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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #272 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(x^{3}cos(y^{4}))}{z}\\&\text{In the direction of }\overrightarrow{v}=(-17,-2,0).\end{align*}$

Possible Answers:

${\frac{(8x^{3}y^{3}sin(y^{4}))}{z}-\frac{ (51x^{2}cos(y^{4}))}{z}}$

${\frac{(205x^{2}y^{3}sin(y^{4}))}{z^{2}}}$

${\frac{(0.48x^{3}y^{3}sin(y^{4}))}{z}-\frac{ (2.97x^{2}cos(y^{4}))}{z}}$

${\frac{(51.4x^{2}cos(y^{4}))}{z}-\frac{ (17.1x^{3}cos(y^{4}))}{z^{2}}-\frac{ (68.5x^{3}y^{3}sin(y^{4}))}{z}}$

Correct answer:

${\frac{(0.48x^{3}y^{3}sin(y^{4}))}{z}-\frac{ (2.97x^{2}cos(y^{4}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-17)^2+(-2)^2+(0)^2}=17.12\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-17}{17.12}=-0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{17.12}=-0.12\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.12}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{(3x^{2}cos(y^{4}))}{z})+(-0.12)(-\frac{(4x^{3}y^{3}sin(y^{4}))}{z})+(0)(-\frac{(x^{3}cos(y^{4}))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.48x^{3}y^{3}sin(y^{4}))}{z}-\frac{ (2.97x^{2}cos(y^{4}))}{z}\end{align*}$

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Example Question #273 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{4}cos(x^{3})sin(z)\\&\text{In the direction of }\overrightarrow{v}=(0,9,-10).\end{align*}$

Possible Answers:

${-161x^{2}y^{3}sin(x^{3})cos(z)}$

${13.4y^{4}cos(x^{3})cos(z) + 53.8y^{3}cos(x^{3})sin(z) - 40.3x^{2}y^{4}sin(x^{3})sin(z)}$

${2.68y^{3}cos(x^{3})sin(z) - 0.74y^{4}cos(x^{3})cos(z)}$

${36y^{3}cos(x^{3})sin(z) - 10y^{4}cos(x^{3})cos(z)}$

Correct answer:

${2.68y^{3}cos(x^{3})sin(z) - 0.74y^{4}cos(x^{3})cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(-10)^2}=13.45\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.45}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{13.45}=0.67\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{13.45}=-0.74\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-3x^{2}y^{4}sin(x^{3})sin(z))+(0.67)(4y^{3}cos(x^{3})sin(z))+(-0.74)(y^{4}cos(x^{3})cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=2.68y^{3}cos(x^{3})sin(z) - 0.74y^{4}cos(x^{3})cos(z)\end{align*}$

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Example Question #272 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4^{(x^{3})}sin(z^{2})e^{(y^{4})} - 4x^{3}z^{4}ln(y^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-5,-3).\end{align*}$

Possible Answers:

${-\frac{ (560x^{2}z^{3})}{y}- 194\cdot 4^{(x^{3})}x^{2}y^{3}zcos(z^{2})e^{(y^{4})}}$

${8.16x^{3}z^{3}ln(y^{2}) +\frac{ (6.88x^{3}z^{4})}{y}+ 3.44\cdot 4^{(x^{3})}y^{3}sin(z^{2})e^{(y^{4})} + 1.02\cdot 4^{(x^{3})}zcos(z^{2})e^{(y^{4})}}$

${- 70x^{2}z^{4}ln(y^{2}) - 93.3x^{3}z^{3}ln(y^{2}) -\frac{ (46.6x^{3}z^{4})}{y}- 24.2\cdot 4^{(x^{3})}x^{2}sin(z^{2})e^{(y^{4})} - 23.3\cdot 4^{(x^{3})}y^{3}sin(z^{2})e^{(y^{4})} - 11.7\cdot 4^{(x^{3})}zcos(z^{2})e^{(y^{4})}}$

${48x^{3}z^{3}ln(y^{2}) +\frac{ (40x^{3}z^{4})}{y}+ 20\cdot 4^{(x^{3})}y^{3}sin(z^{2})e^{(y^{4})} + 6\cdot 4^{(x^{3})}zcos(z^{2})e^{(y^{4})}}$

Correct answer:

${8.16x^{3}z^{3}ln(y^{2}) +\frac{ (6.88x^{3}z^{4})}{y}+ 3.44\cdot 4^{(x^{3})}y^{3}sin(z^{2})e^{(y^{4})} + 1.02\cdot 4^{(x^{3})}zcos(z^{2})e^{(y^{4})}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{5.83}=-0.86\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{5.83}=-0.51\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(- 12x^{2}z^{4}ln(y^{2}) - 4.16\cdot 4^{(x^{3})}x^{2}sin(z^{2})e^{(y^{4})})+(-0.86)(-\frac{ (8x^{3}z^{4})}{y}- 4\cdot 4^{(x^{3})}y^{3}sin(z^{2})e^{(y^{4})})+(-0.51)(- 16x^{3}z^{3}ln(y^{2}) - 2\cdot 4^{(x^{3})}zcos(z^{2})e^{(y^{4})})\\&D_{\overrightarrow{u}}(x,y,z)=8.16x^{3}z^{3}ln(y^{2}) +\frac{ (6.88x^{3}z^{4})}{y}+ 3.44\cdot 4^{(x^{3})}y^{3}sin(z^{2})e^{(y^{4})} + 1.02\cdot 4^{(x^{3})}zcos(z^{2})e^{(y^{4})}\end{align*}$

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Example Question #274 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(4e^{(z)}tan(x))}{y}+ 2^ycos(z^{4})e^{(x)}\\&\text{In the direction of }\overrightarrow{v}=(20,0,17).\end{align*}$

Possible Answers:

${-\frac{ (105e^{(z)}\cdot (tan(x)^{2} + 1))}{y^{2}}- 72.8\cdot 2^yz^{3}sin(z^{4})e^{(x)}}$

${\frac{(105e^{(z)}\cdot (tan(x)^{2} + 1))}{y}+\frac{ (105e^{(z)}tan(x))}{y}-\frac{ (105e^{(z)}tan(x))}{y^{2}}+ 44.4\cdot 2^ycos(z^{4})e^{(x)} - 105\cdot 2^yz^{3}sin(z^{4})e^{(x)}}$

${\frac{(3.04e^{(z)}\cdot (tan(x)^{2} + 1))}{y}+\frac{ (2.6e^{(z)}tan(x))}{y}+ 0.76\cdot 2^ycos(z^{4})e^{(x)} - 2.6\cdot 2^yz^{3}sin(z^{4})e^{(x)}}$

${\frac{(80e^{(z)}\cdot (tan(x)^{2} + 1))}{y}+\frac{ (68e^{(z)}tan(x))}{y}+ 20\cdot 2^ycos(z^{4})e^{(x)} - 68\cdot 2^yz^{3}sin(z^{4})e^{(x)}}$

Correct answer:

${\frac{(3.04e^{(z)}\cdot (tan(x)^{2} + 1))}{y}+\frac{ (2.6e^{(z)}tan(x))}{y}+ 0.76\cdot 2^ycos(z^{4})e^{(x)} - 2.6\cdot 2^yz^{3}sin(z^{4})e^{(x)}}$

Explanation:

$\begin{align*}\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{26.25}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{17}{26.25}=0.65\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.76)(\frac{(4e^{(z)}\cdot (tan(x)^{2} + 1))}{y}+ 2^ycos(z^{4})e^{(x)})+(0)(0.693\cdot 2^ycos(z^{4})e^{(x)} -\frac{ (4e^{(z)}tan(x))}{y^{2}})+(0.65)(\frac{(4e^{(z)}tan(x))}{y}- 4\cdot 2^yz^{3}sin(z^{4})e^{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.04e^{(z)}\cdot (tan(x)^{2} + 1))}{y}+\frac{ (2.6e^{(z)}tan(x))}{y}+ 0.76\cdot 2^ycos(z^{4})e^{(x)} - 2.6\cdot 2^yz^{3}sin(z^{4})e^{(x)}\end{align*}$

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Example Question #275 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^{3}cos(y^{2}) - 2y^{3}e^{(x^{2})}e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(11,0,6).\end{align*}$

Possible Answers:

${37.6z^{2}cos(y^{2}) - 75.2y^{2}e^{(x^{2})}e^{(z^{2})} - 25.1yz^{3}sin(y^{2}) - 50.1xy^{3}e^{(x^{2})}e^{(z^{2})} - 50.1y^{3}ze^{(x^{2})}e^{(z^{2})}}$

${-301xy^{2}ze^{(x^{2})}e^{(z^{2})}}$

${18z^{2}cos(y^{2}) - 44xy^{3}e^{(x^{2})}e^{(z^{2})} - 24y^{3}ze^{(x^{2})}e^{(z^{2})}}$

${1.44z^{2}cos(y^{2}) - 3.52xy^{3}e^{(x^{2})}e^{(z^{2})} - 1.92y^{3}ze^{(x^{2})}e^{(z^{2})}}$

Correct answer:

