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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #3491 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^3sin{(x^2)}sin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(11,-17,0).\end{align*}$

Possible Answers:

${1.08xz^3cos{(x^2)}sin{(y^3)} - 2.52y^2z^3cos{(y^3)}sin{(x^2)}}$

${364xy^2z^2cos{(x^2)}cos{(y^3)}}$

${60.8z^2sin{(x^2)}sin{(y^3)} + 40.5xz^3cos{(x^2)}sin{(y^3)} + 60.8y^2z^3cos{(y^3)}sin{(x^2)}}$

${22xz^3cos{(x^2)}sin{(y^3)} - 51y^2z^3cos{(y^3)}sin{(x^2)}}$

Correct answer:

${1.08xz^3cos{(x^2)}sin{(y^3)} - 2.52y^2z^3cos{(y^3)}sin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(11)^2+(-17)^2+(0)^2}=20.25\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{11}{20.25}=0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{20.25}=-0.84\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.25}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(2xz^3cos{(x^2)}sin{(y^3)})+(-0.84)(3y^2z^3cos{(y^3)}sin{(x^2)})+(0)(3z^2sin{(x^2)}sin{(y^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.08xz^3cos{(x^2)}sin{(y^3)} - 2.52y^2z^3cos{(y^3)}sin{(x^2)}\end{align*}$

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Example Question #3492 : Calculus 3

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

Possible Answers:

${436x^2y^2z^2cos{(y^3)}cos{(z^3)}}$

${2.79x^3z^2cos{(z^3)}sin{(y^3)} - 1.11x^2sin{(y^3)}sin{(z^3)}}$

${45x^3z^2cos{(z^3)}sin{(y^3)} - 18x^2sin{(y^3)}sin{(z^3)}}$

${48.5x^2sin{(y^3)}sin{(z^3)} + 48.5x^3y^2cos{(y^3)}sin{(z^3)} + 48.5x^3z^2cos{(z^3)}sin{(y^3)}}$

Correct answer:

${2.79x^3z^2cos{(z^3)}sin{(y^3)} - 1.11x^2sin{(y^3)}sin{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(0)^2+(15)^2}=16.16\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{16.16}=-0.37\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{16.16}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{16.16}=0.93\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.37)(3x^2sin{(y^3)}sin{(z^3)})+(0)(3x^3y^2cos{(y^3)}sin{(z^3)})+(0.93)(3x^3z^2cos{(z^3)}sin{(y^3)})\\&D_{\overrightarrow{u}}(x,y,z)=2.79x^3z^2cos{(z^3)}sin{(y^3)} - 1.11x^2sin{(y^3)}sin{(z^3)}\end{align*}$

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Example Question #3493 : Calculus 3

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

Possible Answers:

${34x^3yz^2cos{(y^2)} - 6x^3zsin{(y^2)}}$

${207x^2yzcos{(y^2)}}$

${51.8x^2z^2sin{(y^2)} + 34.5x^3zsin{(y^2)} + 34.5x^3yz^2cos{(y^2)}}$

${1.96x^3yz^2cos{(y^2)} - 0.34x^3zsin{(y^2)}}$

Correct answer:

${1.96x^3yz^2cos{(y^2)} - 0.34x^3zsin{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(17)^2+(-3)^2}=17.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.26}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{17.26}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{17.26}=-0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3x^2z^2sin{(y^2)})+(0.98)(2x^3yz^2cos{(y^2)})+(-0.17)(2x^3zsin{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.96x^3yz^2cos{(y^2)} - 0.34x^3zsin{(y^2)}\end{align*}$

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Example Question #61 : Directional Derivatives

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

Possible Answers:

{0}

${8.54y^3cos{(z)} - 34.2zcos{(y^3)} + 25.6y^2sin{(z)} + 51.2y^2z^2sin{(y^3)}}$

${1.4zcos{(y^3)} - 0.35y^3cos{(z)}}$

${12zcos{(y^3)} - 3y^3cos{(z)}}$

Correct answer:

${1.4zcos{(y^3)} - 0.35y^3cos{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(0)^2+(-3)^2}=8.54\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{8.54}=-0.94\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{8.54}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{8.54}=-0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.94)(0.0)+(0)(3y^2sin{(z)} + 6y^2z^2sin{(y^3)})+(-0.35)(y^3cos{(z)} - 4zcos{(y^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.4zcos{(y^3)} - 0.35y^3cos{(z)}\end{align*}$

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Example Question #3495 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(z^3)}e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(1,-11,0).\end{align*}$

Possible Answers:

${11cos{(z^3)}e^{(y)}}$

${33.2z^2sin{(z^3)}e^{(y)} - 11.1cos{(z^3)}e^{(y)}}$

${cos{(z^3)}e^{(y)}}$

{0}

Correct answer:

${cos{(z^3)}e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(-11)^2+(0)^2}=11.05\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{11.05}=0.09\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-11}{11.05}=-1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{11.05}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.09)(0.0)+(-1)(-1cos{(z^3)}e^{(y)})+(0)(3z^2sin{(z^3)}e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=cos{(z^3)}e^{(y)}\end{align*}$

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Example Question #62 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^4e^{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(0,-15,9).\end{align*}$

Possible Answers:

{0}

${-1.72x^4ye^{(y^2)}}$

${70x^3e^{(y^2)} + 35x^4ye^{(y^2)}}$

${-30x^4ye^{(y^2)}}$

Correct answer:

${-1.72x^4ye^{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(9)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{17.49}=-0.86\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^3e^{(y^2)})+(-0.86)(2x^4ye^{(y^2)})+(0.51)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-1.72x^4ye^{(y^2)}\end{align*}$

