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Calculus 3 : Partial Derivatives

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Example Questions

Example Question #1371 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(x^{4})}e^{(y^{2})}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(0,-4,-9).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2^{(x^{4})}e^{(y^{2})}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(0,-4,-9).\end{align*}$

Possible Answers:

${19.7\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z) - 9.85\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) + 27.3\cdot 2^{(x^{4})}x^{3}e^{(y^{2})}cos(z)}$ $\displaystyle {19.7\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z) - 9.85\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) + 27.3\cdot 2^{(x^{4})}x^{3}e^{(y^{2})}cos(z)}$

${0.336\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z) - 0.828\cdot 2^{(x^{4})}e^{(y^{2})}sin(z)}$ $\displaystyle {0.336\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z) - 0.828\cdot 2^{(x^{4})}e^{(y^{2})}sin(z)}$

${-54.6\cdot 2^{(x^{4})}x^{3}ye^{(y^{2})}sin(z)}$ $\displaystyle {-54.6\cdot 2^{(x^{4})}x^{3}ye^{(y^{2})}sin(z)}$

${0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)}$ $\displaystyle {0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)}$

Correct answer:

${0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)}$ $\displaystyle {0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-4)^2+(-9)^2}=9.85\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.85}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{9.85}=-0.41\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-9}{9.85}=-0.91\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2.77\cdot 2^{(x^{4})}x^{3}e^{(y^{2})}cos(z))+(-0.41)(2\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z))+(-0.91)(-1\cdot 2^{(x^{4})}e^{(y^{2})}sin(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-4)^2+(-9)^2}=9.85\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.85}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4}{9.85}=-0.41\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-9}{9.85}=-0.91\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2.77\cdot 2^{(x^{4})}x^{3}e^{(y^{2})}cos(z))+(-0.41)(2\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z))+(-0.91)(-1\cdot 2^{(x^{4})}e^{(y^{2})}sin(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.91\cdot 2^{(x^{4})}e^{(y^{2})}sin(z) - 0.82\cdot 2^{(x^{4})}ye^{(y^{2})}cos(z)\end{align*}$

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Example Question #1372 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(sin(y^{2})e^{(z^{2})})}{x}\\&\text{In the direction of }\overrightarrow{v}=(0,9,1).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(sin(y^{2})e^{(z^{2})})}{x}\\&\text{In the direction of }\overrightarrow{v}=(0,9,1).\end{align*}$

Possible Answers:

${\frac{(18.1ycos(y^{2})e^{(z^{2})})}{x}-\frac{ (9.06sin(y^{2})e^{(z^{2})})}{x^{2}}+\frac{ (18.1zsin(y^{2})e^{(z^{2})})}{x}}$ $\displaystyle {\frac{(18.1ycos(y^{2})e^{(z^{2})})}{x}-\frac{ (9.06sin(y^{2})e^{(z^{2})})}{x^{2}}+\frac{ (18.1zsin(y^{2})e^{(z^{2})})}{x}}$

${-\frac{(36.2yzcos(y^{2})e^{(z^{2})})}{x^{2}}}$ $\displaystyle {-\frac{(36.2yzcos(y^{2})e^{(z^{2})})}{x^{2}}}$

${\frac{(1.98ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.22zsin(y^{2})e^{(z^{2})})}{x}}$ $\displaystyle {\frac{(1.98ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.22zsin(y^{2})e^{(z^{2})})}{x}}$

${\frac{(1.96ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.0242zsin(y^{2})e^{(z^{2})})}{x}}$ $\displaystyle {\frac{(1.96ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.0242zsin(y^{2})e^{(z^{2})})}{x}}$

Correct answer:

${\frac{(1.98ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.22zsin(y^{2})e^{(z^{2})})}{x}}$ $\displaystyle {\frac{(1.98ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.22zsin(y^{2})e^{(z^{2})})}{x}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(1)^2}=9.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.06}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{9.06}=0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{1}{9.06}=0.11\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(sin(y^{2})e^{(z^{2})})}{x^{2}})+(0.99)(\frac{(2ycos(y^{2})e^{(z^{2})})}{x})+(0.11)(\frac{(2zsin(y^{2})e^{(z^{2})})}{x})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.98ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.22zsin(y^{2})e^{(z^{2})})}{x}\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(9)^2+(1)^2}=9.06\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{9.06}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{9.06}=0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{1}{9.06}=0.11\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(sin(y^{2})e^{(z^{2})})}{x^{2}})+(0.99)(\frac{(2ycos(y^{2})e^{(z^{2})})}{x})+(0.11)(\frac{(2zsin(y^{2})e^{(z^{2})})}{x})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.98ycos(y^{2})e^{(z^{2})})}{x}+\frac{ (0.22zsin(y^{2})e^{(z^{2})})}{x}\end{align*}$

