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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

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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Example Question #1248 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(x^{4})}cos(z^{4})sin(y)\\&\text{In the direction of }\overrightarrow{v}=(-6,0,10).\end{align*}$

Possible Answers:

${- 3.44\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 2.24\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${- 40\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 26.4\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${-205\cdot3^{(x^{4})}x^{3}z^{3}sin(z^{4})cos(y)}$

${11.7\cdot3^{(x^{4})}cos(z^{4})cos(y) - 46.6\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 51.2\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Correct answer:

${- 3.44\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 2.24\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-6)^2+(0)^2+(10)^2}=11.66\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-6}{11.66}=-0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.66}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{11.66}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.51)(4.39\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y))+(0)(3^{(x^{4})}cos(z^{4})cos(y))+(0.86)(-4\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=- 3.44\cdot3^{(x^{4})}z^{3}sin(z^{4})sin(y) - 2.24\cdot3^{(x^{4})}x^{3}cos(z^{4})sin(y)\end{align*}$

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Example Question #1249 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{3})}ln(z^{3})sin(x^{2})\\&\text{In the direction of }\overrightarrow{v}=(10,0,7).\end{align*}$

Possible Answers:

${\frac{(1.71\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 1.64\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

${\frac{(241\cdot3^{(y^{3})}xy^{2}cos(x^{2}))}{z}}$

${\frac{(21\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 20\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

${\frac{(36.6\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 40.2\cdot3^{(y^{3})}y^{2}ln(z^{3})sin(x^{2}) + 24.4\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

Correct answer:

${\frac{(1.71\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 1.64\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(0)^2+(7)^2}=12.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{12.21}=0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{12.21}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{12.21}=0.57\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.82)(2\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3}))+(0)(3.3\cdot3^{(y^{3})}y^{2}ln(z^{3})sin(x^{2}))+(0.57)(\frac{(3\cdot3^{(y^{3})}sin(x^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.71\cdot3^{(y^{3})}sin(x^{2}))}{z}+ 1.64\cdot3^{(y^{3})}xcos(x^{2})ln(z^{3})\end{align*}$

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Example Question #1250 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4^{(y^{3})}x^{2}ln(z^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,13,10).\end{align*}$

Possible Answers:

${\frac{(1.22\cdot4^{(y^{3})}x^{2})}{z}+ 3.29\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

${\frac{(32.8\cdot4^{(y^{3})}x^{2})}{z}+ 32.8\cdot4^{(y^{3})}xln(z^{2}) + 68.2\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

${\frac{(273\cdot4^{(y^{3})}xy^{2})}{z}}$

${\frac{(0.744\cdot4^{(y^{3})}x^{2})}{z}+ 2.6\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

Correct answer:

${\frac{(1.22\cdot4^{(y^{3})}x^{2})}{z}+ 3.29\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(13)^2+(10)^2}=16.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{16.4}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{13}{16.4}=0.79\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{16.4}=0.61\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2\cdot4^{(y^{3})}xln(z^{2}))+(0.79)(4.16\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2}))+(0.61)(\frac{(2\cdot4^{(y^{3})}x^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.22\cdot4^{(y^{3})}x^{2})}{z}+ 3.29\cdot4^{(y^{3})}x^{2}y^{2}ln(z^{2})\end{align*}$

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Example Question #1251 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{2}z^{4}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(17,0,3).\end{align*}$

Possible Answers:

${17y^{2}z^{4}sin(x) - 12y^{2}z^{3}cos(x)}$

${0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)}$

${17.3y^{2}z^{4}sin(x) - 69y^{2}z^{3}cos(x) - 34.5yz^{4}cos(x)}$

${138yz^{3}sin(x)}$

Correct answer:

${0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(17)^2+(0)^2+(3)^2}=17.26\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{17}{17.26}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.26}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{3}{17.26}=0.17\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(y^{2}z^{4}sin(x))+(0)(-2yz^{4}cos(x))+(0.17)(-4y^{2}z^{3}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.98y^{2}z^{4}sin(x) - 0.68y^{2}z^{3}cos(x)\end{align*}$

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Example Question #1252 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(y^{4})}e^{(x^{2})}e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(12,14,0).\end{align*}$

Possible Answers:

${0.845\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 2.54\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${18.4\cdot3^{(y^{4})}e^{(x^{2})}e^{(z)} + 36.9\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 81\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

${162\cdot3^{(y^{4})}xy^{3}e^{(x^{2})}e^{(z)}}$

Correct answer:

${1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(12)^2+(14)^2+(0)^2}=18.44\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{12}{18.44}=0.65\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{14}{18.44}=0.76\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{18.44}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.65)(2\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)})+(0.76)(4.39\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)})+(0)(3^{(y^{4})}e^{(x^{2})}e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=1.3\cdot3^{(y^{4})}xe^{(x^{2})}e^{(z)} + 3.34\cdot3^{(y^{4})}y^{3}e^{(x^{2})}e^{(z)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

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

Example Question #1248 : Partial Derivatives

Example Question #1249 : Partial Derivatives

Example Question #1250 : Partial Derivatives

Example Question #1251 : Partial Derivatives

Example Question #1252 : Partial Derivatives

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