We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

Example Question #1101 : Partial Derivatives

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

Possible Answers:

$0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}$

$2.7726\cdot 2^{(x^2)}xy - 17\cdot 2^{(x^2)}$

$22.11\cdot 2^{(x^2)} + 30.651\cdot 2^{(x^2)}xy$

Correct answer:

$0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-17)^2+(-14)^2}=22.11\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{22.11}=0.09\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{22.11}=-0.77\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{22.11}=-0.63\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.09)(1.3863\cdot 2^{(x^2)}xy)+(-0.77)(2^{(x^2)})+(-0.63)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.12477\cdot 2^{(x^2)}xy - 0.77\cdot 2^{(x^2)}\end{align*}$

Report an Error

Example Question #1102 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(z^3)}e^{(y^2)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-18,16,-18).\end{align*}$

Possible Answers:

$-180.42yz^2cos{(z^3)}e^{(y^2)}sin{(x)}$

$60.14ysin{(z^3)}e^{(y^2)}cos{(x)} - 30.07sin{(z^3)}e^{(y^2)}sin{(x)} + 90.21z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

$18sin{(z^3)}e^{(y^2)}sin{(x)} + 32ysin{(z^3)}e^{(y^2)}cos{(x)} - 54z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

$0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

Correct answer:

$0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-18)^2+(16)^2+(-18)^2}=30.07\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-18}{30.07}=-0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{30.07}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-18}{30.07}=-0.6\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.6)(-1sin{(z^3)}e^{(y^2)}sin{(x)})+(0.53)(2ysin{(z^3)}e^{(y^2)}cos{(x)})+(-0.6)(3z^2cos{(z^3)}e^{(y^2)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.6sin{(z^3)}e^{(y^2)}sin{(x)} + 1.06ysin{(z^3)}e^{(y^2)}cos{(x)} - 1.8z^2cos{(z^3)}e^{(y^2)}cos{(x)}\end{align*}$

Report an Error

Example Question #1103 : Partial Derivatives

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

Possible Answers:

$13\cdot 4^{(y^2)}e^{(z)}cos{(x)} + 13.863\cdot 4^{(y^2)}ye^{(z)}cos{(x)}$

$-38.622\cdot 4^{(y^2)}ye^{(z)}sin{(x)}$

$13.93\cdot 4^{(y^2)}e^{(z)}cos{(x)} - 13.93\cdot 4^{(y^2)}e^{(z)}sin{(x)} + 38.622\cdot 4^{(y^2)}ye^{(z)}cos{(x)}$

$0.93\cdot 4^{(y^2)}e^{(z)}cos{(x)} + 0.99813\cdot 4^{(y^2)}ye^{(z)}cos{(x)}$

Correct answer:

$0.93\cdot 4^{(y^2)}e^{(z)}cos{(x)} + 0.99813\cdot 4^{(y^2)}ye^{(z)}cos{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(13)^2}=13.93\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.93}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{13.93}=0.36\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{13}{13.93}=0.93\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-1\cdot 4^{(y^2)}e^{(z)}sin{(x)})+(0.36)(2.7726\cdot 4^{(y^2)}ye^{(z)}cos{(x)})+(0.93)(4^{(y^2)}e^{(z)}cos{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.93\cdot 4^{(y^2)}e^{(z)}cos{(x)} + 0.99813\cdot 4^{(y^2)}ye^{(z)}cos{(x)}\end{align*}$

Report an Error

Example Question #1104 : Partial Derivatives

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

Possible Answers:

$0.14e^{(y^2)}cos{(x)}cos{(z)} - 0.21e^{(y^2)}sin{(x)}sin{(z)} - 1.94ye^{(y^2)}cos{(z)}sin{(x)}$

$2e^{(y^2)}cos{(x)}cos{(z)} - 3e^{(y^2)}sin{(x)}sin{(z)} - 28ye^{(y^2)}cos{(z)}sin{(x)}$

$-28.9ye^{(y^2)}cos{(x)}sin{(z)}$

$14.5e^{(y^2)}cos{(x)}cos{(z)} - 14.5e^{(y^2)}sin{(x)}sin{(z)} + 28.9ye^{(y^2)}cos{(z)}sin{(x)}$

Correct answer:

$0.14e^{(y^2)}cos{(x)}cos{(z)} - 0.21e^{(y^2)}sin{(x)}sin{(z)} - 1.94ye^{(y^2)}cos{(z)}sin{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-14)^2+(3)^2}=14.46\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{14.46}=0.14\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-14}{14.46}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{14.46}=0.21\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.14)(e^{(y^2)}cos{(x)}cos{(z)})+(-0.97)(2ye^{(y^2)}cos{(z)}sin{(x)})+(0.21)(-1e^{(y^2)}sin{(x)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.14e^{(y^2)}cos{(x)}cos{(z)} - 0.21e^{(y^2)}sin{(x)}sin{(z)} - 1.94ye^{(y^2)}cos{(z)}sin{(x)}\end{align*}$

