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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #3601 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-ln(y^{4})sin(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,5,-17).\end{align*}$

Possible Answers:

{0}

${-\frac{ (70.9sin(z^{2}))}{y}- 35.4zcos(z^{2})ln(y^{4})}$

${1.92zcos(z^{2})ln(y^{4}) -\frac{ (1.12sin(z^{2}))}{y}}$

${-\frac{ (0.314sin(z^{2}))}{y}- 1.84zcos(z^{2})ln(y^{4})}$

Correct answer:

${1.92zcos(z^{2})ln(y^{4}) -\frac{ (1.12sin(z^{2}))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(-17)^2}=17.72\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.72}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{17.72}=0.28\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{17.72}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(0.0)+(0.28)(-\frac{(4sin(z^{2}))}{y})+(-0.96)(-2zcos(z^{2})ln(y^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.92zcos(z^{2})ln(y^{4}) -\frac{ (1.12sin(z^{2}))}{y}\end{align*}$

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Example Question #3602 : Calculus 3

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin(y^{3})e^{(x^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(1,9,0).\end{align*}$

Possible Answers:

${109x^{3}y^{2}cos(y^{3})e^{(x^{4})}cos(z)}$

${27y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 4x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

${9.06sin(y^{3})e^{(x^{4})}cos(z) + 27.2y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 36.2x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

${2.97y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 0.44x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

Correct answer:

${2.97y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 0.44x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(1)^2+(9)^2+(0)^2}=9.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{1}{9.06}=0.11\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{9.06}=0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{9.06}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.11)(4x^{3}sin(y^{3})e^{(x^{4})}sin(z))+(0.99)(3y^{2}cos(y^{3})e^{(x^{4})}sin(z))+(0)(sin(y^{3})e^{(x^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=2.97y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 0.44x^{3}sin(y^{3})e^{(x^{4})}sin(z)\end{align*}$

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Example Question #3603 : Calculus 3

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

Possible Answers:

${\frac{(0.06sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (3y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

${\frac{(48.1z^{2}cos(z^{3})e^{(y^{3})})}{x}-\frac{ (16sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (48.1y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

${-\frac{(144y^{2}z^{2}cos(z^{3})e^{(y^{3})})}{x^{2}}}$

${\frac{(sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (48y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

Correct answer:

${\frac{(0.06sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (3y^{2}sin(z^{3})e^{(y^{3})})}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(16)^2+(0)^2}=16.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{16.03}=-0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{16.03}=1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.06)(-\frac{(sin(z^{3})e^{(y^{3})})}{x^{2}})+(1)(\frac{(3y^{2}sin(z^{3})e^{(y^{3})})}{x})+(0)(\frac{(3z^{2}cos(z^{3})e^{(y^{3})})}{x})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.06sin(z^{3})e^{(y^{3})})}{x^{2}}+\frac{ (3y^{2}sin(z^{3})e^{(y^{3})})}{x}\end{align*}$

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Example Question #1241 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(y^{4})}zln(x^{3})\\&\text{In the direction of }\overrightarrow{v}=(0,3,10).\end{align*}$

Possible Answers:

${0.96\cdot2^{(y^{4})}ln(x^{3}) + 0.804\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

${\frac{(86.8\cdot2^{(y^{4})}y^{3})}{x}}$

${10.4\cdot2^{(y^{4})}ln(x^{3}) +\frac{ (31.3\cdot2^{(y^{4})}z)}{x}+ 28.9\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

${0.922\cdot2^{(y^{4})}ln(x^{3}) + 0.233\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

Correct answer:

${0.96\cdot2^{(y^{4})}ln(x^{3}) + 0.804\cdot2^{(y^{4})}y^{3}zln(x^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(3)^2+(10)^2}=10.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{10.44}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{3}{10.44}=0.29\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{10.44}=0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(3\cdot2^{(y^{4})}z)}{x})+(0.29)(2.77\cdot2^{(y^{4})}y^{3}zln(x^{3}))+(0.96)(2^{(y^{4})}ln(x^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=0.96\cdot2^{(y^{4})}ln(x^{3}) + 0.804\cdot2^{(y^{4})}y^{3}zln(x^{3})\end{align*}$

