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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #91 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(x^2)}y^3z^2 - 3x^2e^{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(-15,0,5).\end{align*}$

Possible Answers:

${90xe^{(y^2)} + 10\cdot3^{(x^2)}y^3z - 33\cdot3^{(x^2)}xy^3z^2}$

${31.6\cdot3^{(x^2)}y^3z - 94.9xe^{(y^2)} + 47.4\cdot3^{(x^2)}y^2z^2 - 94.9x^2ye^{(y^2)} + 34.7\cdot3^{(x^2)}xy^3z^2}$

${208\cdot3^{(x^2)}xy^2z}$

${5.7xe^{(y^2)} + 0.64\cdot3^{(x^2)}y^3z - 2.09\cdot3^{(x^2)}xy^3z^2}$

Correct answer:

${5.7xe^{(y^2)} + 0.64\cdot3^{(x^2)}y^3z - 2.09\cdot3^{(x^2)}xy^3z^2}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(0)^2+(5)^2}=15.81\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{15.81}=-0.95\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.81}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{15.81}=0.32\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.95)(2.2\cdot3^{(x^2)}xy^3z^2 - 6xe^{(y^2)})+(0)(3\cdot3^{(x^2)}y^2z^2 - 6x^2ye^{(y^2)})+(0.32)(2\cdot3^{(x^2)}y^3z)\\&D_{\overrightarrow{u}}(x,y,z)=5.7xe^{(y^2)} + 0.64\cdot3^{(x^2)}y^3z - 2.09\cdot3^{(x^2)}xy^3z^2\end{align*}$

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Example Question #92 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^2)}cos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(0,-19,18).\end{align*}$

Possible Answers:

${72.6\cdot4^{(z^2)}zcos{(x^3)} - 78.5\cdot4^{(z^2)}x^2sin{(x^3)}}$

${49.9\cdot4^{(z^2)}zcos{(x^3)}}$

{0}

${1.91\cdot4^{(z^2)}zcos{(x^3)}}$

Correct answer:

${1.91\cdot4^{(z^2)}zcos{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-19)^2+(18)^2}=26.17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{26.17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-19}{26.17}=-0.73\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{26.17}=0.69\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-3\cdot4^{(z^2)}x^2sin{(x^3)})+(-0.73)(0.0)+(0.69)(2.77\cdot4^{(z^2)}zcos{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=1.91\cdot4^{(z^2)}zcos{(x^3)}\end{align*}$

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Example Question #93 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(z^2)}cos{(x^2)}\\&\text{In the direction of }\overrightarrow{v}=(-7,16,0).\end{align*}$

Possible Answers:

${14\cdot4^{(z^2)}xsin{(x^2)}}$

${48.4\cdot4^{(z^2)}zcos{(x^2)} - 34.9\cdot4^{(z^2)}xsin{(x^2)}}$

${0.8\cdot4^{(z^2)}xsin{(x^2)}}$

{0}

Correct answer:

${0.8\cdot4^{(z^2)}xsin{(x^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(16)^2+(0)^2}=17.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{17.46}=-0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{17.46}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.46}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.4)(-2\cdot4^{(z^2)}xsin{(x^2)})+(0.92)(0.0)+(0)(2.77\cdot4^{(z^2)}zcos{(x^2)})\\&D_{\overrightarrow{u}}(x,y,z)=0.8\cdot4^{(z^2)}xsin{(x^2)}\end{align*}$

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Example Question #94 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^3sin{(z^2)}sin{(y)} - 5yz^4\\&\text{In the direction of }\overrightarrow{v}=(0,-7,-13).\end{align*}$

Possible Answers:

${14.8x^3sin{(z^2)}cos{(y)} - 73.8z^4 - 295yz^3 + 44.3x^2sin{(z^2)}sin{(y)} + 29.5x^3zcos{(z^2)}sin{(y)}}$

${88.6x^2zcos{(z^2)}cos{(y)}}$

${260yz^3 + 35z^4 - 7x^3sin{(z^2)}cos{(y)} - 26x^3zcos{(z^2)}sin{(y)}}$

${17.6yz^3 + 2.35z^4 - 0.47x^3sin{(z^2)}cos{(y)} - 1.76x^3zcos{(z^2)}sin{(y)}}$

Correct answer:

${17.6yz^3 + 2.35z^4 - 0.47x^3sin{(z^2)}cos{(y)} - 1.76x^3zcos{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-7)^2+(-13)^2}=14.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.76}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7}{14.76}=-0.47\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-13}{14.76}=-0.88\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3x^2sin{(z^2)}sin{(y)})+(-0.47)(x^3sin{(z^2)}cos{(y)} - 5z^4)+(-0.88)(2x^3zcos{(z^2)}sin{(y)} - 20yz^3)\\&D_{\overrightarrow{u}}(x,y,z)=17.6yz^3 + 2.35z^4 - 0.47x^3sin{(z^2)}cos{(y)} - 1.76x^3zcos{(z^2)}sin{(y)}\end{align*}$

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Example Question #95 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(z^3)}e^{(x)} - 3xz^3cos{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,16,-18).\end{align*}$

Possible Answers:

${2.25z^2sin{(z^3)}e^{(x)} + 6.75xz^2cos{(y)} + 1.98xz^3sin{(y)}}$

${217z^2sin{(y)}}$

${24.1cos{(z^3)}e^{(x)} - 72.2z^3cos{(y)} - 72.2z^2sin{(z^3)}e^{(x)} - 217xz^2cos{(y)} + 72.2xz^3sin{(y)}}$

${54z^2sin{(z^3)}e^{(x)} + 162xz^2cos{(y)} + 48xz^3sin{(y)}}$

Correct answer:

${2.25z^2sin{(z^3)}e^{(x)} + 6.75xz^2cos{(y)} + 1.98xz^3sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(16)^2+(-18)^2}=24.08\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{24.08}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{24.08}=0.66\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{24.08}=-0.75\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(cos{(z^3)}e^{(x)} - 3z^3cos{(y)})+(0.66)(3xz^3sin{(y)})+(-0.75)(- 3z^2sin{(z^3)}e^{(x)} - 9xz^2cos{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.25z^2sin{(z^3)}e^{(x)} + 6.75xz^2cos{(y)} + 1.98xz^3sin{(y)}\end{align*}$

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Example Question #96 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- y^2e^{(x)} - 3z^4e^{(y^2)}\\&\text{In the direction of }\overrightarrow{v}=(-18,0,-19).\end{align*}$

Possible Answers:

${0.69y^2e^{(x)} + 8.76z^3e^{(y^2)}}$

${- 26.2y^2e^{(x)} - 314z^3e^{(y^2)} - 52.3ye^{(x)} - 157yz^4e^{(y^2)}}$

{0}

${18y^2e^{(x)} + 228z^3e^{(y^2)}}$

Correct answer:

${0.69y^2e^{(x)} + 8.76z^3e^{(y^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(0)^2+(-19)^2}=26.17\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{26.17}=-0.69\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{26.17}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{26.17}=-0.73\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.69)(-1y^2e^{(x)})+(0)(- 2ye^{(x)} - 6yz^4e^{(y^2)})+(-0.73)(-12z^3e^{(y^2)})\\&D_{\overrightarrow{u}}(x,y,z)=0.69y^2e^{(x)} + 8.76z^3e^{(y^2)}\end{align*}$

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Example Question #97 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^ye^{(z)} - 5x^4y^3cos{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(-19,0,4).\end{align*}$

Possible Answers:

${0.21\cdot2^ye^{(z)} + 19.6x^3y^3cos{(z^3)} + 3.15x^4y^3z^2sin{(z^3)}}$

${32.9\cdot2^ye^{(z)} - 388x^3y^3cos{(z^3)} - 291x^4y^2cos{(z^3)} + 291x^4y^3z^2sin{(z^3)}}$

${4\cdot2^ye^{(z)} + 380x^3y^3cos{(z^3)} + 60x^4y^3z^2sin{(z^3)}}$

${3500x^3y^2z^2sin{(z^3)}}$

Correct answer:

${0.21\cdot2^ye^{(z)} + 19.6x^3y^3cos{(z^3)} + 3.15x^4y^3z^2sin{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-19)^2+(0)^2+(4)^2}=19.42\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-19}{19.42}=-0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{19.42}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{19.42}=0.21\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.98)(-20x^3y^3cos{(z^3)})+(0)(0.693\cdot2^ye^{(z)} - 15x^4y^2cos{(z^3)})+(0.21)(2^ye^{(z)} + 15x^4y^3z^2sin{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=0.21\cdot2^ye^{(z)} + 19.6x^3y^3cos{(z^3)} + 3.15x^4y^3z^2sin{(z^3)}\end{align*}$

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Example Question #91 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^4cos{(z)}ln{(x)} - x^3cos{(y^3)}sin{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(-16,0,2).\end{align*}$

Possible Answers:

${\frac{{(16.1y^4cos{(z)})}}{x} - 48.4x^2cos{(y^3)}sin{(z^3)} - 16.1y^4ln{(x)}sin{(z)} + 64.5y^3cos{(z)}ln{(x)} - 48.4x^3z^2cos{(y^3)}cos{(z^3)} + 48.4x^3y^2sin{(y^3)}sin{(z^3)}}$

${2.97x^2cos{(y^3)}sin{(z^3)} - 0.12y^4ln{(x)}sin{(z)} -\frac{ {(0.99y^4cos{(z)})}}{x} - 0.36x^3z^2cos{(y^3)}cos{(z^3)}}$

${48x^2cos{(y^3)}sin{(z^3)} - 2y^4ln{(x)}sin{(z)} -\frac{ {(16y^4cos{(z)})}}{x} - 6x^3z^2cos{(y^3)}cos{(z^3)}}$

${435x^2y^2z^2cos{(z^3)}sin{(y^3)} -\frac{ {(64.5y^3sin{(z)})}}{x}}$

Correct answer:

${2.97x^2cos{(y^3)}sin{(z^3)} - 0.12y^4ln{(x)}sin{(z)} -\frac{ {(0.99y^4cos{(z)})}}{x} - 0.36x^3z^2cos{(y^3)}cos{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(0)^2+(2)^2}=16.12\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{16.12}=-0.99\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{16.12}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{16.12}=0.12\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.99)(\frac{{(y^4cos{(z)})}}{x} - 3x^2cos{(y^3)}sin{(z^3)})+(0)(4y^3cos{(z)}ln{(x)} + 3x^3y^2sin{(y^3)}sin{(z^3)})+(0.12)(- y^4ln{(x)}sin{(z)} - 3x^3z^2cos{(y^3)}cos{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=2.97x^2cos{(y^3)}sin{(z^3)} - 0.12y^4ln{(x)}sin{(z)} -\frac{ {(0.99y^4cos{(z)})}}{x} - 0.36x^3z^2cos{(y^3)}cos{(z^3)}\end{align*}$

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Example Question #1161 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=1 - 5ycos{(z^3)}\\&\text{In the direction of }\overrightarrow{v}=(-16,0,-7).\end{align*}$

Possible Answers:

${262yz^2sin{(z^3)} - 87.3cos{(z^3)}}$

${-6yz^2sin{(z^3)}}$

${-105yz^2sin{(z^3)}}$

Correct answer:

${-6yz^2sin{(z^3)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(0)^2+(-7)^2}=17.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{17.46}=-0.92\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.46}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{17.46}=-0.4\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.92)(0.0)+(0)(-5cos{(z^3)})+(-0.4)(15yz^2sin{(z^3)})\\&D_{\overrightarrow{u}}(x,y,z)=-6yz^2sin{(z^3)}\end{align*}$

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Example Question #91 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^2cos{(y^3)}e^{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(0,-5,-12).\end{align*}$

Possible Answers:

${-156xy^2zsin{(y^3)}e^{(z^2)}}$

${26xcos{(y^3)}e^{(z^2)} + 26x^2zcos{(y^3)}e^{(z^2)} - 39x^2y^2sin{(y^3)}e^{(z^2)}}$

${15x^2y^2sin{(y^3)}e^{(z^2)} - 24x^2zcos{(y^3)}e^{(z^2)}}$

${1.14x^2y^2sin{(y^3)}e^{(z^2)} - 1.84x^2zcos{(y^3)}e^{(z^2)}}$

Correct answer:

${1.14x^2y^2sin{(y^3)}e^{(z^2)} - 1.84x^2zcos{(y^3)}e^{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-5)^2+(-12)^2}=13\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{13}=-0.38\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{13}=-0.92\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xcos{(y^3)}e^{(z^2)})+(-0.38)(-3x^2y^2sin{(y^3)}e^{(z^2)})+(-0.92)(2x^2zcos{(y^3)}e^{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=1.14x^2y^2sin{(y^3)}e^{(z^2)} - 1.84x^2zcos{(y^3)}e^{(z^2)}\end{align*}$

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Calculus 3 : Directional Derivatives

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Example Question #91 : Directional Derivatives

Example Question #92 : Directional Derivatives

Example Question #93 : Directional Derivatives

Example Question #94 : Directional Derivatives

Example Question #95 : Directional Derivatives

Example Question #96 : Directional Derivatives

Example Question #97 : Directional Derivatives

Example Question #91 : Directional Derivatives

Example Question #1161 : Partial Derivatives

Example Question #91 : Directional Derivatives

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