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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #24 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2ycos{(x^3)}ln{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(0,15,-6).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2ycos{(x^3)}ln{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(0,15,-6).\end{align*}$

Possible Answers:

${-{(193.92x^2sin{(x^3)})}/z}$ $\displaystyle {-{(193.92x^2sin{(x^3)})}/z}$

${32.32cos{(x^3)}ln{(z^2)} + {(64.64ycos{(x^3)})}/z - 96.96x^2yln{(z^2)}sin{(x^3)}}$ $\displaystyle {32.32cos{(x^3)}ln{(z^2)} + {(64.64ycos{(x^3)})}/z - 96.96x^2yln{(z^2)}sin{(x^3)}}$

${30cos{(x^3)}ln{(z^2)} - {(24ycos{(x^3)})}/z}$ $\displaystyle {30cos{(x^3)}ln{(z^2)} - {(24ycos{(x^3)})}/z}$

${1.86cos{(x^3)}ln{(z^2)} - {(1.48ycos{(x^3)})}/z}$ $\displaystyle {1.86cos{(x^3)}ln{(z^2)} - {(1.48ycos{(x^3)})}/z}$

Correct answer:

${1.86cos{(x^3)}ln{(z^2)} - {(1.48ycos{(x^3)})}/z}$ $\displaystyle {1.86cos{(x^3)}ln{(z^2)} - {(1.48ycos{(x^3)})}/z}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(15)^2+(-6)^2}=16.16\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{16.16}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{16.16}=0.93\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{16.16}=-0.37\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-6x^2yln{(z^2)}sin{(x^3)})+(0.93)(2cos{(x^3)}ln{(z^2)})+(-0.37)({(4ycos{(x^3)})}/z)\\&D_{\overrightarrow{u}}(x,y,z)={1.86cos{(x^3)}ln{(z^2)} - {(1.48ycos{(x^3)})}/z}\end{align*}$ $\displaystyle \begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(15)^2+(-6)^2}=16.16\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{16.16}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{16.16}=0.93\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{16.16}=-0.37\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-6x^2yln{(z^2)}sin{(x^3)})+(0.93)(2cos{(x^3)}ln{(z^2)})+(-0.37)({(4ycos{(x^3)})}/z)\\&D_{\overrightarrow{u}}(x,y,z)={1.86cos{(x^3)}ln{(z^2)} - {(1.48ycos{(x^3)})}/z}\end{align*}$

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Example Question #25 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=6y^2cos{(z^2)}e^{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-20,-17,19).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=6y^2cos{(z^2)}e^{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-20,-17,19).\end{align*}$

Possible Answers:

${388.8ycos{(z^2)}e^{(x^2)} + 388.8xy^2cos{(z^2)}e^{(x^2)} - 388.8y^2zsin{(z^2)}e^{(x^2)}}$ $\displaystyle {388.8ycos{(z^2)}e^{(x^2)} + 388.8xy^2cos{(z^2)}e^{(x^2)} - 388.8y^2zsin{(z^2)}e^{(x^2)}}$

${- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}$ $\displaystyle {- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}$

${- 204ycos{(z^2)}e^{(x^2)} - 240xy^2cos{(z^2)}e^{(x^2)} - 228y^2zsin{(z^2)}e^{(x^2)}}$ $\displaystyle {- 204ycos{(z^2)}e^{(x^2)} - 240xy^2cos{(z^2)}e^{(x^2)} - 228y^2zsin{(z^2)}e^{(x^2)}}$

${-1555.2xyzsin{(z^2)}e^{(x^2)}}$ $\displaystyle {-1555.2xyzsin{(z^2)}e^{(x^2)}}$

Correct answer:

${- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}$ $\displaystyle {- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-17)^2+(19)^2}=32.4\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{32.4}=-0.62\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{32.4}=-0.52\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{32.4}=0.59\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.62)(12xy^2cos{(z^2)}e^{(x^2)})+(-0.52)(12ycos{(z^2)}e^{(x^2)})+(0.59)(-12y^2zsin{(z^2)}e^{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)={- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}\end{align*}$ $\displaystyle \begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(-17)^2+(19)^2}=32.4\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{32.4}=-0.62\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{32.4}=-0.52\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{32.4}=0.59\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.62)(12xy^2cos{(z^2)}e^{(x^2)})+(-0.52)(12ycos{(z^2)}e^{(x^2)})+(0.59)(-12y^2zsin{(z^2)}e^{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)={- 6.24ycos{(z^2)}e^{(x^2)} - 7.44xy^2cos{(z^2)}e^{(x^2)} - 7.08y^2zsin{(z^2)}e^{(x^2)}}\end{align*}$

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Example Question #26 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3z^3\\&\text{In the direction of }\overrightarrow{v}=(16,-17,18).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3z^3\\&\text{In the direction of }\overrightarrow{v}=(16,-17,18).\end{align*}$

Possible Answers:

${54y^3z^2 - 51y^2z^3}$ $\displaystyle {54y^3z^2 - 51y^2z^3}$

${1.83y^3z^2 - 1.74y^2z^3}$ $\displaystyle {1.83y^3z^2 - 1.74y^2z^3}$

${88.44y^2z^3 + 88.44y^3z^2}$ $\displaystyle {88.44y^2z^3 + 88.44y^3z^2}$

$\displaystyle 0$

Correct answer:

${1.83y^3z^2 - 1.74y^2z^3}$ $\displaystyle {1.83y^3z^2 - 1.74y^2z^3}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(16)^2+(-17)^2+(18)^2}=29.48\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{16}{29.48}=0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{29.48}=-0.58\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{29.48}=0.61\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(0.0)+(-0.58)(3y^2z^3)+(0.61)(3y^3z^2)\\&D_{\overrightarrow{u}}(x,y,z)={1.83y^3z^2 - 1.74y^2z^3}\end{align*}$ $\displaystyle \begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(16)^2+(-17)^2+(18)^2}=29.48\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{16}{29.48}=0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{29.48}=-0.58\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{29.48}=0.61\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(0.0)+(-0.58)(3y^2z^3)+(0.61)(3y^3z^2)\\&D_{\overrightarrow{u}}(x,y,z)={1.83y^3z^2 - 1.74y^2z^3}\end{align*}$

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Example Question #27 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3ysin{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-15,-20,2).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3ysin{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-15,-20,2).\end{align*}$

Possible Answers:

${25.08x^3sin{(z^2)} + 75.24x^2ysin{(z^2)} + 50.16x^3yzcos{(z^2)}}$ $\displaystyle {25.08x^3sin{(z^2)} + 75.24x^2ysin{(z^2)} + 50.16x^3yzcos{(z^2)}}$

${4x^3yzcos{(z^2)} - 45x^2ysin{(z^2)} - 20x^3sin{(z^2)}}$ $\displaystyle {4x^3yzcos{(z^2)} - 45x^2ysin{(z^2)} - 20x^3sin{(z^2)}}$

${150.48x^2zcos{(z^2)}}$ $\displaystyle {150.48x^2zcos{(z^2)}}$

${0.16x^3yzcos{(z^2)} - 1.8x^2ysin{(z^2)} - 0.8x^3sin{(z^2)}}$ $\displaystyle {0.16x^3yzcos{(z^2)} - 1.8x^2ysin{(z^2)} - 0.8x^3sin{(z^2)}}$

Correct answer:

${0.16x^3yzcos{(z^2)} - 1.8x^2ysin{(z^2)} - 0.8x^3sin{(z^2)}}$ $\displaystyle {0.16x^3yzcos{(z^2)} - 1.8x^2ysin{(z^2)} - 0.8x^3sin{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(-20)^2+(2)^2}=25.08\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{25.08}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{25.08}=-0.8\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{25.08}=0.08\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(3x^2ysin{(z^2)})+(-0.8)(x^3sin{(z^2)})+(0.08)(2x^3yzcos{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)={0.16x^3yzcos{(z^2)} - 1.8x^2ysin{(z^2)} - 0.8x^3sin{(z^2)}}\end{align*}$ $\displaystyle \begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(-20)^2+(2)^2}=25.08\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{25.08}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{25.08}=-0.8\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{25.08}=0.08\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(3x^2ysin{(z^2)})+(-0.8)(x^3sin{(z^2)})+(0.08)(2x^3yzcos{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)={0.16x^3yzcos{(z^2)} - 1.8x^2ysin{(z^2)} - 0.8x^3sin{(z^2)}}\end{align*}$

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Example Question #28 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(11,-3,19).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(11,-3,19).\end{align*}$

Possible Answers:

$22.16cos{(x^2)}cos{(z)} - 44.32xsin{(x^2)}sin{(z)}$ $\displaystyle 22.16cos{(x^2)}cos{(z)} - 44.32xsin{(x^2)}sin{(z)}$

$0.86cos{(x^2)}cos{(z)} - xsin{(x^2)}sin{(z)}$ $\displaystyle 0.86cos{(x^2)}cos{(z)} - xsin{(x^2)}sin{(z)}$

$\displaystyle 0$

$19cos{(x^2)}cos{(z)} - 22xsin{(x^2)}sin{(z)}$ $\displaystyle 19cos{(x^2)}cos{(z)} - 22xsin{(x^2)}sin{(z)}$

Correct answer:

$0.86cos{(x^2)}cos{(z)} - xsin{(x^2)}sin{(z)}$ $\displaystyle 0.86cos{(x^2)}cos{(z)} - xsin{(x^2)}sin{(z)}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(11)^2+(-3)^2+(19)^2}=22.16\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{11}{22.16}=0.5\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{22.16}=-0.14\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{22.16}=0.86\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.5)(-2xsin{(x^2)}sin{(z)})+(-0.14)(0.0)+(0.86)(cos{(x^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86cos{(x^2)}cos{(z)} - xsin{(x^2)}sin{(z)}\end{align*}$ $\displaystyle \begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(11)^2+(-3)^2+(19)^2}=22.16\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{11}{22.16}=0.5\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{22.16}=-0.14\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{22.16}=0.86\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.5)(-2xsin{(x^2)}sin{(z)})+(-0.14)(0.0)+(0.86)(cos{(x^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86cos{(x^2)}cos{(z)} - xsin{(x^2)}sin{(z)}\end{align*}$

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Example Question #3461 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3cos{(y^2)}cos{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-9,16,13).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3cos{(y^2)}cos{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(-9,16,13).\end{align*}$

Possible Answers:

$67.47x^2cos{(y^2)}cos{(z^2)} - 44.98x^3ycos{(z^2)}sin{(y^2)} - 44.98x^3zcos{(y^2)}sin{(z^2)}$ $\displaystyle 67.47x^2cos{(y^2)}cos{(z^2)} - 44.98x^3ycos{(z^2)}sin{(y^2)} - 44.98x^3zcos{(y^2)}sin{(z^2)}$

$- 1.2x^2cos{(y^2)}cos{(z^2)} - 1.42x^3ycos{(z^2)}sin{(y^2)} - 1.16x^3zcos{(y^2)}sin{(z^2)}$ $\displaystyle - 1.2x^2cos{(y^2)}cos{(z^2)} - 1.42x^3ycos{(z^2)}sin{(y^2)} - 1.16x^3zcos{(y^2)}sin{(z^2)}$

$- 27x^2cos{(y^2)}cos{(z^2)} - 32x^3ycos{(z^2)}sin{(y^2)} - 26x^3zcos{(y^2)}sin{(z^2)}$ $\displaystyle - 27x^2cos{(y^2)}cos{(z^2)} - 32x^3ycos{(z^2)}sin{(y^2)} - 26x^3zcos{(y^2)}sin{(z^2)}$

$269.88x^2yzsin{(y^2)}sin{(z^2)}$ $\displaystyle 269.88x^2yzsin{(y^2)}sin{(z^2)}$

Correct answer:

$- 1.2x^2cos{(y^2)}cos{(z^2)} - 1.42x^3ycos{(z^2)}sin{(y^2)} - 1.16x^3zcos{(y^2)}sin{(z^2)}$ $\displaystyle - 1.2x^2cos{(y^2)}cos{(z^2)} - 1.42x^3ycos{(z^2)}sin{(y^2)} - 1.16x^3zcos{(y^2)}sin{(z^2)}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-9)^2+(16)^2+(13)^2}=22.49\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-9}{22.49}=-0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{22.49}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{13}{22.49}=0.58\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.4)(3x^2cos{(y^2)}cos{(z^2)})+(0.71)(-2x^3ycos{(z^2)}sin{(y^2)})+(0.58)(-2x^3zcos{(y^2)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.2x^2cos{(y^2)}cos{(z^2)} - 1.42x^3ycos{(z^2)}sin{(y^2)} - 1.16x^3zcos{(y^2)}sin{(z^2)}\end{align*}$ $\displaystyle \begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-9)^2+(16)^2+(13)^2}=22.49\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-9}{22.49}=-0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{22.49}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{13}{22.49}=0.58\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.4)(3x^2cos{(y^2)}cos{(z^2)})+(0.71)(-2x^3ycos{(z^2)}sin{(y^2)})+(0.58)(-2x^3zcos{(y^2)}sin{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.2x^2cos{(y^2)}cos{(z^2)} - 1.42x^3ycos{(z^2)}sin{(y^2)} - 1.16x^3zcos{(y^2)}sin{(z^2)}\end{align*}$

