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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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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Example Question #1161 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- z^3ln{(x)} - y^2cos{(z^2)}sin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(1,-17,-4).\end{align*}$

Possible Answers:

${12z^2ln{(x)} -\frac{ {(1z^3)}}{x} + 34ycos{(z^2)}sin{(x^3)} - 8y^2zsin{(x^3)}sin{(z^2)} - 3x^2y^2cos{(x^3)}cos{(z^2)}}$

${35y^2zsin{(x^3)}sin{(z^2)} - 52.5z^2ln{(x)} - 35ycos{(z^2)}sin{(x^3)} -\frac{ {(17.5z^3)}}{x} - 52.5x^2y^2cos{(x^3)}cos{(z^2)}}$

${0.69z^2ln{(x)} -\frac{ {(0.06z^3)}}{x} + 1.94ycos{(z^2)}sin{(x^3)} - 0.46y^2zsin{(x^3)}sin{(z^2)} - 0.18x^2y^2cos{(x^3)}cos{(z^2)}}$

${210x^2yzcos{(x^3)}sin{(z^2)}}$

Correct answer:

${0.69z^2ln{(x)} -\frac{ {(0.06z^3)}}{x} + 1.94ycos{(z^2)}sin{(x^3)} - 0.46y^2zsin{(x^3)}sin{(z^2)} - 0.18x^2y^2cos{(x^3)}cos{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(-17)^2+(-4)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{17.49}=0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{17.49}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-4}{17.49}=-0.23\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.06)(-\frac{ {(1z^3)}}{x} - 3x^2y^2cos{(x^3)}cos{(z^2)})+(-0.97)(-2ycos{(z^2)}sin{(x^3)})+(-0.23)(2y^2zsin{(x^3)}sin{(z^2)} - 3z^2ln{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.69z^2ln{(x)} -\frac{ {(0.06z^3)}}{x} + 1.94ycos{(z^2)}sin{(x^3)} - 0.46y^2zsin{(x^3)}sin{(z^2)} - 0.18x^2y^2cos{(x^3)}cos{(z^2)}\end{align*}$

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Example Question #102 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4cos{(y^2)}sin{(x)} + e^{(y^2)}cos{(x)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-13,-10,12).\end{align*}$

Possible Answers:

${-40.6ye^{(y^2)}cos{(z)}sin{(x)}}$

${81.3cos{(y^2)}cos{(x)} - 163ysin{(y^2)}sin{(x)} + 20.3e^{(y^2)}cos{(x)}cos{(z)} - 20.3e^{(y^2)}sin{(x)}sin{(z)} + 40.6ye^{(y^2)}cos{(x)}sin{(z)}}$

${3.92ysin{(y^2)}sin{(x)} - 2.56cos{(y^2)}cos{(x)} + 0.59e^{(y^2)}cos{(x)}cos{(z)} + 0.64e^{(y^2)}sin{(x)}sin{(z)} - 0.98ye^{(y^2)}cos{(x)}sin{(z)}}$

${80ysin{(y^2)}sin{(x)} - 52cos{(y^2)}cos{(x)} + 12e^{(y^2)}cos{(x)}cos{(z)} + 13e^{(y^2)}sin{(x)}sin{(z)} - 20ye^{(y^2)}cos{(x)}sin{(z)}}$

Correct answer:

${3.92ysin{(y^2)}sin{(x)} - 2.56cos{(y^2)}cos{(x)} + 0.59e^{(y^2)}cos{(x)}cos{(z)} + 0.64e^{(y^2)}sin{(x)}sin{(z)} - 0.98ye^{(y^2)}cos{(x)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-10)^2+(12)^2}=20.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{20.32}=-0.64\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{20.32}=-0.49\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{20.32}=0.59\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.64)(4cos{(y^2)}cos{(x)} - e^{(y^2)}sin{(x)}sin{(z)})+(-0.49)(2ye^{(y^2)}cos{(x)}sin{(z)} - 8ysin{(y^2)}sin{(x)})+(0.59)(e^{(y^2)}cos{(x)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=3.92ysin{(y^2)}sin{(x)} - 2.56cos{(y^2)}cos{(x)} + 0.59e^{(y^2)}cos{(x)}cos{(z)} + 0.64e^{(y^2)}sin{(x)}sin{(z)} - 0.98ye^{(y^2)}cos{(x)}sin{(z)}\end{align*}$

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Example Question #103 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(4y^4ln{(z)})}}{x} - cos{(y^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-15,-15,-17).\end{align*}$

Possible Answers:

${17cos{(y^2)}cos{(z)} - 30ysin{(y^2)}sin{(z)} -\frac{ {(68y^4)}}{{(xz)}} -\frac{ {(240y^3ln{(z)})}}{x} +\frac{ {(60y^4ln{(z)})}}{x^2}}$

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

${54.4ysin{(y^2)}sin{(z)} - 27.2cos{(y^2)}cos{(z)} +\frac{ {(109y^4)}}{{(xz)}} +\frac{ {(435y^3ln{(z)})}}{x} -\frac{ {(109y^4ln{(z)})}}{x^2}}$

${0.63cos{(y^2)}cos{(z)} - 1.1ysin{(y^2)}sin{(z)} -\frac{ {(2.52y^4)}}{{(xz)}} -\frac{ {(8.8y^3ln{(z)})}}{x} +\frac{ {(2.2y^4ln{(z)})}}{x^2}}$

Correct answer:

${0.63cos{(y^2)}cos{(z)} - 1.1ysin{(y^2)}sin{(z)} -\frac{ {(2.52y^4)}}{{(xz)}} -\frac{ {(8.8y^3ln{(z)})}}{x} +\frac{ {(2.2y^4ln{(z)})}}{x^2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(-15)^2+(-17)^2}=27.18\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{27.18}=-0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{27.18}=-0.55\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{27.18}=-0.63\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.55)(\frac{-{(4y^4ln{(z)})}}{x^2})+(-0.55)(2ysin{(y^2)}sin{(z)} +\frac{ {(16y^3ln{(z)})}}{x})+(-0.63)(\frac{{(4y^4)}}{{(xz)}} - cos{(y^2)}cos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.63cos{(y^2)}cos{(z)} - 1.1ysin{(y^2)}sin{(z)} -\frac{ {(2.52y^4)}}{{(xz)}} -\frac{ {(8.8y^3ln{(z)})}}{x} +\frac{ {(2.2y^4ln{(z)})}}{x^2}\end{align*}$

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Example Question #101 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- yz - cos{(z^2)}sin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-7,-8,-12).\end{align*}$

Possible Answers:

${0.75y + 0.5z - 1.5zsin{(x^3)}sin{(z^2)} + 1.32x^2cos{(x^3)}cos{(z^2)}}$

${32.1zsin{(x^3)}sin{(z^2)} - 16z - 16y - 48.1x^2cos{(x^3)}cos{(z^2)}}$

${12y + 8z - 24zsin{(x^3)}sin{(z^2)} + 21x^2cos{(x^3)}cos{(z^2)}}$

Correct answer:

${0.75y + 0.5z - 1.5zsin{(x^3)}sin{(z^2)} + 1.32x^2cos{(x^3)}cos{(z^2)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(-8)^2+(-12)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{16.03}=-0.44\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{16.03}=-0.5\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{16.03}=-0.75\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.44)(-3x^2cos{(x^3)}cos{(z^2)})+(-0.5)(-1z)+(-0.75)(2zsin{(x^3)}sin{(z^2)} - y)\\&D_{\overrightarrow{u}}(x,y,z)=0.75y + 0.5z - 1.5zsin{(x^3)}sin{(z^2)} + 1.32x^2cos{(x^3)}cos{(z^2)}\end{align*}$

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Example Question #102 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=xy^2\\&\text{In the direction of }\overrightarrow{v}=(20,-14,-10).\end{align*}$

Possible Answers:

{0}

${20y^2 - 28xy}$

${52.8xy + 26.4y^2}$

${0.76y^2 - 1.06xy}$

Correct answer:

${0.76y^2 - 1.06xy}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(-14)^2+(-10)^2}=26.38\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{26.38}=0.76\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{26.38}=-0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-10}{26.38}=-0.38\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.76)(y^2)+(-0.53)(2xy)+(-0.38)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.76y^2 - 1.06xy\end{align*}$

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Calculus 3 : Calculus 3

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