Report an Error

Example Question #1105 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^4cos{(x^3)}sin{(z^2)}\\&\text{In the direction of }\overrightarrow{v}=(8,18,12).\end{align*}$

Possible Answers:

$-554x^2y^3zcos{(z^2)}sin{(x^3)}$

$72y^3cos{(x^3)}sin{(z^2)} + 24y^4zcos{(x^3)}cos{(z^2)} - 24x^2y^4sin{(x^3)}sin{(z^2)}$

$92.3y^3cos{(x^3)}sin{(z^2)} + 46.1y^4zcos{(x^3)}cos{(z^2)} - 69.2x^2y^4sin{(x^3)}sin{(z^2)}$

$3.12y^3cos{(x^3)}sin{(z^2)} + 1.04y^4zcos{(x^3)}cos{(z^2)} - 1.05x^2y^4sin{(x^3)}sin{(z^2)}$

Correct answer:

$3.12y^3cos{(x^3)}sin{(z^2)} + 1.04y^4zcos{(x^3)}cos{(z^2)} - 1.05x^2y^4sin{(x^3)}sin{(z^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8)^2+(18)^2+(12)^2}=23.07\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8}{23.07}=0.35\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{23.07}=0.78\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{23.07}=0.52\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.35)(-3x^2y^4sin{(x^3)}sin{(z^2)})+(0.78)(4y^3cos{(x^3)}sin{(z^2)})+(0.52)(2y^4zcos{(x^3)}cos{(z^2)})\\&D_{\overrightarrow{u}}(x,y,z)=3.12y^3cos{(x^3)}sin{(z^2)} + 1.04y^4zcos{(x^3)}cos{(z^2)} - 1.05x^2y^4sin{(x^3)}sin{(z^2)}\end{align*}$

Report an Error

Example Question #1106 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin{(y^2)}cos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-17,7,0).\end{align*}$

Possible Answers:

$17sin{(y^2)}sin{(x)} + 14ycos{(y^2)}cos{(x)}$

$36.8ycos{(y^2)}cos{(x)} - 18.4sin{(y^2)}sin{(x)}$

$0.92sin{(y^2)}sin{(x)} + 0.76ycos{(y^2)}cos{(x)}$

Correct answer:

$0.92sin{(y^2)}sin{(x)} + 0.76ycos{(y^2)}cos{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-17)^2+(7)^2+(0)^2}=18.38\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-17}{18.38}=-0.92\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{7}{18.38}=0.38\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.38}=0\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.92)(-1sin{(y^2)}sin{(x)})+(0.38)(2ycos{(y^2)}cos{(x)})+(0)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=0.92sin{(y^2)}sin{(x)} + 0.76ycos{(y^2)}cos{(x)}\end{align*}$

Report an Error

Example Question #1107 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(x^2)}cos{(y)}\\&\text{In the direction of }\overrightarrow{v}=(14,5,1).\end{align*}$

Possible Answers:

$- 0.34cos{(x^2)}sin{(y)} - 1.88xsin{(x^2)}cos{(y)}$

$- 5cos{(x^2)}sin{(y)} - 28xsin{(x^2)}cos{(y)}$

$- 14.9cos{(x^2)}sin{(y)} - 29.8xsin{(x^2)}cos{(y)}$

Correct answer:

$- 0.34cos{(x^2)}sin{(y)} - 1.88xsin{(x^2)}cos{(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(14)^2+(5)^2+(1)^2}=14.9\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{14}{14.9}=0.94\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{14.9}=0.34\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{1}{14.9}=0.07\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.94)(-2xsin{(x^2)}cos{(y)})+(0.34)(-1cos{(x^2)}sin{(y)})+(0.07)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=- 0.34cos{(x^2)}sin{(y)} - 1.88xsin{(x^2)}cos{(y)}\end{align*}$

Report an Error

Example Question #1108 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=zsin{(y^3)}\\&\text{In the direction of }\overrightarrow{v}=(17,20,0).\end{align*}$

Possible Answers:

$2.28y^2zcos{(y^3)}$

$26.2sin{(y^3)} + 78.8y^2zcos{(y^3)}$

$60y^2zcos{(y^3)}$

Correct answer:

$2.28y^2zcos{(y^3)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17)^2+(20)^2+(0)^2}=26.25\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17}{26.25}=0.65\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{26.25}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{26.25}=0\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.65)(0.0)+(0.76)(3y^2zcos{(y^3)})+(0)(sin{(y^3)})\\&D_{\overrightarrow{u}}(x,y,z)=2.28y^2zcos{(y^3)}\end{align*}$