${1.44z^{2}cos(y^{2}) - 3.52xy^{3}e^{(x^{2})}e^{(z^{2})} - 1.92y^{3}ze^{(x^{2})}e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(11)^2+(0)^2+(6)^2}=12.53\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{11}{12.53}=0.88\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{12.53}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{12.53}=0.48\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.88)(-4xy^{3}e^{(x^{2})}e^{(z^{2})})+(0)(- 6y^{2}e^{(x^{2})}e^{(z^{2})} - 2yz^{3}sin(y^{2}))+(0.48)(3z^{2}cos(y^{2}) - 4y^{3}ze^{(x^{2})}e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.44z^{2}cos(y^{2}) - 3.52xy^{3}e^{(x^{2})}e^{(z^{2})} - 1.92y^{3}ze^{(x^{2})}e^{(z^{2})}\end{align*}$

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Example Question #281 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 5xln(y^{2})ln(z) -\frac{ (z^{3}cos(x^{3}))}{y^{2}}\\&\text{In the direction of }\overrightarrow{v}=(0,19,-16).\end{align*}$

Possible Answers:

${\frac{(1.16z^{3}cos(x^{3}))}{y^{3}}-\frac{ (1.23z^{2}cos(x^{3}))}{y^{2}}-\frac{ (5.78xln(z))}{y}-\frac{ (2.05xln(y^{2}))}{z}}$

${\frac{(49.7z^{3}cos(x^{3}))}{y^{3}}-\frac{ (74.5z^{2}cos(x^{3}))}{y^{2}}- 124ln(y^{2})ln(z) -\frac{ (248xln(z))}{y}-\frac{ (124xln(y^{2}))}{z}+\frac{ (74.5x^{2}z^{3}sin(x^{3}))}{y^{2}}}$

${\frac{(1.92z^{2}cos(x^{3}))}{y^{2}}+\frac{ (1.52z^{3}cos(x^{3}))}{y^{3}}-\frac{ (7.6xln(z))}{y}+\frac{ (3.2xln(y^{2}))}{z}}$

${-\frac{ 248.0}{(yz)}-\frac{ (447x^{2}z^{2}sin(x^{3}))}{y^{3}}}$

Correct answer:

${\frac{(1.92z^{2}cos(x^{3}))}{y^{2}}+\frac{ (1.52z^{3}cos(x^{3}))}{y^{3}}-\frac{ (7.6xln(z))}{y}+\frac{ (3.2xln(y^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(19)^2+(-16)^2}=24.84\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{24.84}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{19}{24.84}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{24.84}=-0.64\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(3x^{2}z^{3}sin(x^{3}))}{y^{2}}- 5ln(y^{2})ln(z))+(0.76)(\frac{(2z^{3}cos(x^{3}))}{y^{3}}-\frac{ (10xln(z))}{y})+(-0.64)(-\frac{ (3z^{2}cos(x^{3}))}{y^{2}}-\frac{ (5xln(y^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.92z^{2}cos(x^{3}))}{y^{2}}+\frac{ (1.52z^{3}cos(x^{3}))}{y^{3}}-\frac{ (7.6xln(z))}{y}+\frac{ (3.2xln(y^{2}))}{z}\end{align*}$

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Example Question #282 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^xcos(y)\\&\text{In the direction of }\overrightarrow{v}=(4,0,-6).\end{align*}$

Possible Answers:

${-0.604\cdot 3^xcos(y)}$

${7.21\cdot 3^xsin(y) - 7.92\cdot 3^xcos(y)}$

${-0.332\cdot 3^xcos(y)}$

Correct answer:

${-0.604\cdot 3^xcos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(4)^2+(0)^2+(-6)^2}=7.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{4}{7.21}=0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{7.21}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{7.21}=-0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(-1.1\cdot 3^xcos(y))+(0)(3^xsin(y))+(-0.83)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-0.604\cdot 3^xcos(y)\end{align*}$

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Example Question #283 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2^xln(y^{2}) - 3y^{4}z^{3}ln(x^{2})\\&\text{In the direction of }\overrightarrow{v}=(-2,10,0).\end{align*}$

Possible Answers:

${0.139\cdot 2^xln(y^{2}) -\frac{ (1.96\cdot 2^x)}{y}- 11.8y^{3}z^{3}ln(x^{2}) +\frac{ (1.2y^{4}z^{3})}{x}}$

${-\frac{(734y^{3}z^{2})}{x}}$

${-\frac{ (20.4\cdot 2^x)}{y}- 7.07\cdot 2^xln(y^{2}) - 122y^{3}z^{3}ln(x^{2}) - 91.8y^{4}z^{2}ln(x^{2}) -\frac{ (61.2y^{4}z^{3})}{x}}$