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Example Question #1242 : Partial Derivatives

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

Possible Answers:

${269\cdot4^{(z^{4})}x^{2}y^{2}z^{3}cos(x^{3})cos(y^{3})}$

${2.05\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}) - 2.79\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3})}$

${2.59\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3}) + 0.759\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3})}$

${16.2\cdot4^{(z^{4})}x^{2}cos(x^{3})sin(y^{3}) + 16.2\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3}) + 29.9\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3})}$

Correct answer:

${2.05\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}) - 2.79\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-5)^2+(2)^2}=5.39\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{5.39}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{5.39}=-0.93\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{5.39}=0.37\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3\cdot4^{(z^{4})}x^{2}cos(x^{3})sin(y^{3}))+(-0.93)(3\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3}))+(0.37)(5.55\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=2.05\cdot4^{(z^{4})}z^{3}sin(x^{3})sin(y^{3}) - 2.79\cdot4^{(z^{4})}y^{2}cos(y^{3})sin(x^{3})\end{align*}$

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Example Question #1243 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^{4}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(12,-9,0).\end{align*}$

Possible Answers:

${-9z^{4}cos(y)}$

${15z^{4}cos(y) + 60z^{3}sin(y)}$

${-0.6z^{4}cos(y)}$

{0}

Correct answer:

${-0.6z^{4}cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(12)^2+(-9)^2+(0)^2}=15\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{12}{15}=0.8\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-9}{15}=-0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.8)(0.0)+(-0.6)(z^{4}cos(y))+(0)(4z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=-0.6z^{4}cos(y)\end{align*}$

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Example Question #1244 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(x^{3})}ln(z^{3})\\&\text{In the direction of }\overrightarrow{v}=(-9,0,7).\end{align*}$

Possible Answers:

${-\frac{ (34.2\cdot2^{(x^{3})})}{z}- 23.7\cdot2^{(x^{3})}x^{2}ln(z^{3})}$

${18.7\cdot2^{(x^{3})}x^{2}ln(z^{3}) -\frac{ (21\cdot2^{(x^{3})})}{z}}$

${1.64\cdot2^{(x^{3})}x^{2}ln(z^{3}) -\frac{ (1.83\cdot2^{(x^{3})})}{z}}$

Correct answer:

${1.64\cdot2^{(x^{3})}x^{2}ln(z^{3}) -\frac{ (1.83\cdot2^{(x^{3})})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-9)^2+(0)^2+(7)^2}=11.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-9}{11.4}=-0.79\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.4}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{11.4}=0.61\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.79)(-2.08\cdot2^{(x^{3})}x^{2}ln(z^{3}))+(0)(0.0)+(0.61)(-\frac{(3\cdot2^{(x^{3})})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.64\cdot2^{(x^{3})}x^{2}ln(z^{3}) -\frac{ (1.83\cdot2^{(x^{3})})}{z}\end{align*}$

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Example Question #1245 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(y^{4})}ze^{(x^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,-16,5).\end{align*}$

Possible Answers:

${2.63\cdot2^{(y^{4})}y^{3}ze^{(x^{2})} - 0.3\cdot2^{(y^{4})}e^{(x^{2})}}$

${44.4\cdot2^{(y^{4})}y^{3}ze^{(x^{2})} - 5\cdot2^{(y^{4})}e^{(x^{2})}}$

${44.4\cdot2^{(y^{4})}y^{3}ze^{(x^{2})} + 5\cdot2^{(y^{4})}e^{(x^{2})}}$

Correct answer:

