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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #21 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^2sin{(x^2)}e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-13,-7,-12).\end{align*}$

Possible Answers:

${76.12xycos{(x^2)}e^{(z)}}$

${38.06ysin{(x^2)}e^{(z)} + 19.03y^2sin{(x^2)}e^{(z)} + 38.06xy^2cos{(x^2)}e^{(z)}}$

${- 0.74ysin{(x^2)}e^{(z)} - 0.63y^2sin{(x^2)}e^{(z)} - 1.36xy^2cos{(x^2)}e^{(z)}}$

${- 14ysin{(x^2)}e^{(z)} - 12y^2sin{(x^2)}e^{(z)} - 26xy^2cos{(x^2)}e^{(z)}}$

Correct answer:

${- 0.74ysin{(x^2)}e^{(z)} - 0.63y^2sin{(x^2)}e^{(z)} - 1.36xy^2cos{(x^2)}e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-7)^2+(-12)^2}=19.03\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{19.03}=-0.68\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{19.03}=-0.37\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{19.03}=-0.63\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.68)(2xy^2cos{(x^2)}e^{(z)})+(-0.37)(2ysin{(x^2)}e^{(z)})+(-0.63)(y^2sin{(x^2)}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)={- 0.74ysin{(x^2)}e^{(z)} - 0.63y^2sin{(x^2)}e^{(z)} - 1.36xy^2cos{(x^2)}e^{(z)}}\end{align*}$

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Example Question #3451 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(16,-15,-5).\end{align*}$

Possible Answers:

{112.45cos{(y)}}

{-3.35cos{(y)}}

{-75cos{(y)}}

Correct answer:

{-3.35cos{(y)}}

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(16)^2+(-15)^2+(-5)^2}=22.49\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{16}{22.49}=0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{22.49}=-0.67\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-5}{22.49}=-0.22\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.71)(0)+(-0.67)(5cos{(y)})+(-0.22)(0)\\&D_{\overrightarrow{u}}(x,y,z)={-3.35cos{(y)}}\end{align*}$

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Example Question #23 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2y^4e^{(z)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(10,14,-7).\end{align*}$

Possible Answers:

${112y^3e^{(z)}cos{(x)} - 20y^4e^{(z)}sin{(x)} - 14y^4e^{(z)}cos{(x)}}$

${6y^3e^{(z)}cos{(x)} - 1.08y^4e^{(z)}sin{(x)} - 0.76y^4e^{(z)}cos{(x)}}$

${148.56y^3e^{(z)}cos{(x)} - 37.14y^4e^{(z)}sin{(x)} + 37.14y^4e^{(z)}cos{(x)}}$

${-148.56y^3e^{(z)}sin{(x)}}$

Correct answer:

${6y^3e^{(z)}cos{(x)} - 1.08y^4e^{(z)}sin{(x)} - 0.76y^4e^{(z)}cos{(x)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(14)^2+(-7)^2}=18.57\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{18.57}=0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{18.57}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{18.57}=-0.38\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(-2y^4e^{(z)}sin{(x)})+(0.75)(8y^3e^{(z)}cos{(x)})+(-0.38)(2y^4e^{(z)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)={6y^3e^{(z)}cos{(x)} - 1.08y^4e^{(z)}sin{(x)} - 0.76y^4e^{(z)}cos{(x)}}\end{align*}$

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Example Question #24 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=7x^4z^4cos{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-7,14,10).\end{align*}$

Possible Answers:

${280x^4z^3cos{(y)} - 196x^3z^4cos{(y)} - 98x^4z^4sin{(y)}}$

${-2079.84x^3z^3sin{(y)}}$

${519.96x^3z^4cos{(y)} + 519.96x^4z^3cos{(y)} - 129.99x^4z^4sin{(y)}}$

${15.12x^4z^3cos{(y)} - 10.64x^3z^4cos{(y)} - 5.25x^4z^4sin{(y)}}$

Correct answer:

${15.12x^4z^3cos{(y)} - 10.64x^3z^4cos{(y)} - 5.25x^4z^4sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(14)^2+(10)^2}=18.57\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{18.57}=-0.38\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{18.57}=0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{18.57}=0.54\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.38)(28x^3z^4cos{(y)})+(0.75)(-7x^4z^4sin{(y)})+(0.54)(28x^4z^3cos{(y)})\\&D_{\overrightarrow{u}}(x,y,z)={15.12x^4z^3cos{(y)} - 10.64x^3z^4cos{(y)} - 5.25x^4z^4sin{(y)}}\end{align*}$

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Example Question #25 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=7y^3z^2cos{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-6,-2,-11).\end{align*}$

Possible Answers:

${6.58xy^3z^2sin{(x^2)} - 12.18y^3zcos{(x^2)} - 3.36y^2z^2cos{(x^2)}}$

${-1065.96xy^2zsin{(x^2)}}$

${266.49y^2z^2cos{(x^2)} + 177.66y^3zcos{(x^2)} - 177.66xy^3z^2sin{(x^2)}}$

${84xy^3z^2sin{(x^2)} - 154y^3zcos{(x^2)} - 42y^2z^2cos{(x^2)}}$

Correct answer:

${6.58xy^3z^2sin{(x^2)} - 12.18y^3zcos{(x^2)} - 3.36y^2z^2cos{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(-2)^2+(-11)^2}=12.69\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{12.69}=-0.47\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{12.69}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-11}{12.69}=-0.87\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.47)(-14xy^3z^2sin{(x^2)})+(-0.16)(21y^2z^2cos{(x^2)})+(-0.87)(14y^3zcos{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)={6.58xy^3z^2sin{(x^2)} - 12.18y^3zcos{(x^2)} - 3.36y^2z^2cos{(x^2)}}\end{align*}$

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Example Question #26 : Directional Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #21 : Directional Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #28 : Directional Derivatives

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #29 : Directional Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #30 : Directional Derivatives

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

Possible Answers:

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

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

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Directional Derivatives

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Example Question #21 : Directional Derivatives

Example Question #3451 : Calculus 3

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Example Question #25 : Directional Derivatives

Example Question #26 : Directional Derivatives

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