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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1221 : Partial Derivatives

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

Possible Answers:

${\frac{{(0.99zcos{(x^2)})}}{y} + 0.3xzsin{(x^2)}ln{(y)}}$

${\frac{{(26.3xsin{(x^2)})}}{y}}$

${0.045xzsin{(x^2)}ln{(y)} -\frac{ {(0.98zcos{(x^2)})}}{y}}$

${26.3xzsin{(x^2)}ln{(y)} -\frac{ {(13.1zcos{(x^2)})}}{y} - 13.1cos{(x^2)}ln{(y)}}$

Correct answer:

${\frac{{(0.99zcos{(x^2)})}}{y} + 0.3xzsin{(x^2)}ln{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(-13)^2+(0)^2}=13.15\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{13.15}=0.15\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{13.15}=-0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{13.15}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.15)(2xzsin{(x^2)}ln{(y)})+(-0.99)(\frac{-{(1zcos{(x^2)})}}{y})+(0)(-1cos{(x^2)}ln{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(0.99zcos{(x^2)})}}{y} + 0.3xzsin{(x^2)}ln{(y)}\end{align*}$

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Example Question #1222 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(x^3)}ln{(z)}sin{(y)}\\&\text{In the direction of }\overrightarrow{v}=(5,0,10).\end{align*}$

Possible Answers:

${\frac{{(33.5x^2sin{(x^3)}cos{(y)})}}{z}}$

${33.5x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(11.2cos{(x^3)}sin{(y)})}}{z} - 11.2cos{(x^3)}cos{(y)}ln{(z)}}$

${1.35x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.89cos{(x^3)}sin{(y)})}}{z}}$

${0.607x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.792cos{(x^3)}sin{(y)})}}{z}}$

Correct answer:

${1.35x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.89cos{(x^3)}sin{(y)})}}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(0)^2+(10)^2}=11.18\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{11.18}=0.45\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{11.18}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{10}{11.18}=0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.45)(3x^2sin{(x^3)}ln{(z)}sin{(y)})+(0)(-1cos{(x^3)}cos{(y)}ln{(z)})+(0.89)(\frac{-{(1cos{(x^3)}sin{(y)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=1.35x^2sin{(x^3)}ln{(z)}sin{(y)} -\frac{ {(0.89cos{(x^3)}sin{(y)})}}{z}\end{align*}$

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Example Question #1222 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos{(x^2)}cos{(y)}ln{(z)}\\&\text{In the direction of }\overrightarrow{v}=(6,-8,0).\end{align*}$

Possible Answers:

${\frac{{(20xsin{(x^2)}sin{(y)})}}{z}}$

${0.8cos{(x^2)}ln{(z)}sin{(y)} - 1.2xsin{(x^2)}cos{(y)}ln{(z)}}$

${8cos{(x^2)}ln{(z)}sin{(y)} - 12xsin{(x^2)}cos{(y)}ln{(z)}}$

${\frac{{(10cos{(x^2)}cos{(y)})}}{z} - 10cos{(x^2)}ln{(z)}sin{(y)} - 20xsin{(x^2)}cos{(y)}ln{(z)}}$

Correct answer:

${0.8cos{(x^2)}ln{(z)}sin{(y)} - 1.2xsin{(x^2)}cos{(y)}ln{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(6)^2+(-8)^2+(0)^2}=10\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{6}{10}=0.6\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-8}{10}=-0.8\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{10}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.6)(-2xsin{(x^2)}cos{(y)}ln{(z)})+(-0.8)(-1cos{(x^2)}ln{(z)}sin{(y)})+(0)(\frac{{(cos{(x^2)}cos{(y)})}}{z})\\&D_{\overrightarrow{u}}(x,y,z)=0.8cos{(x^2)}ln{(z)}sin{(y)} - 1.2xsin{(x^2)}cos{(y)}ln{(z)}\end{align*}$

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Example Question #1223 : Partial Derivatives

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

Possible Answers:

${-36.6x^2}$

${- 7x^3y - 30x^2yz}$

${- 12.2x^3y - 12.2x^3z - 36.6x^2yz}$

${- 0.57x^3y - 2.46x^2yz}$

Correct answer:

${- 0.57x^3y - 2.46x^2yz}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(10)^2+(0)^2+(7)^2}=12.21\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{10}{12.21}=0.82\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{12.21}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{12.21}=0.57\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.82)(-3x^2yz)+(0)(-1x^3z)+(0.57)(-1x^3y)\\&D_{\overrightarrow{u}}(x,y,z)=- 0.57x^3y - 2.46x^2yz\end{align*}$