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Example Question #1373 : Partial Derivatives

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

Possible Answers:

${0.925x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 1.39xy^{2}z^{2} - 1.07xyz^{3} + 3.2x^{2}y^{2}tan(z)}$ $\displaystyle {0.925x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 1.39xy^{2}z^{2} - 1.07xyz^{3} + 3.2x^{2}y^{2}tan(z)}$

${212xy^{2}\cdot (tan(z)^{2} + 1) - 106yz^{2}}$ $\displaystyle {212xy^{2}\cdot (tan(z)^{2} + 1) - 106yz^{2}}$

${70.8xy^{3}tan(z) - 35.4xyz^{3} - 53.1xy^{2}z^{2} - 17.7y^{2}z^{3} + 35.4x^{2}y^{3}\cdot (tan(z)^{2} + 1) + 106x^{2}y^{2}tan(z)}$ $\displaystyle {70.8xy^{3}tan(z) - 35.4xyz^{3} - 53.1xy^{2}z^{2} - 17.7y^{2}z^{3} + 35.4x^{2}y^{3}\cdot (tan(z)^{2} + 1) + 106x^{2}y^{2}tan(z)}$

${1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)}$ $\displaystyle {1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(-12)^2}=17.69\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.69}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{17.69}=-0.73\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{17.69}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4xy^{3}tan(z) - y^{2}z^{3})+(-0.73)(6x^{2}y^{2}tan(z) - 2xyz^{3})+(-0.68)(2x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 3xy^{2}z^{2})\\&D_{\overrightarrow{u}}(x,y,z)=1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(-12)^2}=17.69\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.69}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{17.69}=-0.73\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{17.69}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4xy^{3}tan(z) - y^{2}z^{3})+(-0.73)(6x^{2}y^{2}tan(z) - 2xyz^{3})+(-0.68)(2x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 3xy^{2}z^{2})\\&D_{\overrightarrow{u}}(x,y,z)=1.46xyz^{3} + 2.04xy^{2}z^{2} - 1.36x^{2}y^{3}\cdot (tan(z)^{2} + 1) - 4.38x^{2}y^{2}tan(z)\end{align*}$

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Example Question #1374 : Partial Derivatives

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

Possible Answers:

${\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)}$ $\displaystyle {\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)}$

${-\frac{(161\cdot 2^{(y^{2})}xy)}{z}}$ $\displaystyle {-\frac{(161\cdot 2^{(y^{2})}xy)}{z}}$

${\frac{(29.1sin(z^{2}))}{y}-\frac{ (58.2\cdot 2^{(y^{2})}x^{2})}{z}- 29.1\cdot 2^{(y^{2})}xln(z^{4}) + 58.2zcos(z^{2})ln(y) - 20.2\cdot 2^{(y^{2})}x^{2}yln(z^{4})}$ $\displaystyle {\frac{(29.1sin(z^{2}))}{y}-\frac{ (58.2\cdot 2^{(y^{2})}x^{2})}{z}- 29.1\cdot 2^{(y^{2})}xln(z^{4}) + 58.2zcos(z^{2})ln(y) - 20.2\cdot 2^{(y^{2})}x^{2}yln(z^{4})}$

