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Calculus 3 : Directional Derivatives

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

Example Questions

Example Question #1318 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5z^{4}ln(x^{2})e^{(y^{2})} - x^{2}cos(z^{2})ln(y^{2})\\&\text{In the direction of }\overrightarrow{v}=(0,-15,-4).\end{align*}$

Possible Answers:

${\frac{(1.94x^{2}cos(z^{2}))}{y}- 5.2z^{3}ln(x^{2})e^{(y^{2})} - 9.7yz^{4}ln(x^{2})e^{(y^{2})} - 0.52x^{2}zln(y^{2})sin(z^{2})}$

${\frac{(124xzsin(z^{2}))}{y}+\frac{ (1244yz^{3}e^{(y^{2})})}{x}}$

${\frac{(155z^{4}e^{(y^{2})})}{x}-\frac{ (31x^{2}cos(z^{2}))}{y}- 31xcos(z^{2})ln(y^{2}) + 310z^{3}ln(x^{2})e^{(y^{2})} + 155yz^{4}ln(x^{2})e^{(y^{2})} + 31x^{2}zln(y^{2})sin(z^{2})}$

${1.35z^{3}ln(x^{2})e^{(y^{2})} -\frac{ (1.88x^{2}cos(z^{2}))}{y}+ 9.41yz^{4}ln(x^{2})e^{(y^{2})} + 0.135x^{2}zln(y^{2})sin(z^{2})}$

Correct answer:

${\frac{(1.94x^{2}cos(z^{2}))}{y}- 5.2z^{3}ln(x^{2})e^{(y^{2})} - 9.7yz^{4}ln(x^{2})e^{(y^{2})} - 0.52x^{2}zln(y^{2})sin(z^{2})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-15)^2+(-4)^2}=15.52\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{15.52}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{15.52}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-4}{15.52}=-0.26\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(\frac{(10z^{4}e^{(y^{2})})}{x}- 2xcos(z^{2})ln(y^{2}))+(-0.97)(10yz^{4}ln(x^{2})e^{(y^{2})} -\frac{ (2x^{2}cos(z^{2}))}{y})+(-0.26)(20z^{3}ln(x^{2})e^{(y^{2})} + 2x^{2}zln(y^{2})sin(z^{2}))\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.94x^{2}cos(z^{2}))}{y}- 5.2z^{3}ln(x^{2})e^{(y^{2})} - 9.7yz^{4}ln(x^{2})e^{(y^{2})} - 0.52x^{2}zln(y^{2})sin(z^{2})\end{align*}$

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Example Question #1319 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{4}sin(y^{3})sin(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(18,9,0).\end{align*}$

Possible Answers:

${966x^{3}y^{2}z^{3}cos(y^{3})cos(z^{4})}$

${72x^{3}sin(y^{3})sin(z^{4}) + 27x^{4}y^{2}cos(y^{3})sin(z^{4})}$

${3.56x^{3}sin(y^{3})sin(z^{4}) + 1.35x^{4}y^{2}cos(y^{3})sin(z^{4})}$

${80.5x^{3}sin(y^{3})sin(z^{4}) + 60.4x^{4}y^{2}cos(y^{3})sin(z^{4}) + 80.5x^{4}z^{3}cos(z^{4})sin(y^{3})}$

Correct answer:

${3.56x^{3}sin(y^{3})sin(z^{4}) + 1.35x^{4}y^{2}cos(y^{3})sin(z^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(9)^2+(0)^2}=20.12\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{20.12}=0.89\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{9}{20.12}=0.45\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{20.12}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.89)(4x^{3}sin(y^{3})sin(z^{4}))+(0.45)(3x^{4}y^{2}cos(y^{3})sin(z^{4}))+(0)(4x^{4}z^{3}cos(z^{4})sin(y^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=3.56x^{3}sin(y^{3})sin(z^{4}) + 1.35x^{4}y^{2}cos(y^{3})sin(z^{4})\end{align*}$