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Example Question #1091 : Partial Derivatives

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

Possible Answers:

$29.667\cdot 2^{(x^2)}xsin{(y)}sin{(z)}$ $\displaystyle 29.667\cdot 2^{(x^2)}xsin{(y)}sin{(z)}$

$0.14\cdot 2^{(x^2)}cos{(y)}sin{(z)} - 0.33\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 1.2893\cdot 2^{(x^2)}xcos{(y)}cos{(z)}$ $\displaystyle 0.14\cdot 2^{(x^2)}cos{(y)}sin{(z)} - 0.33\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 1.2893\cdot 2^{(x^2)}xcos{(y)}cos{(z)}$

$3\cdot 2^{(x^2)}cos{(y)}sin{(z)} - 7\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 27.726\cdot 2^{(x^2)}xcos{(y)}cos{(z)}$ $\displaystyle 3\cdot 2^{(x^2)}cos{(y)}sin{(z)} - 7\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 27.726\cdot 2^{(x^2)}xcos{(y)}cos{(z)}$

$29.667\cdot 2^{(x^2)}xcos{(y)}cos{(z)} - 21.4\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 21.4\cdot 2^{(x^2)}cos{(y)}sin{(z)}$ $\displaystyle 29.667\cdot 2^{(x^2)}xcos{(y)}cos{(z)} - 21.4\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 21.4\cdot 2^{(x^2)}cos{(y)}sin{(z)}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(7)^2+(-3)^2}=21.4\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{21.4}=-0.93\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{21.4}=0.33\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{21.4}=-0.14\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.93)(1.3863\cdot 2^{(x^2)}xcos{(y)}cos{(z)})+(0.33)(-1.0\cdot 2^{(x^2)}cos{(z)}sin{(y)})+(-0.14)(-1.0\cdot 2^{(x^2)}cos{(y)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.14\cdot 2^{(x^2)}cos{(y)}sin{(z)} - 0.33\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 1.2893\cdot 2^{(x^2)}xcos{(y)}cos{(z)}\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-20)^2+(7)^2+(-3)^2}=21.4\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-20}{21.4}=-0.93\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{21.4}=0.33\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{21.4}=-0.14\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.93)(1.3863\cdot 2^{(x^2)}xcos{(y)}cos{(z)})+(0.33)(-1.0\cdot 2^{(x^2)}cos{(z)}sin{(y)})+(-0.14)(-1.0\cdot 2^{(x^2)}cos{(y)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.14\cdot 2^{(x^2)}cos{(y)}sin{(z)} - 0.33\cdot 2^{(x^2)}cos{(z)}sin{(y)} - 1.2893\cdot 2^{(x^2)}xcos{(y)}cos{(z)}\end{align*}$

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Example Question #33 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^4y^2z^4\\&\text{In the direction of }\overrightarrow{v}=(-4,1,6).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^4y^2z^4\\&\text{In the direction of }\overrightarrow{v}=(-4,1,6).\end{align*}$

Possible Answers:

$\displaystyle 0.28x^4yz^4 - 2.2x^3y^2z^4 + 3.28x^4y^2z^3$

$\displaystyle 232.96x^3yz^3$

$\displaystyle 2x^4yz^4 - 16x^3y^2z^4 + 24x^4y^2z^3$

$\displaystyle 14.56x^4yz^4 + 29.12x^3y^2z^4 + 29.12x^4y^2z^3$

Correct answer:

$\displaystyle 0.28x^4yz^4 - 2.2x^3y^2z^4 + 3.28x^4y^2z^3$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(1)^2+(6)^2}=7.28\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{7.28}=-0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{7.28}=0.14\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{7.28}=0.82\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.55)(4x^3y^2z^4)+(0.14)(2x^4yz^4)+(0.82)(4x^4y^2z^3)\\&D_{\overrightarrow{u}}(x,y,z)=0.28x^4yz^4 - 2.2x^3y^2z^4 + 3.28x^4y^2z^3\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(1)^2+(6)^2}=7.28\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{7.28}=-0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{1}{7.28}=0.14\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{7.28}=0.82\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.55)(4x^3y^2z^4)+(0.14)(2x^4yz^4)+(0.82)(4x^4y^2z^3)\\&D_{\overrightarrow{u}}(x,y,z)=0.28x^4yz^4 - 2.2x^3y^2z^4 + 3.28x^4y^2z^3\end{align*}$

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Example Question #31 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(x^2)}y\\&\text{In the direction of }\overrightarrow{v}=(2,-17,-14).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(x^2)}y\\&\text{In the direction of }\overrightarrow{v}=(2,-17,-14).\end{align*}$

Possible Answers:

$0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}$ $\displaystyle 0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}$

$2.7726\cdot 2^{(x^2)}xy - 17\cdot 2^{(x^2)}$ $\displaystyle 2.7726\cdot 2^{(x^2)}xy - 17\cdot 2^{(x^2)}$

$22.11\cdot 2^{(x^2)} + 30.651\cdot 2^{(x^2)}xy$ $\displaystyle 22.11\cdot 2^{(x^2)} + 30.651\cdot 2^{(x^2)}xy$

$\displaystyle 0$

Correct answer:

$0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}$ $\displaystyle 0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-17)^2+(-14)^2}=22.11\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{22.11}=0.09\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{22.11}=-0.77\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{22.11}=-0.63\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.09)(1.3863\cdot 2^{(x^2)}xy)+(-0.77)(2^{(x^2)})+(-0.63)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-17)^2+(-14)^2}=22.11\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{22.11}=0.09\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{22.11}=-0.77\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{22.11}=-0.63\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.09)(1.3863\cdot 2^{(x^2)}xy)+(-0.77)(2^{(x^2)})+(-0.63)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}\end{align*}$

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Example Question #31 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(z^3)}e^{(y^2)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-18,16,-18).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(z^3)}e^{(y^2)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-18,16,-18).\end{align*}$

Possible Answers:

$-180.42yz^2cos{(z^3)}e^{(y^2)}sin{(x)}$ $\displaystyle -180.42yz^2cos{(z^3)}e^{(y^2)}sin{(x)}$

$18sin{(z^3)}e^{(y^2)}sin{(x)} + 32ysin{(z^3)}e^{(y^2)}cos{(x)} - 54z^2cos{(z^3)}e^{(y^2)}cos{(x)}$ $\displaystyle 18sin{(z^3)}e^{(y^2)}sin{(x)} + 32ysin{(z^3)}e^{(y^2)}cos{(x)} - 54z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

$60.14ysin{(z^3)}e^{(y^2)}cos{(x)} - 30.07sin{(z^3)}e^{(y^2)}sin{(x)} + 90.21z^2cos{(z^3)}e^{(y^2)}cos{(x)}$ $\displaystyle 60.14ysin{(z^3)}e^{(y^2)}cos{(x)} - 30.07sin{(z^3)}e^{(y^2)}sin{(x)} + 90.21z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

$0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}$ $\displaystyle 0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(16)^2+(-18)^2}=30.07\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{30.07}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{30.07}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{30.07}=-0.6\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(-1sin{(z^3)}e^{(y^2)}sin{(x)})+(0.53)(2ysin{(z^3)}e^{(y^2)}cos{(x)})+(-0.6)(3z^2cos{(z^3)}e^{(y^2)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(16)^2+(-18)^2}=30.07\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{30.07}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{30.07}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{30.07}=-0.6\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(-1sin{(z^3)}e^{(y^2)}sin{(x)})+(0.53)(2ysin{(z^3)}e^{(y^2)}cos{(x)})+(-0.6)(3z^2cos{(z^3)}e^{(y^2)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}\end{align*}$

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Calculus 3 : Calculus 3

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