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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1218 : Partial Derivatives

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

Possible Answers:

${-29.7x^2e^{(y)}}$

${2.13x^2ze^{(y)} - 0.71x^3e^{(y)}}$

${- 9.9x^3e^{(y)} - 29.7x^2ze^{(y)} - 9.9x^3ze^{(y)}}$

${21x^2ze^{(y)} - 7x^3e^{(y)}}$

Correct answer:

${2.13x^2ze^{(y)} - 0.71x^3e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(0)^2+(7)^2}=9.9\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{9.9}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{9.9}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7}{9.9}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-3x^2ze^{(y)})+(0)(-1x^3ze^{(y)})+(0.71)(-1x^3e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=2.13x^2ze^{(y)} - 0.71x^3e^{(y)}\end{align*}$

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Example Question #1219 : Partial Derivatives

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

Possible Answers:

${-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}}$

${-\frac{ {(17.5cos{(x)})}}{{(y^2z^2)}} -\frac{ {(35cos{(x)})}}{{(y^3z)}} -\frac{ {(17.5sin{(x)})}}{{(y^2z)}}}$

${-\frac{ {(0.74cos{(x)})}}{{(y^2z^2)}} -\frac{ {(0.52cos{(x)})}}{{(y^3z)}}}$

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

Correct answer:

${-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(15)^2}=17.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.49}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{17.49}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{15}{17.49}=0.86\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{-{(1sin{(x)})}}{{(y^2z)}})+(0.51)(\frac{-{(2cos{(x)})}}{{(y^3z)}})+(0.86)(\frac{-{(1cos{(x)})}}{{(y^2z^2)}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ {(0.86cos{(x)})}}{{(y^2z^2)}} -\frac{ {(1.02cos{(x)})}}{{(y^3z)}}\end{align*}$

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Example Question #1220 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^3z^3sin{(x)}\\&\text{In the direction of }\overrightarrow{v}=(20,12,0).\end{align*}$

Possible Answers:

${20y^3z^3cos{(x)} + 36y^2z^3sin{(x)}}$

${210y^2z^2cos{(x)}}$

${23.3y^3z^3cos{(x)} + 70y^2z^3sin{(x)} + 70y^3z^2sin{(x)}}$

${0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}}$

Correct answer:

${0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(20)^2+(12)^2+(0)^2}=23.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{20}{23.32}=0.86\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12}{23.32}=0.51\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.86)(y^3z^3cos{(x)})+(0.51)(3y^2z^3sin{(x)})+(0)(3y^3z^2sin{(x)})\\&D_{\overrightarrow{u}}(x,y,z)=0.86y^3z^3cos{(x)} + 1.53y^2z^3sin{(x)}\end{align*}$

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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 #151 : Directional 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:

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

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

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

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

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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Calculus 3 : Directional Derivatives

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Example Questions

Example Question #1218 : Partial Derivatives

Example Question #1219 : Partial Derivatives

Example Question #1220 : Partial Derivatives

Example Question #1221 : Partial Derivatives

Example Question #151 : Directional 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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