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Example Question #63 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(x^3)}e^{(y)}e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-1,0,5).\end{align*}$

Possible Answers:

${- 0.98cos{(x^3)}e^{(y)}e^{(z)} - 0.6x^2sin{(x^3)}e^{(y)}e^{(z)}}$

${15.3x^2sin{(x^3)}e^{(y)}e^{(z)} - 10.2cos{(x^3)}e^{(y)}e^{(z)}}$

${- 5cos{(x^3)}e^{(y)}e^{(z)} - 3x^2sin{(x^3)}e^{(y)}e^{(z)}}$

${15.3x^2sin{(x^3)}e^{(y)}e^{(z)}}$

Correct answer:

${- 0.98cos{(x^3)}e^{(y)}e^{(z)} - 0.6x^2sin{(x^3)}e^{(y)}e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(0)^2+(5)^2}=5.1\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{5.1}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{5.1}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{5.1}=0.98\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(3x^2sin{(x^3)}e^{(y)}e^{(z)})+(0)(-1cos{(x^3)}e^{(y)}e^{(z)})+(0.98)(-1cos{(x^3)}e^{(y)}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.98cos{(x^3)}e^{(y)}e^{(z)} - 0.6x^2sin{(x^3)}e^{(y)}e^{(z)}\end{align*}$

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Example Question #64 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-xy^2e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(0,-18,7).\end{align*}$

Possible Answers:

${- 19.3y^2e^{(z)} - 38.6xye^{(z)} - 19.3xy^2e^{(z)}}$

${-38.6ye^{(z)}}$

${1.86xye^{(z)} - 0.36xy^2e^{(z)}}$

${36xye^{(z)} - 7xy^2e^{(z)}}$

Correct answer:

${1.86xye^{(z)} - 0.36xy^2e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(7)^2}=19.31\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.31}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{19.31}=-0.93\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{19.31}=0.36\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1y^2e^{(z)})+(-0.93)(-2xye^{(z)})+(0.36)(-1xy^2e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=1.86xye^{(z)} - 0.36xy^2e^{(z)}\end{align*}$

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Example Question #65 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(x^2)}\cdot3^{(y^2)}cos{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(0,-17,-15).\end{align*}$

Possible Answers:

${37.4\cdot3^{(x^2)}\cdot3^{(y^2)}ycos{(z^3)} - 45\cdot3^{(x^2)}\cdot3^{(y^2)}z^2sin{(z^3)}}$

${68\cdot3^{(x^2)}\cdot3^{(y^2)}z^2sin{(z^3)} - 49.8\cdot3^{(x^2)}\cdot3^{(y^2)}ycos{(z^3)} - 49.8\cdot3^{(x^2)}\cdot3^{(y^2)}xcos{(z^3)}}$

${1.65\cdot3^{(x^2)}\cdot3^{(y^2)}ycos{(z^3)} - 1.98\cdot3^{(x^2)}\cdot3^{(y^2)}z^2sin{(z^3)}}$

${328\cdot3^{(x^2)}\cdot3^{(y^2)}xyz^2sin{(z^3)}}$

Correct answer:

${1.65\cdot3^{(x^2)}\cdot3^{(y^2)}ycos{(z^3)} - 1.98\cdot3^{(x^2)}\cdot3^{(y^2)}z^2sin{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-17)^2+(-15)^2}=22.67\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.67}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{22.67}=-0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-15}{22.67}=-0.66\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2.2\cdot3^{(x^2)}\cdot3^{(y^2)}xcos{(z^3)})+(-0.75)(-2.2\cdot3^{(x^2)}\cdot3^{(y^2)}ycos{(z^3)})+(-0.66)(3\cdot3^{(x^2)}\cdot3^{(y^2)}z^2sin{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.65\cdot3^{(x^2)}\cdot3^{(y^2)}ycos{(z^3)} - 1.98\cdot3^{(x^2)}\cdot3^{(y^2)}z^2sin{(z^3)}\end{align*}$

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Example Question #66 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4y^3sin{(z)} - sin{(z^3)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(20,0,-16).\end{align*}$

Possible Answers:

${64y^3cos{(z)} + 48z^2cos{(z^3)}sin{(y)}}$

${2.48y^3cos{(z)} + 1.86z^2cos{(z^3)}sin{(y)}}$

${- 102y^3cos{(z)} - 307y^2sin{(z)} - 25.6sin{(z^3)}cos{(y)} - 76.8z^2cos{(z^3)}sin{(y)}}$

Correct answer:

${2.48y^3cos{(z)} + 1.86z^2cos{(z^3)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(0)^2+(-16)^2}=25.61\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{25.61}=0.78\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{25.61}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{25.61}=-0.62\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.78)(0.0)+(0)(- 12y^2sin{(z)} - sin{(z^3)}cos{(y)})+(-0.62)(- 4y^3cos{(z)} - 3z^2cos{(z^3)}sin{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.48y^3cos{(z)} + 1.86z^2cos{(z^3)}sin{(y)}\end{align*}$

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Calculus 3 : Directional Derivatives

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #3491 : Calculus 3

Example Question #3492 : Calculus 3

Example Question #3493 : Calculus 3

Example Question #61 : Directional Derivatives

Example Question #3495 : Calculus 3

Example Question #62 : Directional Derivatives

Example Question #63 : Directional Derivatives

Example Question #64 : Directional Derivatives

Example Question #65 : Directional Derivatives

Example Question #66 : Directional Derivatives

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