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Calculus 3 : Calculus 3

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Example Questions

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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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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:

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

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Example Question #1111 : Partial Derivatives

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

Possible Answers:

$- 16y^4sin{(x)} - 8y^4zcos{(x)} - 12y^3zsin{(x)}$

$- 0.88y^4sin{(x)} - 0.44y^4zcos{(x)} - 0.68y^3zsin{(x)}$

$18.1y^4sin{(x)} + 18.1y^4zcos{(x)} + 72.6y^3zsin{(x)}$

$72.6y^3cos{(x)}$

Correct answer:

$- 0.88y^4sin{(x)} - 0.44y^4zcos{(x)} - 0.68y^3zsin{(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-8)^2+(-3)^2+(-16)^2}=18.14\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-8}{18.14}=-0.44\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-3}{18.14}=-0.17\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{18.14}=-0.88\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.44)(y^4zcos{(x)})+(-0.17)(4y^3zsin{(x)})+(-0.88)(y^4sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.88y^4sin{(x)} - 0.44y^4zcos{(x)} - 0.68y^3zsin{(x)}\end{align*}$

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Example Question #1112 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^xsin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-10,-13,-15).\end{align*}$

Possible Answers:

$- 15\cdot 2^xcos{(z)} - 6.93\cdot 2^xsin{(z)}$

$22.2\cdot 2^xcos{(z)} + 15.4\cdot 2^xsin{(z)}$

$- 0.67\cdot 2^xcos{(z)} - 0.312\cdot 2^xsin{(z)}$

Correct answer:

$- 0.67\cdot 2^xcos{(z)} - 0.312\cdot 2^xsin{(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-10)^2+(-13)^2+(-15)^2}=22.23\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-10}{22.23}=-0.45\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{22.23}=-0.58\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-15}{22.23}=-0.67\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.45)(0.693\cdot 2^xsin{(z)})+(-0.58)(0.0)+(-0.67)(2^xcos{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.67\cdot 2^xcos{(z)} - 0.312\cdot 2^xsin{(z)}\end{align*}$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #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

Example Question #1111 : Partial Derivatives

Example Question #1112 : Partial Derivatives

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