${\frac{(56\cdot 2^{(y^{2})}x^{2})}{z}+ 8\cdot 2^{(y^{2})}xln(z^{4}) - 56zcos(z^{2})ln(y)}$ $\displaystyle {\frac{(56\cdot 2^{(y^{2})}x^{2})}{z}+ 8\cdot 2^{(y^{2})}xln(z^{4}) - 56zcos(z^{2})ln(y)}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(-14)^2}=14.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{14.56}=-0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{14.56}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.27)(-2\cdot 2^{(y^{2})}xln(z^{4}))+(0)(\frac{(2sin(z^{2}))}{y}- 1.39\cdot 2^{(y^{2})}x^{2}yln(z^{4}))+(-0.96)(4zcos(z^{2})ln(y) -\frac{ (4\cdot 2^{(y^{2})}x^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(0)^2+(-14)^2}=14.56\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{14.56}=-0.27\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{14.56}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{14.56}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.27)(-2\cdot 2^{(y^{2})}xln(z^{4}))+(0)(\frac{(2sin(z^{2}))}{y}- 1.39\cdot 2^{(y^{2})}x^{2}yln(z^{4}))+(-0.96)(4zcos(z^{2})ln(y) -\frac{ (4\cdot 2^{(y^{2})}x^{2})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.84\cdot 2^{(y^{2})}x^{2})}{z}+ 0.54\cdot 2^{(y^{2})}xln(z^{4}) - 3.84zcos(z^{2})ln(y)\end{align*}$

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Example Question #1375 : Partial Derivatives

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

Possible Answers:

${12\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 8.79\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$ $\displaystyle {12\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 8.79\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${-63.5\cdot 3^{(x^{4})}x^{3}z^{3}sin(z^{4})cos(y)}$ $\displaystyle {-63.5\cdot 3^{(x^{4})}x^{3}z^{3}sin(z^{4})cos(y)}$

${3.61\cdot 3^{(x^{4})}cos(z^{4})cos(y) - 14.4\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 15.9\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$ $\displaystyle {3.61\cdot 3^{(x^{4})}cos(z^{4})cos(y) - 14.4\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 15.9\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

${3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$ $\displaystyle {3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Correct answer:

${3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$ $\displaystyle {3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(0)^2+(-3)^2}=3.61\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{3.61}=0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{3.61}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{3.61}=-0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(4.39\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y))+(0)(3^{(x^{4})}cos(z^{4})cos(y))+(-0.83)(-4\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(2)^2+(0)^2+(-3)^2}=3.61\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{3.61}=0.55\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{3.61}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3}{3.61}=-0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.55)(4.39\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y))+(0)(3^{(x^{4})}cos(z^{4})cos(y))+(-0.83)(-4\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=3.32\cdot 3^{(x^{4})}z^{3}sin(z^{4})sin(y) + 2.42\cdot 3^{(x^{4})}x^{3}cos(z^{4})sin(y)\end{align*}$

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Example Question #1376 : Partial Derivatives

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

Possible Answers:

${\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$ $\displaystyle {\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$

${-\frac{ (97.4ln(y^{2})ln(z^{2}))}{x}-\frac{ (97.4ln(x^{2})ln(z^{2}))}{y}-\frac{ (97.4ln(x^{2})ln(y^{2}))}{z}-\frac{ (36.5\cdot 2^{(x^{3})}ln(z^{3}))}{y}-\frac{ (36.5\cdot 2^{(x^{3})}ln(y^{3}))}{z}- 25.3\cdot 2^{(x^{3})}x^{2}ln(y^{3})ln(z^{3})}$ $\displaystyle {-\frac{ (97.4ln(y^{2})ln(z^{2}))}{x}-\frac{ (97.4ln(x^{2})ln(z^{2}))}{y}-\frac{ (97.4ln(x^{2})ln(y^{2}))}{z}-\frac{ (36.5\cdot 2^{(x^{3})}ln(z^{3}))}{y}-\frac{ (36.5\cdot 2^{(x^{3})}ln(y^{3}))}{z}- 25.3\cdot 2^{(x^{3})}x^{2}ln(y^{3})ln(z^{3})}$

${-\frac{ (0.205ln(x^{2})ln(z^{2}))}{y}-\frac{ (7.84ln(x^{2})ln(y^{2}))}{z}-\frac{ (0.0768\cdot 2^{(x^{3})}ln(z^{3}))}{y}-\frac{ (2.94\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$ $\displaystyle {-\frac{ (0.205ln(x^{2})ln(z^{2}))}{y}-\frac{ (7.84ln(x^{2})ln(y^{2}))}{z}-\frac{ (0.0768\cdot 2^{(x^{3})}ln(z^{3}))}{y}-\frac{ (2.94\cdot 2^{(x^{3})}ln(y^{3}))}{z}}$

${-\frac{ 389.0}{(xyz)}-\frac{ (228\cdot 2^{(x^{3})}x^{2})}{(yz)}}$ $\displaystyle {-\frac{ 389.0}{(xyz)}-\frac{ (228\cdot 2^{(x^{3})}x^{2})}{(yz)}}$

Correct answer:

Explanation:

$\begin{align*}\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{12.17}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{12.17}=-0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{ (8ln(y^{2})ln(z^{2}))}{x}- 2.08\cdot 2^{(x^{3})}x^{2}ln(y^{3})ln(z^{3}))+(-0.16)(-\frac{ (8ln(x^{2})ln(z^{2}))}{y}-\frac{ (3\cdot 2^{(x^{3})}ln(z^{3}))}{y})+(-0.99)(-\frac{ (8ln(x^{2})ln(y^{2}))}{z}-\frac{ (3\cdot 2^{(x^{3})}ln(y^{3}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}\end{align*}$ $\displaystyle \begin{align*}\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{12.17}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-2}{12.17}=-0.16\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{12.17}=-0.99\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{ (8ln(y^{2})ln(z^{2}))}{x}- 2.08\cdot 2^{(x^{3})}x^{2}ln(y^{3})ln(z^{3}))+(-0.16)(-\frac{ (8ln(x^{2})ln(z^{2}))}{y}-\frac{ (3\cdot 2^{(x^{3})}ln(z^{3}))}{y})+(-0.99)(-\frac{ (8ln(x^{2})ln(y^{2}))}{z}-\frac{ (3\cdot 2^{(x^{3})}ln(y^{3}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.28ln(x^{2})ln(z^{2}))}{y}+\frac{ (7.92ln(x^{2})ln(y^{2}))}{z}+\frac{ (0.48\cdot 2^{(x^{3})}ln(z^{3}))}{y}+\frac{ (2.97\cdot 2^{(x^{3})}ln(y^{3}))}{z}\end{align*}$

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Example Question #1377 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ln(x^{2})e^{(z^{2})}sin(y) - 3^{(x^{2})}zsin(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,-13,19).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=ln(x^{2})e^{(z^{2})}sin(y) - 3^{(x^{2})}zsin(y^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,-13,19).\end{align*}$

Possible Answers:

${\frac{(92.1ze^{(z^{2})}cos(y))}{x}- 202\cdot 3^{(x^{2})}xy^{3}cos(y^{4})}$ $\displaystyle {\frac{(92.1ze^{(z^{2})}cos(y))}{x}- 202\cdot 3^{(x^{2})}xy^{3}cos(y^{4})}$

${0.314ln(x^{2})e^{(z^{2})}cos(y) - 0.689\cdot 3^{(x^{2})}sin(y^{4}) + 1.38zln(x^{2})e^{(z^{2})}sin(y) - 1.25\cdot 3^{(x^{2})}y^{3}zcos(y^{4})}$ $\displaystyle {0.314ln(x^{2})e^{(z^{2})}cos(y) - 0.689\cdot 3^{(x^{2})}sin(y^{4}) + 1.38zln(x^{2})e^{(z^{2})}sin(y) - 1.25\cdot 3^{(x^{2})}y^{3}zcos(y^{4})}$

${\frac{(46e^{(z^{2})}sin(y))}{x}- 23\cdot 3^{(x^{2})}sin(y^{4}) + 23ln(x^{2})e^{(z^{2})}cos(y) - 50.6\cdot 3^{(x^{2})}xzsin(y^{4}) + 46zln(x^{2})e^{(z^{2})}sin(y) - 92.1\cdot 3^{(x^{2})}y^{3}zcos(y^{4})}$ $\displaystyle {\frac{(46e^{(z^{2})}sin(y))}{x}- 23\cdot 3^{(x^{2})}sin(y^{4}) + 23ln(x^{2})e^{(z^{2})}cos(y) - 50.6\cdot 3^{(x^{2})}xzsin(y^{4}) + 46zln(x^{2})e^{(z^{2})}sin(y) - 92.1\cdot 3^{(x^{2})}y^{3}zcos(y^{4})}$

${1.66zln(x^{2})e^{(z^{2})}sin(y) - 0.56ln(x^{2})e^{(z^{2})}cos(y) - 0.83\cdot 3^{(x^{2})}sin(y^{4}) + 2.24\cdot 3^{(x^{2})}y^{3}zcos(y^{4})}$ $\displaystyle {1.66zln(x^{2})e^{(z^{2})}sin(y) - 0.56ln(x^{2})e^{(z^{2})}cos(y) - 0.83\cdot 3^{(x^{2})}sin(y^{4}) + 2.24\cdot 3^{(x^{2})}y^{3}zcos(y^{4})}$

Correct answer:

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(19)^2}=23.02\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{23.02}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{23.02}=-0.56\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{23.02}=0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(2e^{(z^{2})}sin(y))}{x}- 2.2\cdot 3^{(x^{2})}xzsin(y^{4}))+(-0.56)(ln(x^{2})e^{(z^{2})}cos(y) - 4\cdot 3^{(x^{2})}y^{3}zcos(y^{4}))+(0.83)(2zln(x^{2})e^{(z^{2})}sin(y) - 1\cdot 3^{(x^{2})}sin(y^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.66zln(x^{2})e^{(z^{2})}sin(y) - 0.56ln(x^{2})e^{(z^{2})}cos(y) - 0.83\cdot 3^{(x^{2})}sin(y^{4}) + 2.24\cdot 3^{(x^{2})}y^{3}zcos(y^{4})\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(19)^2}=23.02\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{23.02}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{23.02}=-0.56\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19}{23.02}=0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(2e^{(z^{2})}sin(y))}{x}- 2.2\cdot 3^{(x^{2})}xzsin(y^{4}))+(-0.56)(ln(x^{2})e^{(z^{2})}cos(y) - 4\cdot 3^{(x^{2})}y^{3}zcos(y^{4}))+(0.83)(2zln(x^{2})e^{(z^{2})}sin(y) - 1\cdot 3^{(x^{2})}sin(y^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.66zln(x^{2})e^{(z^{2})}sin(y) - 0.56ln(x^{2})e^{(z^{2})}cos(y) - 0.83\cdot 3^{(x^{2})}sin(y^{4}) + 2.24\cdot 3^{(x^{2})}y^{3}zcos(y^{4})\end{align*}$

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Example Question #311 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(5cos(y))}{x}+ x^{4}z^{4}cos(y^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-18,18).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(5cos(y))}{x}+ x^{4}z^{4}cos(y^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-18,18).\end{align*}$

Possible Answers:

${-815x^{3}yz^{3}sin(y^{2})}$ $\displaystyle {-815x^{3}yz^{3}sin(y^{2})}$

${2.02x^{4}z^{3}cos(y^{2}) -\frac{ (2.52sin(y))}{x}- 1.01x^{4}yz^{4}sin(y^{2})}$ $\displaystyle {2.02x^{4}z^{3}cos(y^{2}) -\frac{ (2.52sin(y))}{x}- 1.01x^{4}yz^{4}sin(y^{2})}$

${\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})}$ $\displaystyle {\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})}$

${102x^{3}z^{4}cos(y^{2}) -\frac{ (127sin(y))}{x}-\frac{ (127cos(y))}{x^{2}}+ 102x^{4}z^{3}cos(y^{2}) - 50.9x^{4}yz^{4}sin(y^{2})}$ $\displaystyle {102x^{3}z^{4}cos(y^{2}) -\frac{ (127sin(y))}{x}-\frac{ (127cos(y))}{x^{2}}+ 102x^{4}z^{3}cos(y^{2}) - 50.9x^{4}yz^{4}sin(y^{2})}$

Correct answer:

${\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})}$ $\displaystyle {\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(18)^2}=25.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{25.46}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{25.46}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{25.46}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}z^{4}cos(y^{2}) -\frac{ (5cos(y))}{x^{2}})+(-0.71)(-\frac{ (5sin(y))}{x}- 2x^{4}yz^{4}sin(y^{2}))+(0.71)(4x^{4}z^{3}cos(y^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(18)^2}=25.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{25.46}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{25.46}=-0.71\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{18}{25.46}=0.71\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}z^{4}cos(y^{2}) -\frac{ (5cos(y))}{x^{2}})+(-0.71)(-\frac{ (5sin(y))}{x}- 2x^{4}yz^{4}sin(y^{2}))+(0.71)(4x^{4}z^{3}cos(y^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.55sin(y))}{x}+ 2.84x^{4}z^{3}cos(y^{2}) + 1.42x^{4}yz^{4}sin(y^{2})\end{align*}$

