^2)})+(0.17)(4yz^3e^{(x^2)}e^{(y^2)} + 2\cdot3^xyzsin{(y^2)})+(0.7)(6z^2e^{(x^2)}e^{(y^2)} - 1\cdot3^xcos{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=4.2z^2e^{(x^2)}e^{(y^2)} - 0.769\cdot3^xzcos{(y^2)} - 0.7\cdot3^xcos{(y^2)} + 2.8xz^3e^{(x^2)}e^{(y^2)} + 0.68yz^3e^{(x^2)}e^{(y^2)} + 0.34\cdot3^xyzsin{(y^2)}\end{align*}$

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Example Question #1194 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(z^2)}sin{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(4,15,18).\end{align*}$

Possible Answers:

${47.5ycos{(y^2)}cos{(z^2)} - 47.5zsin{(y^2)}sin{(z^2)}}$

${1.26ycos{(y^2)}cos{(z^2)} - 1.52zsin{(y^2)}sin{(z^2)}}$

${0.794ycos{(y^2)}cos{(z^2)} - 1.16zsin{(y^2)}sin{(z^2)}}$

Correct answer:

${1.26ycos{(y^2)}cos{(z^2)} - 1.52zsin{(y^2)}sin{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(4)^2+(15)^2+(18)^2}=23.77\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{4}{23.77}=0.17\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{23.77}=0.63\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{23.77}=0.76\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.17)(0.0)+(0.63)(2ycos{(y^2)}cos{(z^2)})+(0.76)(-2zsin{(y^2)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.26ycos{(y^2)}cos{(z^2)} - 1.52zsin{(y^2)}sin{(z^2)}\end{align*}$

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Example Question #1195 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-x^4ln{(y^5)}e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(3,-7,-2).\end{align*}$

Possible Answers:

${\frac{-{(157x^3e^{(z)})}}{y}}$

${- 0.578x^3ln{(y^5)}e^{(z)} - 0.0625x^4ln{(y^5)}e^{(z)} -\frac{ {(3.96x^4e^{(z)})}}{y}}$

${- 31.5x^3ln{(y^5)}e^{(z)} - 7.87x^4ln{(y^5)}e^{(z)} -\frac{ {(39.3x^4e^{(z)})}}{y}}$

${0.25x^4ln{(y^5)}e^{(z)} - 1.52x^3ln{(y^5)}e^{(z)} +\frac{ {(4.45x^4e^{(z)})}}{y}}$

Correct answer:

${0.25x^4ln{(y^5)}e^{(z)} - 1.52x^3ln{(y^5)}e^{(z)} +\frac{ {(4.45x^4e^{(z)})}}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(3)^2+(-7)^2+(-2)^2}=7.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{3}{7.87}=0.38\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{7.87}=-0.89\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{7.87}=-0.25\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.38)(-4x^3ln{(y^5)}e^{(z)})+(-0.89)(\frac{-{(5x^4e^{(z)})}}{y})+(-0.25)(-1x^4ln{(y^5)}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.25x^4ln{(y^5)}e^{(z)} - 1.52x^3ln{(y^5)}e^{(z)} +\frac{ {(4.45x^4e^{(z)})}}{y}\end{align*}$

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Example Question #1196 : Partial Derivatives

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

Possible Answers:

${18.5x^2sin{(x^3)}sin{(y^2)}sin{(z^2)} - 12.3ycos{(x^3)}cos{(y^2)}sin{(z^2)} - 12.3zcos{(x^3)}cos{(z^2)}sin{(y^2)}}$

${- 2.43x^2sin{(x^3)}sin{(y^2)}sin{(z^2)} - 0.64ycos{(x^3)}cos{(y^2)}sin{(z^2)} - 0.98zcos{(x^3)}cos{(z^2)}sin{(y^2)}}$

${- 15x^2sin{(x^3)}sin{(y^2)}sin{(z^2)} - 4ycos{(x^3)}cos{(y^2)}sin{(z^2)} - 6zcos{(x^3)}cos{(z^2)}sin{(y^2)}}$

${73.9x^2yzcos{(y^2)}cos{(z^2)}sin{(x^3)}}$

Correct answer:

${- 2.43x^2sin{(x^3)}sin{(y^2)}sin{(z^2)} - 0.64ycos{(x^3)}cos{(y^2)}sin{(z^2)} - 0.98zcos{(x^3)}cos{(z^2)}sin{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5)^2+(2)^2+(3)^2}=6.16\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5}{6.16}=-0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{2}{6.16}=0.32\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{6.16}=0.49\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.81)(3x^2sin{(x^3)}sin{(y^2)}sin{(z^2)})+(0.32)(-2ycos{(x^3)}cos{(y^2)}sin{(z^2)})+(0.49)(-2zcos{(x^3)}cos{(z^2)}sin{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=- 2.43x^2sin{(x^3)}sin{(y^2)}sin{(z^2)} - 0.64ycos{(x^3)}cos{(y^2)}sin{(z^2)} - 0.98zcos{(x^3)}cos{(z^2)}sin{(y^2)}\end{align*}$

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Example Question #1197 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^yln{(x^6)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-8,-20,2).\end{align*}$

Possible Answers:

${\frac{{(90\cdot2^ycos{(z)})}}{x}}$

${0.09\cdot2^yln{(x^6)}cos{(z)} -\frac{ {(2.22\cdot2^ysin{(z)})}}{x} - 0.638\cdot2^yln{(x^6)}sin{(z)}}$

${2\cdot2^yln{(x^6)}cos{(z)} -\frac{ {(48\cdot2^ysin{(z)})}}{x} - 13.9\cdot2^yln{(x^6)}sin{(z)}}$

${\frac{{(130\cdot2^ysin{(z)})}}{x} + 21.6\cdot2^yln{(x^6)}cos{(z)} + 15\cdot2^yln{(x^6)}sin{(z)}}$

Correct answer:

${0.09\cdot2^yln{(x^6)}cos{(z)} -\frac{ {(2.22\cdot2^ysin{(z)})}}{x} - 0.638\cdot2^yln{(x^6)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(-20)^2+(2)^2}=21.63\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{21.63}=-0.37\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{21.63}=-0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{21.63}=0.09\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.37)(\frac{{(6\cdot2^ysin{(z)})}}{x})+(-0.92)(0.693\cdot2^yln{(x^6)}sin{(z)})+(0.09)(2^yln{(x^6)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.09\cdot2^yln{(x^6)}cos{(z)} -\frac{ {(2.22\cdot2^ysin{(z)})}}{x} - 0.638\cdot2^yln{(x^6)}sin{(z)}\end{align*}$

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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:

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

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

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

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

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:

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

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

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

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

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:

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

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

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

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

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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Calculus 3 : Partial Derivatives

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Example Question #1191 : Partial Derivatives

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