${2.63\cdot2^{(y^{4})}y^{3}ze^{(x^{2})} - 0.3\cdot2^{(y^{4})}e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-16)^2+(5)^2}=16.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{16.76}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{16.76}=-0.95\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{5}{16.76}=0.3\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2\cdot2^{(y^{4})}xze^{(x^{2})})+(-0.95)(-2.77\cdot2^{(y^{4})}y^{3}ze^{(x^{2})})+(0.3)(-1\cdot2^{(y^{4})}e^{(x^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=2.63\cdot2^{(y^{4})}y^{3}ze^{(x^{2})} - 0.3\cdot2^{(y^{4})}e^{(x^{2})}\end{align*}$

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Example Question #1246 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-x^{3}sin(y^{2})sin(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(-13,-12,0).\end{align*}$

Possible Answers:

${-212x^{2}yzcos(y^{2})cos(z^{2})}$

${- 53.1x^{2}sin(y^{2})sin(z^{2}) - 35.4x^{3}ycos(y^{2})sin(z^{2}) - 35.4x^{3}zcos(z^{2})sin(y^{2})}$

${- 1.6x^{2}sin(y^{2})sin(z^{2}) - 0.925x^{3}ycos(y^{2})sin(z^{2})}$

${2.19x^{2}sin(y^{2})sin(z^{2}) + 1.36x^{3}ycos(y^{2})sin(z^{2})}$

Correct answer:

${2.19x^{2}sin(y^{2})sin(z^{2}) + 1.36x^{3}ycos(y^{2})sin(z^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(-12)^2+(0)^2}=17.69\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{17.69}=-0.73\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-12}{17.69}=-0.68\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.69}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.73)(-3x^{2}sin(y^{2})sin(z^{2}))+(-0.68)(-2x^{3}ycos(y^{2})sin(z^{2}))+(0)(-2x^{3}zcos(z^{2})sin(y^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=2.19x^{2}sin(y^{2})sin(z^{2}) + 1.36x^{3}ycos(y^{2})sin(z^{2})\end{align*}$

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Example Question #1247 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(x^{2})}z^{2}ln(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(-4,0,-14).\end{align*}$

Possible Answers:

${-\frac{(256\cdot3^{(x^{2})}xz)}{y}}$

${-\frac{ (58.2\cdot3^{(x^{2})}z^{2})}{y}- 29.1\cdot3^{(x^{2})}zln(y^{4}) - 32\cdot3^{(x^{2})}xz^{2}ln(y^{4})}$

${28\cdot3^{(x^{2})}zln(y^{4}) + 8.79\cdot3^{(x^{2})}xz^{2}ln(y^{4})}$

${1.92\cdot3^{(x^{2})}zln(y^{4}) + 0.593\cdot3^{(x^{2})}xz^{2}ln(y^{4})}$

Correct answer:

${1.92\cdot3^{(x^{2})}zln(y^{4}) + 0.593\cdot3^{(x^{2})}xz^{2}ln(y^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(-14)^2}=14.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{14.56}=-0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{14.56}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.27)(-2.2\cdot3^{(x^{2})}xz^{2}ln(y^{4}))+(0)(-\frac{(4\cdot3^{(x^{2})}z^{2})}{y})+(-0.96)(-2\cdot3^{(x^{2})}zln(y^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.92\cdot3^{(x^{2})}zln(y^{4}) + 0.593\cdot3^{(x^{2})}xz^{2}ln(y^{4})\end{align*}$

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Calculus 3 : Directional Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #3601 : Calculus 3

Example Question #3602 : Calculus 3

Example Question #3603 : Calculus 3

Example Question #1241 : Partial Derivatives

Example Question #1242 : Partial Derivatives

Example Question #1243 : Partial Derivatives

Example Question #1244 : Partial Derivatives

Example Question #1245 : Partial Derivatives

Example Question #1246 : Partial Derivatives

Example Question #1247 : Partial Derivatives

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