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Calculus 3 : Calculus 3

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Example Questions

Example Question #336 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2x^{4}e^{(z^{2})} - z^{3}e^{(x^{3})}e^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,-11,7).\end{align*}$

Possible Answers:

${2.16x^{4}ze^{(z^{2})} - 1.62z^{2}e^{(x^{3})}e^{(y^{2})} + 1.68yz^{3}e^{(x^{3})}e^{(y^{2})}}$

${104x^{3}e^{(z^{2})} - 39.1z^{2}e^{(x^{3})}e^{(y^{2})} + 52.2x^{4}ze^{(z^{2})} - 26.1yz^{3}e^{(x^{3})}e^{(y^{2})} - 39.1x^{2}z^{3}e^{(x^{3})}e^{(y^{2})}}$

${1.17x^{4}ze^{(z^{2})} - 0.875z^{2}e^{(x^{3})}e^{(y^{2})} - 1.41yz^{3}e^{(x^{3})}e^{(y^{2})}}$

${-235x^{2}yz^{2}e^{(x^{3})}e^{(y^{2})}}$

Correct answer:

${2.16x^{4}ze^{(z^{2})} - 1.62z^{2}e^{(x^{3})}e^{(y^{2})} + 1.68yz^{3}e^{(x^{3})}e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-11)^2+(7)^2}=13.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.04}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-11}{13.04}=-0.84\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{13.04}=0.54\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(8x^{3}e^{(z^{2})} - 3x^{2}z^{3}e^{(x^{3})}e^{(y^{2})})+(-0.84)(-2yz^{3}e^{(x^{3})}e^{(y^{2})})+(0.54)(4x^{4}ze^{(z^{2})} - 3z^{2}e^{(x^{3})}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=2.16x^{4}ze^{(z^{2})} - 1.62z^{2}e^{(x^{3})}e^{(y^{2})} + 1.68yz^{3}e^{(x^{3})}e^{(y^{2})}\end{align*}$

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Example Question #337 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(y^{4}ln(z^{4}))}{x^{2}}\\&\text{In the direction of }\overrightarrow{v}=(18,-16,15).\end{align*}$

Possible Answers:

${\frac{(113y^{3}ln(z^{4}))}{x^{2}}-\frac{ (56.7y^{4}ln(z^{4}))}{x^{3}}+\frac{ (113y^{4})}{(x^{2}z)}}$

${\frac{(60y^{4})}{(x^{2}z)}-\frac{ (36y^{4}ln(z^{4}))}{x^{3}}-\frac{ (64y^{3}ln(z^{4}))}{x^{2}}}$

${-\frac{(908y^{3})}{(x^{3}z)}}$

${\frac{(2.12y^{4})}{(x^{2}z)}-\frac{ (1.26y^{4}ln(z^{4}))}{x^{3}}-\frac{ (2.24y^{3}ln(z^{4}))}{x^{2}}}$

Correct answer:

${\frac{(2.12y^{4})}{(x^{2}z)}-\frac{ (1.26y^{4}ln(z^{4}))}{x^{3}}-\frac{ (2.24y^{3}ln(z^{4}))}{x^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(-16)^2+(15)^2}=28.37\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{28.37}=0.63\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{28.37}=-0.56\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{28.37}=0.53\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.63)(-\frac{(2y^{4}ln(z^{4}))}{x^{3}})+(-0.56)(\frac{(4y^{3}ln(z^{4}))}{x^{2}})+(0.53)(\frac{(4y^{4})}{(x^{2}z)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(2.12y^{4})}{(x^{2}z)}-\frac{ (1.26y^{4}ln(z^{4}))}{x^{3}}-\frac{ (2.24y^{3}ln(z^{4}))}{x^{2}}\end{align*}$

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Example Question #338 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 2y^{3}sin(z) - z^{2}e^{(x^{3})}e^{(y^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-7,5,-8).\end{align*}$

Possible Answers:

${- 23.5y^{3}cos(z) - 70.5y^{2}sin(z) - 23.5ze^{(x^{3})}e^{(y^{2})} - 23.5yz^{2}e^{(x^{3})}e^{(y^{2})} - 35.2x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

${16y^{3}cos(z) - 30y^{2}sin(z) + 16ze^{(x^{3})}e^{(y^{2})} - 10yz^{2}e^{(x^{3})}e^{(y^{2})} + 21x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

${-141x^{2}yze^{(x^{3})}e^{(y^{2})}}$

${1.36y^{3}cos(z) - 2.58y^{2}sin(z) + 1.36ze^{(x^{3})}e^{(y^{2})} - 0.86yz^{2}e^{(x^{3})}e^{(y^{2})} + 1.8x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

Correct answer:

${1.36y^{3}cos(z) - 2.58y^{2}sin(z) + 1.36ze^{(x^{3})}e^{(y^{2})} - 0.86yz^{2}e^{(x^{3})}e^{(y^{2})} + 1.8x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(5)^2+(-8)^2}=11.75\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{11.75}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{11.75}=0.43\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{11.75}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(-3x^{2}z^{2}e^{(x^{3})}e^{(y^{2})})+(0.43)(- 6y^{2}sin(z) - 2yz^{2}e^{(x^{3})}e^{(y^{2})})+(-0.68)(- 2y^{3}cos(z) - 2ze^{(x^{3})}e^{(y^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.36y^{3}cos(z) - 2.58y^{2}sin(z) + 1.36ze^{(x^{3})}e^{(y^{2})} - 0.86yz^{2}e^{(x^{3})}e^{(y^{2})} + 1.8x^{2}z^{2}e^{(x^{3})}e^{(y^{2})}\end{align*}$

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Example Question #339 : Directional Derivatives

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

Possible Answers:

${0.75y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 1.88xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 0.75z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

${12y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 30xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 12z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

${-289xy^{2}z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

${- 32.1xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 48.1y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 48.1z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

Correct answer:

${0.75y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 1.88xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 0.75z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(15)^2+(-4)^2+(4)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{15}{16.03}=0.94\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{16.03}=-0.25\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{16.03}=0.25\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.94)(-2xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})})+(-0.25)(-3y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})})+(0.25)(-3z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})})\\&D_{\overrightarrow{u}}(x,y,z)=0.75y^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 1.88xe^{(x^{2})}e^{(y^{3})}e^{(z^{3})} - 0.75z^{2}e^{(x^{2})}e^{(y^{3})}e^{(z^{3})}\end{align*}$

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Example Question #340 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(x^{4})sin(z^{4})ln(y)\\&\text{In the direction of }\overrightarrow{v}=(17.01,15.74,-1.27).\end{align*}$

Possible Answers:

${5.08z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (15.7ln(x^{4})sin(z^{4}))}{y}-\frac{ (68sin(z^{4})ln(y))}{x}}$

${-\frac{ (92.8sin(z^{4})ln(y))}{x}-\frac{ (23.2ln(x^{4})sin(z^{4}))}{y}- 92.8z^{3}cos(z^{4})ln(x^{4})ln(y)}$

${-\frac{(371z^{3}cos(z^{4}))}{(xy)}}$

${0.2z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (0.68ln(x^{4})sin(z^{4}))}{y}-\frac{ (2.92sin(z^{4})ln(y))}{x}}$

Correct answer:

${0.2z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (0.68ln(x^{4})sin(z^{4}))}{y}-\frac{ (2.92sin(z^{4})ln(y))}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17.01)^2+(15.74)^2+(-1.27)^2}=23.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17.01}{23.21}=0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15.74}{23.21}=0.68\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-1.27}{23.21}=-0.05\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.73)(-\frac{(4sin(z^{4})ln(y))}{x})+(0.68)(-\frac{(ln(x^{4})sin(z^{4}))}{y})+(-0.05)(-4z^{3}cos(z^{4})ln(x^{4})ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.2z^{3}cos(z^{4})ln(x^{4})ln(y) -\frac{ (0.68ln(x^{4})sin(z^{4}))}{y}-\frac{ (2.92sin(z^{4})ln(y))}{x}\end{align*}$

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Example Question #1406 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4xcos(z^{2}) + cos(z^{2})e^{(x^{4})}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(-11.32,-2.46,7.54).\end{align*}$

Possible Answers:

${-111x^{3}zsin(z^{2})e^{(x^{4})}cos(y)}$

${55.3cos(z^{2}) - 111xzsin(z^{2}) + 13.8cos(z^{2})e^{(x^{4})}cos(y) - 27.6zsin(z^{2})e^{(x^{4})}sin(y) + 55.3x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

${- 3.28cos(z^{2}) - 4.4xzsin(z^{2}) - 0.18cos(z^{2})e^{(x^{4})}cos(y) - 1.1zsin(z^{2})e^{(x^{4})}sin(y) - 3.28x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

${2.69cos(z^{2}) - 2.42xzsin(z^{2}) + 0.0324cos(z^{2})e^{(x^{4})}cos(y) - 0.605zsin(z^{2})e^{(x^{4})}sin(y) + 2.69x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

Correct answer:

${- 3.28cos(z^{2}) - 4.4xzsin(z^{2}) - 0.18cos(z^{2})e^{(x^{4})}cos(y) - 1.1zsin(z^{2})e^{(x^{4})}sin(y) - 3.28x^{3}cos(z^{2})e^{(x^{4})}sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11.32)^2+(-2.46)^2+(7.54)^2}=13.82\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11.32}{13.82}=-0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2.46}{13.82}=-0.18\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7.54}{13.82}=0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.82)(4cos(z^{2}) + 4x^{3}cos(z^{2})e^{(x^{4})}sin(y))+(-0.18)(cos(z^{2})e^{(x^{4})}cos(y))+(0.55)(- 8xzsin(z^{2}) - 2zsin(z^{2})e^{(x^{4})}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=- 3.28cos(z^{2}) - 4.4xzsin(z^{2}) - 0.18cos(z^{2})e^{(x^{4})}cos(y) - 1.1zsin(z^{2})e^{(x^{4})}sin(y) - 3.28x^{3}cos(z^{2})e^{(x^{4})}sin(y)\end{align*}$

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Example Question #341 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^xe^{(z^{2})}ln(y)\\&\text{In the direction of }\overrightarrow{v}=(5.49,17.87,-11.29).\end{align*}$

Possible Answers:

${\frac{(30.3\cdot 2^xze^{(z^{2})})}{y}}$

${\frac{(0.82\cdot 2^xe^{(z^{2})})}{y}+ 0.173\cdot 2^xe^{(z^{2})}ln(y) - 1.04\cdot 2^xze^{(z^{2})}ln(y)}$

${\frac{(21.8\cdot 2^xe^{(z^{2})})}{y}+ 15.1\cdot 2^xe^{(z^{2})}ln(y) + 43.7\cdot 2^xze^{(z^{2})}ln(y)}$

${\frac{(0.672\cdot 2^xe^{(z^{2})})}{y}+ 0.0433\cdot 2^xe^{(z^{2})}ln(y) + 0.541\cdot 2^xze^{(z^{2})}ln(y)}$

Correct answer:

${\frac{(0.82\cdot 2^xe^{(z^{2})})}{y}+ 0.173\cdot 2^xe^{(z^{2})}ln(y) - 1.04\cdot 2^xze^{(z^{2})}ln(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5.49)^2+(17.87)^2+(-11.29)^2}=21.84\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5.49}{21.84}=0.25\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17.87}{21.84}=0.82\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11.29}{21.84}=-0.52\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.25)(0.693\cdot 2^xe^{(z^{2})}ln(y))+(0.82)(\frac{(2^xe^{(z^{2})})}{y})+(-0.52)(2\cdot 2^xze^{(z^{2})}ln(y))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.82\cdot 2^xe^{(z^{2})})}{y}+ 0.173\cdot 2^xe^{(z^{2})}ln(y) - 1.04\cdot 2^xze^{(z^{2})}ln(y)\end{align*}$

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Example Question #342 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3yln(z^{2}) + ln(y^{4})e^{(x^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(-15.17,11.6,3.25).\end{align*}$

Possible Answers:

${1.08ln(z^{2}) +\frac{ (0.173y)}{z}+\frac{ (1.44e^{(x^{4})}sin(z))}{y}+ 0.0289ln(y^{4})e^{(x^{4})}cos(z) + 2.43x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

${1.8ln(z^{2}) +\frac{ (1.02y)}{z}+\frac{ (2.4e^{(x^{4})}sin(z))}{y}+ 0.17ln(y^{4})e^{(x^{4})}cos(z) - 3.12x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

${\frac{(310x^{3}e^{(x^{4})}cos(z))}{y}}$

${58.1ln(z^{2}) +\frac{ (116y)}{z}+\frac{ (77.5e^{(x^{4})}sin(z))}{y}+ 19.4ln(y^{4})e^{(x^{4})}cos(z) + 77.5x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

Correct answer:

${1.8ln(z^{2}) +\frac{ (1.02y)}{z}+\frac{ (2.4e^{(x^{4})}sin(z))}{y}+ 0.17ln(y^{4})e^{(x^{4})}cos(z) - 3.12x^{3}ln(y^{4})e^{(x^{4})}sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15.17)^2+(11.6)^2+(3.25)^2}=19.37\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15.17}{19.37}=-0.78\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11.6}{19.37}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3.25}{19.37}=0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.78)(4x^{3}ln(y^{4})e^{(x^{4})}sin(z))+(0.6)(3ln(z^{2}) +\frac{ (4e^{(x^{4})}sin(z))}{y})+(0.17)(\frac{(6y)}{z}+ ln(y^{4})e^{(x^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=1.8ln(z^{2}) +\frac{ (1.02y)}{z}+\frac{ (2.4e^{(x^{4})}sin(z))}{y}+ 0.17ln(y^{4})e^{(x^{4})}cos(z) - 3.12x^{3}ln(y^{4})e^{(x^{4})}sin(z)\end{align*}$

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Example Question #341 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(3^{(x^{3})}cos(y^{3}))}{z}\\&\text{In the direction of }\overrightarrow{v}=(-12.37,-17.98,8.28).\end{align*}$

Possible Answers:

${\frac{(0.35\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (1.75\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (2.31\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

${-\frac{(231\cdot 3^{(x^{3})}x^{2}y^{2}sin(y^{3}))}{z^{2}}}$

${\frac{(8.28\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (40.8\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (53.9\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

${\frac{(23.3\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}-\frac{ (76.9\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}+\frac{ (70\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

Correct answer:

${\frac{(0.35\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (1.75\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (2.31\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12.37)^2+(-17.98)^2+(8.28)^2}=23.34\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12.37}{23.34}=-0.53\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17.98}{23.34}=-0.77\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{8.28}{23.34}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.53)(-\frac{(3.3\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z})+(-0.77)(\frac{(3\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z})+(0.35)(\frac{(3^{(x^{3})}cos(y^{3}))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.35\cdot 3^{(x^{3})}cos(y^{3}))}{z^{2}}+\frac{ (1.75\cdot 3^{(x^{3})}x^{2}cos(y^{3}))}{z}-\frac{ (2.31\cdot 3^{(x^{3})}y^{2}sin(y^{3}))}{z}\end{align*}$

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Example Question #342 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^{4}cos(x^{4})cos(y)\\&\text{In the direction of }\overrightarrow{v}=(19.26,-17.85,-17.23).\end{align*}$

Possible Answers:

${126z^{3}cos(x^{4})cos(y) - 31.4z^{4}cos(x^{4})sin(y) - 126x^{3}z^{4}sin(x^{4})cos(y)}$

${0.57z^{4}cos(x^{4})sin(y) - 2.2z^{3}cos(x^{4})cos(y) - 2.44x^{3}z^{4}sin(x^{4})cos(y)}$

${17.9z^{4}cos(x^{4})sin(y) - 68.9z^{3}cos(x^{4})cos(y) - 77x^{3}z^{4}sin(x^{4})cos(y)}$

${503x^{3}z^{3}sin(x^{4})sin(y)}$

Correct answer:

${0.57z^{4}cos(x^{4})sin(y) - 2.2z^{3}cos(x^{4})cos(y) - 2.44x^{3}z^{4}sin(x^{4})cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(19.26)^2+(-17.85)^2+(-17.23)^2}=31.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{19.26}{31.41}=0.61\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17.85}{31.41}=-0.57\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17.23}{31.41}=-0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.61)(-4x^{3}z^{4}sin(x^{4})cos(y))+(-0.57)(-1z^{4}cos(x^{4})sin(y))+(-0.55)(4z^{3}cos(x^{4})cos(y))\\&D_{\overrightarrow{u}}(x,y,z)=0.57z^{4}cos(x^{4})sin(y) - 2.2z^{3}cos(x^{4})cos(y) - 2.44x^{3}z^{4}sin(x^{4})cos(y)\end{align*}$

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