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Example Question #1224 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-sin{(z^2)}e^{(x)}ln{(y)}\\&\text{In the direction of }\overrightarrow{v}=(5,0,1).\end{align*}$

Possible Answers:

${- 5sin{(z^2)}e^{(x)}ln{(y)} - 2zcos{(z^2)}e^{(x)}ln{(y)}}$

${-\frac{ {(5.1sin{(z^2)}e^{(x)})}}{y} - 5.1sin{(z^2)}e^{(x)}ln{(y)} - 10.2zcos{(z^2)}e^{(x)}ln{(y)}}$

${- 0.98sin{(z^2)}e^{(x)}ln{(y)} - 0.4zcos{(z^2)}e^{(x)}ln{(y)}}$

${\frac{-{(10.2zcos{(z^2)}e^{(x)})}}{y}}$

Correct answer:

${- 0.98sin{(z^2)}e^{(x)}ln{(y)} - 0.4zcos{(z^2)}e^{(x)}ln{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(5)^2+(0)^2+(1)^2}=5.1\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{5}{5.1}=0.98\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{5.1}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{1}{5.1}=0.2\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.98)(-1sin{(z^2)}e^{(x)}ln{(y)})+(0)(\frac{-{(1sin{(z^2)}e^{(x)})}}{y})+(0.2)(-2zcos{(z^2)}e^{(x)}ln{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.98sin{(z^2)}e^{(x)}ln{(y)} - 0.4zcos{(z^2)}e^{(x)}ln{(y)}\end{align*}$

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Example Question #1226 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{{(cos{(x^2)}ln{(z^2)})}}{y^2}\\&\text{In the direction of }\overrightarrow{v}=(-15,0,9).\end{align*}$

Possible Answers:

${\frac{{(35cos{(x^2)})}}{{(y^2z)}} -\frac{ {(35cos{(x^2)}ln{(z^2)})}}{y^3} -\frac{ {(35xln{(z^2)}sin{(x^2)})}}{y^2}}$

${\frac{{(1.02cos{(x^2)})}}{{(y^2z)}} +\frac{ {(1.72xln{(z^2)}sin{(x^2)})}}{y^2}}$

${\frac{{(140xsin{(x^2)})}}{{(y^3z)}}}$

${\frac{{(0.52cos{(x^2)})}}{{(y^2z)}} -\frac{ {(1.48xln{(z^2)}sin{(x^2)})}}{y^2}}$

Correct answer:

${\frac{{(1.02cos{(x^2)})}}{{(y^2z)}} +\frac{ {(1.72xln{(z^2)}sin{(x^2)})}}{y^2}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-15)^2+(0)^2+(9)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-15}{17.49}=-0.86\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.86)(\frac{-{(2xln{(z^2)}sin{(x^2)})}}{y^2})+(0)(\frac{-{(2cos{(x^2)}ln{(z^2)})}}{y^3})+(0.51)(\frac{{(2cos{(x^2)})}}{{(y^2z)}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{{(1.02cos{(x^2)})}}{{(y^2z)}} +\frac{ {(1.72xln{(z^2)}sin{(x^2)})}}{y^2}\end{align*}$

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Example Question #1225 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{-4^{(z^2)}}{x}\\&\text{In the direction of }\overrightarrow{v}=(-11,0,11).\end{align*}$

Possible Answers:

${-\frac{ {(0.71\cdot4^{(z^2)})}}{x^2} -\frac{ {(1.97\cdot4^{(z^2)}z)}}{x}}$

${-\frac{ {(11\cdot4^{(z^2)})}}{x^2} -\frac{ {(30.5\cdot4^{(z^2)}z)}}{x}}$

${\frac{{(15.6\cdot4^{(z^2)})}}{x^2} -\frac{ {(43.1\cdot4^{(z^2)}z)}}{x}}$

Correct answer:

${-\frac{ {(0.71\cdot4^{(z^2)})}}{x^2} -\frac{ {(1.97\cdot4^{(z^2)}z)}}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-11)^2+(0)^2+(11)^2}=15.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-11}{15.56}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{15.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{11}{15.56}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(\frac{4^{(z^2)}}{x^2})+(0)(0.0)+(0.71)(\frac{-{(2.77\cdot4^{(z^2)}z)}}{x})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(0.71\cdot4^{(z^2)})}}{x^2} -\frac{ {(1.97\cdot4^{(z^2)}z)}}{x}\end{align*}$

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Example Question #1228 : Partial Derivatives

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

Possible Answers:

${0.78\cdot3^ye^{(x)}cos{(z)} - 0.71\cdot3^ye^{(x)}sin{(z)}}$

${59.3\cdot3^ye^{(x)}cos{(z)} - 28.3\cdot3^ye^{(x)}sin{(z)}}$

${-31.1\cdot3^ye^{(x)}sin{(z)}}$

${22\cdot3^ye^{(x)}cos{(z)} - 20\cdot3^ye^{(x)}sin{(z)}}$

Correct answer:

${0.78\cdot3^ye^{(x)}cos{(z)} - 0.71\cdot3^ye^{(x)}sin{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(20)^2+(20)^2}=28.28\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{28.28}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{28.28}=0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{20}{28.28}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3^ye^{(x)}cos{(z)})+(0.71)(1.1\cdot3^ye^{(x)}cos{(z)})+(0.71)(-1\cdot3^ye^{(x)}sin{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=0.78\cdot3^ye^{(x)}cos{(z)} - 0.71\cdot3^ye^{(x)}sin{(z)}\end{align*}$

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Example Question #1229 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos{(z^4)}cos{(x)}ln{(y)}\\&\text{In the direction of }\overrightarrow{v}=(-1,15,0).\end{align*}$

Possible Answers:

${15cos{(z^4)}ln{(y)}sin{(x)} -\frac{ {(15cos{(z^4)}cos{(x)})}}{y} + 60.1z^3sin{(z^4)}cos{(x)}ln{(y)}}$

${-\frac{ {(cos{(z^4)}cos{(x)})}}{y} - 0.07cos{(z^4)}ln{(y)}sin{(x)}}$

${0.0049cos{(z^4)}ln{(y)}sin{(x)} -\frac{ {(cos{(z^4)}cos{(x)})}}{y}}$

${\frac{-{(60.1z^3sin{(z^4)}sin{(x)})}}{y}}$

Correct answer:

${-\frac{ {(cos{(z^4)}cos{(x)})}}{y} - 0.07cos{(z^4)}ln{(y)}sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(15)^2+(0)^2}=15.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{15.03}=-0.07\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{15}{15.03}=1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.07)(cos{(z^4)}ln{(y)}sin{(x)})+(1)(\frac{-{(cos{(z^4)}cos{(x)})}}{y})+(0)(4z^3sin{(z^4)}cos{(x)}ln{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(1cos{(z^4)}cos{(x)})}}{y} - 0.07cos{(z^4)}ln{(y)}sin{(x)}\end{align*}$

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Example Question #1230 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-sin{(x^2)}sin{(y^4)}sin{(z^4)}\\&\text{In the direction of }\overrightarrow{v}=(-7,11,0).\end{align*}$

Possible Answers:

${- 52.2y^3cos{(y^4)}sin{(x^2)}sin{(z^4)} - 52.2z^3cos{(z^4)}sin{(x^2)}sin{(y^4)} - 26.1xcos{(x^2)}sin{(y^4)}sin{(z^4)}}$

${-417xy^3z^3cos{(x^2)}cos{(y^4)}cos{(z^4)}}$

${1.08xcos{(x^2)}sin{(y^4)}sin{(z^4)} - 3.36y^3cos{(y^4)}sin{(x^2)}sin{(z^4)}}$

${- 2.82y^3cos{(y^4)}sin{(x^2)}sin{(z^4)} - 0.583xcos{(x^2)}sin{(y^4)}sin{(z^4)}}$

Correct answer:

${1.08xcos{(x^2)}sin{(y^4)}sin{(z^4)} - 3.36y^3cos{(y^4)}sin{(x^2)}sin{(z^4)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(11)^2+(0)^2}=13.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{13.04}=-0.54\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{11}{13.04}=0.84\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{13.04}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.54)(-2xcos{(x^2)}sin{(y^4)}sin{(z^4)})+(0.84)(-4y^3cos{(y^4)}sin{(x^2)}sin{(z^4)})+(0)(-4z^3cos{(z^4)}sin{(x^2)}sin{(y^4)})\\&D_{\overrightarrow{u}}(x,y,z)=1.08xcos{(x^2)}sin{(y^4)}sin{(z^4)} - 3.36y^3cos{(y^4)}sin{(x^2)}sin{(z^4)}\end{align*}$

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Calculus 3 : Partial Derivatives

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Example Question #1221 : Partial Derivatives

Example Question #1222 : Partial Derivatives

Example Question #1222 : Partial Derivatives

Example Question #1223 : Partial Derivatives

Example Question #1224 : Partial Derivatives

Example Question #1226 : Partial Derivatives

Example Question #1225 : Partial Derivatives

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