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Example Question #312 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(y^{2})}zln(x^{2})\\&\text{In the direction of }\overrightarrow{v}=(-4,20,0).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-2^{(y^{2})}zln(x^{2})\\&\text{In the direction of }\overrightarrow{v}=(-4,20,0).\end{align*}$

Possible Answers:

${-\frac{(56.6\cdot 2^{(y^{2})}y)}{x}}$ $\displaystyle {-\frac{(56.6\cdot 2^{(y^{2})}y)}{x}}$

${- 20.4\cdot 2^{(y^{2})}ln(x^{2}) -\frac{ (40.8\cdot 2^{(y^{2})}z)}{x}- 28.3\cdot 2^{(y^{2})}yzln(x^{2})}$ $\displaystyle {- 20.4\cdot 2^{(y^{2})}ln(x^{2}) -\frac{ (40.8\cdot 2^{(y^{2})}z)}{x}- 28.3\cdot 2^{(y^{2})}yzln(x^{2})}$

${\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})}$ $\displaystyle {\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})}$

${-\frac{ (0.08\cdot 2^{(y^{2})}z)}{x}- 1.33\cdot 2^{(y^{2})}yzln(x^{2})}$ $\displaystyle {-\frac{ (0.08\cdot 2^{(y^{2})}z)}{x}- 1.33\cdot 2^{(y^{2})}yzln(x^{2})}$

Correct answer:

${\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})}$ $\displaystyle {\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(20)^2+(0)^2}=20.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{20.4}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{20.4}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.4}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(-\frac{(2\cdot 2^{(y^{2})}z)}{x})+(0.98)(-1.39\cdot 2^{(y^{2})}yzln(x^{2}))+(0)(-1\cdot 2^{(y^{2})}ln(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(20)^2+(0)^2}=20.4\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{20.4}=-0.2\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{20.4}=0.98\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.4}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.2)(-\frac{(2\cdot 2^{(y^{2})}z)}{x})+(0.98)(-1.39\cdot 2^{(y^{2})}yzln(x^{2}))+(0)(-1\cdot 2^{(y^{2})}ln(x^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.4\cdot 2^{(y^{2})}z)}{x}- 1.36\cdot 2^{(y^{2})}yzln(x^{2})\end{align*}$

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Example Question #313 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(z^{4})}x^{4}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(0,-17,9).\end{align*}$ $\displaystyle \begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=3^{(z^{4})}x^{4}sin(y)\\&\text{In the direction of }\overrightarrow{v}=(0,-17,9).\end{align*}$

Possible Answers:

${19.2\cdot 3^{(z^{4})}x^{4}cos(y) + 77\cdot 3^{(z^{4})}x^{3}sin(y) + 84.5\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$ $\displaystyle {19.2\cdot 3^{(z^{4})}x^{4}cos(y) + 77\cdot 3^{(z^{4})}x^{3}sin(y) + 84.5\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${0.774\cdot 3^{(z^{4})}x^{4}cos(y) + 0.971\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$ $\displaystyle {0.774\cdot 3^{(z^{4})}x^{4}cos(y) + 0.971\cdot 3^{(z^{4})}x^{4}z^{3}sin(y)}$

${338\cdot 3^{(z^{4})}x^{3}z^{3}cos(y)}$ $\displaystyle {338\cdot 3^{(z^{4})}x^{3}z^{3}cos(y)}$

${2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)}$ $\displaystyle {2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)}$

Correct answer:

${2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)}$ $\displaystyle {2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-17)^2+(9)^2}=19.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.24}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{19.24}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{19.24}=0.47\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4\cdot 3^{(z^{4})}x^{3}sin(y))+(-0.88)(3^{(z^{4})}x^{4}cos(y))+(0.47)(4.39\cdot 3^{(z^{4})}x^{4}z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)\end{align*}$ $\displaystyle \begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-17)^2+(9)^2}=19.24\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{19.24}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{19.24}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{9}{19.24}=0.47\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4\cdot 3^{(z^{4})}x^{3}sin(y))+(-0.88)(3^{(z^{4})}x^{4}cos(y))+(0.47)(4.39\cdot 3^{(z^{4})}x^{4}z^{3}sin(y))\\&D_{\overrightarrow{u}}(x,y,z)=2.07\cdot 3^{(z^{4})}x^{4}z^{3}sin(y) - 0.88\cdot 3^{(z^{4})}x^{4}cos(y)\end{align*}$

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Calculus 3 : Partial Derivatives

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