Report an Error

Example Question #1109 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^ycos{(x)}\\&\text{In the direction of }\overrightarrow{v}=(-2,18,6).\end{align*}$

Possible Answers:

$1.03\cdot 3^ycos{(x)} + 0.1\cdot 3^ysin{(x)}$

$21\cdot 3^ycos{(x)} - 19.1\cdot 3^ysin{(x)}$

$19.8\cdot 3^ycos{(x)} + 2\cdot 3^ysin{(x)}$

Correct answer:

$1.03\cdot 3^ycos{(x)} + 0.1\cdot 3^ysin{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(18)^2+(6)^2}=19.08\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{19.08}=-0.1\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{18}{19.08}=0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{19.08}=0.31\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.1)(-1\cdot 3^ysin{(x)})+(0.94)(1.1\cdot 3^ycos{(x)})+(0.31)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=1.03\cdot 3^ycos{(x)} + 0.1\cdot 3^ysin{(x)}\end{align*}$

Report an Error

Example Question #1110 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^4cos{(y^2)}sin{(x^3)}\\&\text{In the direction of }\overrightarrow{v}=(-16,-2,-8).\end{align*}$

Possible Answers:

$-432x^2yz^3cos{(x^3)}sin{(y^2)}$

$4yz^4sin{(x^3)}sin{(y^2)} - 32z^3cos{(y^2)}sin{(x^3)} - 48x^2z^4cos{(x^3)}cos{(y^2)}$

$0.22yz^4sin{(x^3)}sin{(y^2)} - 1.76z^3cos{(y^2)}sin{(x^3)} - 2.67x^2z^4cos{(x^3)}cos{(y^2)}$

$72z^3cos{(y^2)}sin{(x^3)} - 36yz^4sin{(x^3)}sin{(y^2)} + 54x^2z^4cos{(x^3)}cos{(y^2)}$

Correct answer:

$0.22yz^4sin{(x^3)}sin{(y^2)} - 1.76z^3cos{(y^2)}sin{(x^3)} - 2.67x^2z^4cos{(x^3)}cos{(y^2)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-16)^2+(-2)^2+(-8)^2}=18\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-16}{18}=-0.89\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{18}=-0.11\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-8}{18}=-0.44\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.89)(3x^2z^4cos{(x^3)}cos{(y^2)})+(-0.11)(-2yz^4sin{(x^3)}sin{(y^2)})+(-0.44)(4z^3cos{(y^2)}sin{(x^3)})\\&D_{\overrightarrow{u}}(x,y,z)=0.22yz^4sin{(x^3)}sin{(y^2)} - 1.76z^3cos{(y^2)}sin{(x^3)} - 2.67x^2z^4cos{(x^3)}cos{(y^2)}\end{align*}$

Report an Error

View Calculus 3 Tutors

Dalton
Certified Tutor

University of Massachusetts Amherst, Doctor of Philosophy, Mathematics.

View Calculus 3 Tutors

Satweka
Certified Tutor

The University of Texas at Dallas, Master of Science, Biomedical Engineering.

View Calculus 3 Tutors

Robert Edward
Certified Tutor

University of Calgary, Master of Engineering, Chemistry. University of Ottawa, Bachelor of Engineering, Civil Engineering.

All Calculus 3 Resources

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

Popular Subjects

GRE Tutors in San Diego, LSAT Tutors in Miami, ACT Tutors in San Diego, Physics Tutors in Atlanta, Math Tutors in Los Angeles, SSAT Tutors in Chicago, Chemistry Tutors in Houston, Spanish Tutors in San Diego, LSAT Tutors in San Diego, Statistics Tutors in Houston

Popular Courses & Classes

SAT Courses & Classes in Philadelphia, ACT Courses & Classes in Phoenix, ACT Courses & Classes in New York City, LSAT Courses & Classes in Dallas Fort Worth, LSAT Courses & Classes in Los Angeles, SSAT Courses & Classes in Phoenix, GRE Courses & Classes in Chicago, GRE Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in Boston, Spanish Courses & Classes in San Francisco-Bay Area

Popular Test Prep

MCAT Test Prep in San Francisco-Bay Area, LSAT Test Prep in Miami, ACT Test Prep in Phoenix, SSAT Test Prep in Boston, ACT Test Prep in Denver, GRE Test Prep in Chicago, MCAT Test Prep in Houston, SSAT Test Prep in Washington DC, GMAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Boston

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1101 : Partial Derivatives

Example Question #1102 : Partial Derivatives

Example Question #1103 : Partial Derivatives

Example Question #1104 : Partial Derivatives

Example Question #1105 : Partial Derivatives

Example Question #1106 : Partial Derivatives

Example Question #1107 : Partial Derivatives

Example Question #1108 : Partial Derivatives

Example Question #1109 : Partial Derivatives

Example Question #1110 : Partial Derivatives

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: