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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 #16 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ycos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(7,9,6).\end{align*}$

Possible Answers:

${(12.88)(cos{(x^3)} - 3x^2ysin{(x^3)})}$

$(-21x^2ysin{(x^3)})+(9cos{(x^3)})$

$(-1.62x^2ysin{(x^3)})+(0.7cos{(x^3)})$

Correct answer:

$(-1.62x^2ysin{(x^3)})+(0.7cos{(x^3)})$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(9)^2+(6)^2}=12.88\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{12.88}=0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{12.88}=0.7\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{12.88}=0.47\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.54)(-3x^2ysin{(x^3)})+(0.7)(cos{(x^3)})+(0.47)(0)\end{align*}$

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Example Question #17 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5cos{(y^2)}ln{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-12,-13,-19).\end{align*}$

Possible Answers:

$130ysin{(y^2)}ln{(z)}-{(95cos{(y^2)})}/z$

${(129.8cos{(y^2)})}/z - 259.6ysin{(y^2)}ln{(z)}$

$5ysin{(y^2)}ln{(z)}-{3.65cos{(y^2})}/z$

Correct answer:

$5ysin{(y^2)}ln{(z)}-{3.65cos{(y^2})}/z$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12)^2+(-13)^2+(-19)^2}=25.96\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12}{25.96}=-0.46\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{25.96}=-0.5\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-19}{25.96}=-0.73\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.46)(0)+(-0.5)(-10ysin{(y^2)}ln{(z)})+(-0.73)({(5cos{(y^2)})}/z)\end{align*}$

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Example Question #18 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3y^2zsin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(14,17,11).\end{align*}$

Possible Answers:

${73.86y^2sin{(x)} + 147.72yzsin{(x)} + 73.86y^2zcos{(x)}}$

${1.35y^2sin{(x)} + 4.14yzsin{(x)} + 1.71y^2zcos{(x)}}$

${33y^2sin{(x)} + 102yzsin{(x)} + 42y^2zcos{(x)}}$

{147.72ycos{(x)}}

Correct answer:

${1.35y^2sin{(x)} + 4.14yzsin{(x)} + 1.71y^2zcos{(x)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(17)^2+(11)^2}=24.62\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{24.62}=0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{17}{24.62}=0.69\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{24.62}=0.45\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.57)(3y^2zcos{(x)})+(0.69)(6yzsin{(x)})+(0.45)(3y^2sin{(x)})\end{align*}$

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Example Question #19 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=7y^2z^3cos{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(14,11,18).\end{align*}$

Possible Answers:

${-3190.32x^2yz^2sin{(x^3)}}$

${14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}$

${531.72y^2z^2cos{(x^3)} + 354.48yz^3cos{(x^3)} - 531.72x^2y^2z^3sin{(x^3)}}$

${378y^2z^2cos{(x^3)} + 154yz^3cos{(x^3)} - 294x^2y^2z^3sin{(x^3)}}$

Correct answer:

${14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(11)^2+(18)^2}=25.32\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{25.32}=0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{25.32}=0.43\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{25.32}=0.71\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(-21x^2y^2z^3sin{(x^3)})+(0.43)(14yz^3cos{(x^3)})+(0.71)(21y^2z^2cos{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)={14.91y^2z^2cos{(x^3)} + 6.02yz^3cos{(x^3)} - 11.55x^2y^2z^3sin{(x^3)}}\end{align*}$

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Example Question #20 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=9x^4e^{(z^2)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(7,-10,-8).\end{align*}$

Possible Answers:

${131.31x^4e^{(z^2)}cos{(y)} + 525.24x^3e^{(z^2)}sin{(y)} + 262.62x^4ze^{(z^2)}sin{(y)}}$

${17.28x^3e^{(z^2)}sin{(y)} - 6.21x^4e^{(z^2)}cos{(y)} - 9.9x^4ze^{(z^2)}sin{(y)}}$

${1050.48x^3ze^{(z^2)}cos{(y)}}$

${252x^3e^{(z^2)}sin{(y)} - 90x^4e^{(z^2)}cos{(y)} - 144x^4ze^{(z^2)}sin{(y)}}$

Correct answer:

${17.28x^3e^{(z^2)}sin{(y)} - 6.21x^4e^{(z^2)}cos{(y)} - 9.9x^4ze^{(z^2)}sin{(y)}}$

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(7)^2+(-10)^2+(-8)^2}=14.59\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{7}{14.59}=0.48\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-10}{14.59}=-0.69\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{14.59}=-0.55\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.48)(36x^3e^{(z^2)}sin{(y)})+(-0.69)(9x^4e^{(z^2)}cos{(y)})+(-0.55)(18x^4ze^{(z^2)}sin{(y)})\end{align*}$

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Example Question #1081 : Partial 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:

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

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

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

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

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 #22 : Directional Derivatives

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

Possible Answers:

{-3.35cos{(y)}}

{112.45cos{(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:

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

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

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

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

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:

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

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

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

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

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:

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

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

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

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

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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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #16 : Directional Derivatives

Example Question #17 : Directional Derivatives

Example Question #18 : Directional Derivatives

Example Question #19 : Directional Derivatives

Example Question #20 : Directional Derivatives

Example Question #1081 : Partial Derivatives

Example Question #22 : Directional Derivatives

Example Question #23 : Directional Derivatives

Example Question #24 : Directional Derivatives

Example Question #25 : Directional Derivatives

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