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Calculus 3 : Partial Derivatives

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

Example Questions

Example Question #1291 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-4^{(y^{3})}e^{(x^{2})}\\&\text{In the direction of }\overrightarrow{v}=(0,10,-20).\end{align*}$

Possible Answers:

${-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

${- 44.7\cdot 4^{(y^{3})}xe^{(x^{2})} - 93\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

${-0.842\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

Correct answer:

${-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(-20)^2}=22.36\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{22.36}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{22.36}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{22.36}=-0.89\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-2\cdot 4^{(y^{3})}xe^{(x^{2})})+(0.45)(-4.16\cdot 4^{(y^{3})}y^{2}e^{(x^{2})})+(-0.89)(0.0)\\&D_{\overrightarrow{u}}(x,y,z)=-1.87\cdot 4^{(y^{3})}y^{2}e^{(x^{2})}\end{align*}$

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Example Question #1292 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=e^{(y^{3})}e^{(x)}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(0,10,-16).\end{align*}$

Possible Answers:

${18.9e^{(y^{3})}e^{(x)}cos(z) + 18.9e^{(y^{3})}e^{(x)}sin(z) + 56.6y^{2}e^{(y^{3})}e^{(x)}sin(z)}$

${1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)}$

${0.723e^{(y^{3})}e^{(x)}cos(z) + 0.843y^{2}e^{(y^{3})}e^{(x)}sin(z)}$

${56.6y^{2}e^{(y^{3})}e^{(x)}cos(z)}$

Correct answer:

${1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(10)^2+(-16)^2}=18.87\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.87}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{10}{18.87}=0.53\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-16}{18.87}=-0.85\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(e^{(y^{3})}e^{(x)}sin(z))+(0.53)(3y^{2}e^{(y^{3})}e^{(x)}sin(z))+(-0.85)(e^{(y^{3})}e^{(x)}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=1.59y^{2}e^{(y^{3})}e^{(x)}sin(z) - 0.85e^{(y^{3})}e^{(x)}cos(z)\end{align*}$

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Example Question #1293 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=4zln(y)tan(x) -\frac{ e^{(x)}}{(yz)}\\&\text{In the direction of }\overrightarrow{v}=(0,-16,6).\end{align*}$

Possible Answers:

${1.4ln(y)tan(x) -\frac{ (3.76ztan(x))}{y}+\frac{ (0.35e^{(x)})}{(yz^{2})}-\frac{ (0.94e^{(x)})}{(y^{2}z)}}$

${68.4ln(y)tan(x) +\frac{ (68.4ztan(x))}{y}+ 68.4zln(y)\cdot (tan(x)^{2} + 1) -\frac{ (17.1e^{(x)})}{(yz)}+\frac{ (17.1e^{(x)})}{(yz^{2})}+\frac{ (17.1e^{(x)})}{(y^{2}z)}}$

${0.49ln(y)tan(x) +\frac{ (3.53ztan(x))}{y}+\frac{ (0.122e^{(x)})}{(yz^{2})}+\frac{ (0.884e^{(x)})}{(y^{2}z)}}$

${\frac{(68.4\cdot (tan(x)^{2} + 1))}{y}-\frac{ (17.1e^{(x)})}{(y^{2}z^{2})}}$

Correct answer:

${1.4ln(y)tan(x) -\frac{ (3.76ztan(x))}{y}+\frac{ (0.35e^{(x)})}{(yz^{2})}-\frac{ (0.94e^{(x)})}{(y^{2}z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-16)^2+(6)^2}=17.09\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.09}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{17.09}=-0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{6}{17.09}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4zln(y)\cdot (tan(x)^{2} + 1) -\frac{ (e^{(x)})}{(yz)})+(-0.94)(\frac{(4ztan(x))}{y}+\frac{ e^{(x)}}{(y^{2}z)})+(0.35)(4ln(y)tan(x) +\frac{ e^{(x)}}{(yz^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=1.4ln(y)tan(x) -\frac{ (3.76ztan(x))}{y}+\frac{ (0.35e^{(x)})}{(yz^{2})}-\frac{ (0.94e^{(x)})}{(y^{2}z)}\end{align*}$

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Example Question #1294 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}ln(z^{4})e^{(y)}\\&\text{In the direction of }\overrightarrow{v}=(0,-5,-12).\end{align*}$

Possible Answers:

${0.144x^{4}ln(z^{4})e^{(y)} +\frac{ (3.39x^{4}e^{(y)})}{z}}$

${52x^{3}ln(z^{4})e^{(y)} + 13x^{4}ln(z^{4})e^{(y)} +\frac{ (52x^{4}e^{(y)})}{z}}$

${- 0.38x^{4}ln(z^{4})e^{(y)} -\frac{ (3.68x^{4}e^{(y)})}{z}}$

${\frac{(208x^{3}e^{(y)})}{z}}$

Correct answer:

${- 0.38x^{4}ln(z^{4})e^{(y)} -\frac{ (3.68x^{4}e^{(y)})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-5)^2+(-12)^2}=13\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-5}{13}=-0.38\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-12}{13}=-0.92\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}ln(z^{4})e^{(y)})+(-0.38)(x^{4}ln(z^{4})e^{(y)})+(-0.92)(\frac{(4x^{4}e^{(y)})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=- 0.38x^{4}ln(z^{4})e^{(y)} -\frac{ (3.68x^{4}e^{(y)})}{z}\end{align*}$

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Example Question #1295 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 4^{(y^{3})}x^{4}cos(z^{3}) - x^{4}y^{3}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(-4,-16,0).\end{align*}$

Possible Answers:

${198x^{3}y^{2}sin(z) + 823\cdot 4^{(y^{3})}x^{3}y^{2}z^{2}sin(z^{3})}$

${0.96\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 0.96x^{3}y^{3}cos(z) + 2.91x^{4}y^{2}cos(z) + 4.03\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})}$

${16.5x^{4}y^{3}sin(z) - 66x^{3}y^{3}cos(z) - 49.5x^{4}y^{2}cos(z) - 66\cdot 4^{(y^{3})}x^{3}cos(z^{3}) - 68.6\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3}) + 49.5\cdot 4^{(y^{3})}x^{4}z^{2}sin(z^{3})}$

${16\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 16x^{3}y^{3}cos(z) + 48x^{4}y^{2}cos(z) + 66.5\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})}$

Correct answer:

${0.96\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 0.96x^{3}y^{3}cos(z) + 2.91x^{4}y^{2}cos(z) + 4.03\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-16)^2+(0)^2}=16.49\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{16.49}=-0.24\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16}{16.49}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{16.49}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.24)(- 4\cdot 4^{(y^{3})}x^{3}cos(z^{3}) - 4x^{3}y^{3}cos(z))+(-0.97)(- 3x^{4}y^{2}cos(z) - 4.16\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3}))+(0)(x^{4}y^{3}sin(z) + 3\cdot 4^{(y^{3})}x^{4}z^{2}sin(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=0.96\cdot 4^{(y^{3})}x^{3}cos(z^{3}) + 0.96x^{3}y^{3}cos(z) + 2.91x^{4}y^{2}cos(z) + 4.03\cdot 4^{(y^{3})}x^{4}y^{2}cos(z^{3})\end{align*}$

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Example Question #1296 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1297 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1298 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1299 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #1300 : Partial Derivatives

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Calculus 3 : Partial Derivatives

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