${1.39\cdot 2^xln(y^{2}) -\frac{ (20\cdot 2^x)}{y}- 120y^{3}z^{3}ln(x^{2}) +\frac{ (12y^{4}z^{3})}{x}}$

Correct answer:

${0.139\cdot 2^xln(y^{2}) -\frac{ (1.96\cdot 2^x)}{y}- 11.8y^{3}z^{3}ln(x^{2}) +\frac{ (1.2y^{4}z^{3})}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(10)^2+(0)^2}=10.2\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{10.2}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{10.2}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{10.2}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(- 0.693\cdot 2^xln(y^{2}) -\frac{ (6y^{4}z^{3})}{x})+(0.98)(-\frac{ (2\cdot 2^x)}{y}- 12y^{3}z^{3}ln(x^{2}))+(0)(-9y^{4}z^{2}ln(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=0.139\cdot 2^xln(y^{2}) -\frac{ (1.96\cdot 2^x)}{y}- 11.8y^{3}z^{3}ln(x^{2}) +\frac{ (1.2y^{4}z^{3})}{x}\end{align*}$

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Example Question #284 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(y^{2})}xsin(z^{4}) - 4y^{2}ln(z)\\&\text{In the direction of }\overrightarrow{v}=(-13,-19,0).\end{align*}$

Possible Answers:

${6.64yln(z) - 0.56\cdot 2^{(y^{2})}sin(z^{4}) - 1.15\cdot 2^{(y^{2})}xysin(z^{4})}$

${152yln(z) - 13\cdot 2^{(y^{2})}sin(z^{4}) - 26.3\cdot 2^{(y^{2})}xysin(z^{4})}$

${128\cdot 2^{(y^{2})}yz^{3}cos(z^{4})}$

${23\cdot 2^{(y^{2})}sin(z^{4}) -\frac{ (92.1y^{2})}{z}- 184yln(z) + 31.9\cdot 2^{(y^{2})}xysin(z^{4}) + 92.1\cdot 2^{(y^{2})}xz^{3}cos(z^{4})}$

Correct answer:

${6.64yln(z) - 0.56\cdot 2^{(y^{2})}sin(z^{4}) - 1.15\cdot 2^{(y^{2})}xysin(z^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-19)^2+(0)^2}=23.02\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{23.02}=-0.56\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-19}{23.02}=-0.83\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.02}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.56)(2^{(y^{2})}sin(z^{4}))+(-0.83)(1.39\cdot 2^{(y^{2})}xysin(z^{4}) - 8yln(z))+(0)(4\cdot 2^{(y^{2})}xz^{3}cos(z^{4}) -\frac{ (4y^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=6.64yln(z) - 0.56\cdot 2^{(y^{2})}sin(z^{4}) - 1.15\cdot 2^{(y^{2})}xysin(z^{4})\end{align*}$

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Example Question #285 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2cos(z^{3})tan(y^{2}) - y^{4}ln(x^{3})ln(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,-14,-11).\end{align*}$

Possible Answers:

${2.31z^{2}tan(y^{2})sin(z^{3}) - 2.5ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) - 2.5y^{3}ln(x^{3})ln(z^{4}) -\frac{ (1.54y^{4}ln(x^{3}))}{z}}$

${\frac{(2.48y^{4}ln(x^{3}))}{z}+ 3.16ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) + 3.16y^{3}ln(x^{3})ln(z^{4}) - 3.72z^{2}tan(y^{2})sin(z^{3})}$

${107z^{2}tan(y^{2})sin(z^{3}) -\frac{ (53.4y^{4}ln(z^{4}))}{x}- 71.2ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) - 71.2y^{3}ln(x^{3})ln(z^{4}) -\frac{ (71.2y^{4}ln(x^{3}))}{z}}$

${-\frac{(854y^{3})}{(xz)}}$

Correct answer:

${\frac{(2.48y^{4}ln(x^{3}))}{z}+ 3.16ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) + 3.16y^{3}ln(x^{3})ln(z^{4}) - 3.72z^{2}tan(y^{2})sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-14)^2+(-11)^2}=17.8\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.8}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{17.8}=-0.79\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{17.8}=-0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(3y^{4}ln(z^{4}))}{x})+(-0.79)(- 4ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) - 4y^{3}ln(x^{3})ln(z^{4}))+(-0.62)(6z^{2}tan(y^{2})sin(z^{3}) -\frac{ (4y^{4}ln(x^{3}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(2.48y^{4}ln(x^{3}))}{z}+ 3.16ycos(z^{3})\cdot (tan(y^{2})^{2} + 1) + 3.16y^{3}ln(x^{3})ln(z^{4}) - 3.72z^{2}tan(y^{2})sin(z^{3})\end{align*}$

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Calculus 3 : Calculus 3

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