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Example Question #1320 : Partial Derivatives

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

Possible Answers:

${- 21.7y^{2}zsin(y^{3})e^{(z^{2})} -\frac{ (28.9xycos(x^{2})cos(y^{2}))}{z^{3}}}$

${3cos(y^{3})e^{(z^{2})} -\frac{ (4sin(x^{2})sin(y^{2}))}{z^{3}}+\frac{ (6xcos(x^{2})sin(y^{2}))}{z^{2}}+ 4xzcos(y^{3})e^{(z^{2})}}$

${0.83cos(y^{3})e^{(z^{2})} -\frac{ (1.1sin(x^{2})sin(y^{2}))}{z^{3}}+\frac{ (1.66xcos(x^{2})sin(y^{2}))}{z^{2}}+ 1.1xzcos(y^{3})e^{(z^{2})}}$

${3.61cos(y^{3})e^{(z^{2})} -\frac{ (7.22sin(x^{2})sin(y^{2}))}{z^{3}}+\frac{ (7.22xcos(x^{2})sin(y^{2}))}{z^{2}}+\frac{ (7.22ycos(y^{2})sin(x^{2}))}{z^{2}}- 10.8xy^{2}sin(y^{3})e^{(z^{2})} + 7.22xzcos(y^{3})e^{(z^{2})}}$

Correct answer:

${0.83cos(y^{3})e^{(z^{2})} -\frac{ (1.1sin(x^{2})sin(y^{2}))}{z^{3}}+\frac{ (1.66xcos(x^{2})sin(y^{2}))}{z^{2}}+ 1.1xzcos(y^{3})e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(3)^2+(0)^2+(2)^2}=3.61\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{3}{3.61}=0.83\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{3.61}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2}{3.61}=0.55\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.83)(cos(y^{3})e^{(z^{2})} +\frac{ (2xcos(x^{2})sin(y^{2}))}{z^{2}})+(0)(\frac{(2ycos(y^{2})sin(x^{2}))}{z^{2}}- 3xy^{2}sin(y^{3})e^{(z^{2})})+(0.55)(2xzcos(y^{3})e^{(z^{2})} -\frac{ (2sin(x^{2})sin(y^{2}))}{z^{3}})\\&D_{\overrightarrow{u}}(x,y,z)=0.83cos(y^{3})e^{(z^{2})} -\frac{ (1.1sin(x^{2})sin(y^{2}))}{z^{3}}+\frac{ (1.66xcos(x^{2})sin(y^{2}))}{z^{2}}+ 1.1xzcos(y^{3})e^{(z^{2})}\end{align*}$

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Example Question #1321 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=2e^{(z^{2})}e^{(y)} +\frac{ (cos(z)ln(x))}{y}\\&\text{In the direction of }\overrightarrow{v}=(-13,0,16).\end{align*}$

Possible Answers:

${\frac{(0.397cos(z))}{(xy)}-\frac{ (0.608ln(x)sin(z))}{y}+ 2.43ze^{(z^{2})}e^{(y)}}$

${3.12ze^{(z^{2})}e^{(y)} -\frac{ (0.63cos(z))}{(xy)}-\frac{ (0.78ln(x)sin(z))}{y}}$

${41.2e^{(z^{2})}e^{(y)} -\frac{ (20.6ln(x)sin(z))}{y}+\frac{ (20.6cos(z))}{(xy)}+ 82.5ze^{(z^{2})}e^{(y)} -\frac{ (20.6cos(z)ln(x))}{y^{2}}}$

${\frac{(20.6sin(z))}{(xy^{2})}}$

Correct answer:

${3.12ze^{(z^{2})}e^{(y)} -\frac{ (0.63cos(z))}{(xy)}-\frac{ (0.78ln(x)sin(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-13)^2+(0)^2+(16)^2}=20.62\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-13}{20.62}=-0.63\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{20.62}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{16}{20.62}=0.78\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.63)(\frac{cos(z)}{(xy)})+(0)(2e^{(z^{2})}e^{(y)} -\frac{ (cos(z)ln(x))}{y^{2}})+(0.78)(4ze^{(z^{2})}e^{(y)} -\frac{ (ln(x)sin(z))}{y})\\&D_{\overrightarrow{u}}(x,y,z)=3.12ze^{(z^{2})}e^{(y)} -\frac{ (0.63cos(z))}{(xy)}-\frac{ (0.78ln(x)sin(z))}{y}\end{align*}$

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Example Question #1322 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(e^{(x)}ln(z))}{y}- 5e^{(x^{2})}e^{(y)}cos(z)\\&\text{In the direction of }\overrightarrow{v}=(-4,-15,0).\end{align*}$

Possible Answers:

${75e^{(x^{2})}e^{(y)}cos(z) -\frac{ (4e^{(x)}ln(z))}{y}+\frac{ (15e^{(x)}ln(z))}{y^{2}}+ 40xe^{(x^{2})}e^{(y)}cos(z)}$

${155xe^{(x^{2})}e^{(y)}sin(z) -\frac{ (15.5e^{(x)})}{(y^{2}z)}}$

${77.6e^{(x^{2})}e^{(y)}sin(z) - 77.6e^{(x^{2})}e^{(y)}cos(z) +\frac{ (15.5e^{(x)})}{(yz)}+\frac{ (15.5e^{(x)}ln(z))}{y}-\frac{ (15.5e^{(x)}ln(z))}{y^{2}}- 155xe^{(x^{2})}e^{(y)}cos(z)}$

${4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)}$

Correct answer:

${4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-4)^2+(-15)^2+(0)^2}=15.52\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-4}{15.52}=-0.26\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-15}{15.52}=-0.97\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{15.52}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.26)(\frac{(e^{(x)}ln(z))}{y}- 10xe^{(x^{2})}e^{(y)}cos(z))+(-0.97)(- 5e^{(x^{2})}e^{(y)}cos(z) -\frac{ (e^{(x)}ln(z))}{y^{2}})+(0)(5e^{(x^{2})}e^{(y)}sin(z) +\frac{ e^{(x)}}{(yz)})\\&D_{\overrightarrow{u}}(x,y,z)=4.85e^{(x^{2})}e^{(y)}cos(z) -\frac{ (0.26e^{(x)}ln(z))}{y}+\frac{ (0.97e^{(x)}ln(z))}{y^{2}}+ 2.6xe^{(x^{2})}e^{(y)}cos(z)\end{align*}$

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Example Question #1323 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(4^{(x^{3})}e^{(z^{4})})}{y^{2}}\\&\text{In the direction of }\overrightarrow{v}=(-7,16,0).\end{align*}$

Possible Answers:

${\frac{(72.6\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}-\frac{ (34.9\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}+\frac{ (69.8\cdot 4^{(x^{3})}z^{3}e^{(z^{4})})}{y^{2}}}$

${-\frac{ (1.84\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}-\frac{ (1.66\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}}$

${\frac{(0.665\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}-\frac{ (1.69\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}}$

${-\frac{(581\cdot 4^{(x^{3})}x^{2}z^{3}e^{(z^{4})})}{y^{3}}}$

Correct answer:

${-\frac{ (1.84\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}-\frac{ (1.66\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(16)^2+(0)^2}=17.46\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{17.46}=-0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{16}{17.46}=0.92\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.46}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.4)(\frac{(4.16\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}})+(0.92)(-\frac{(2\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}})+(0)(\frac{(4\cdot 4^{(x^{3})}z^{3}e^{(z^{4})})}{y^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=-\frac{ (1.84\cdot 4^{(x^{3})}e^{(z^{4})})}{y^{3}}-\frac{ (1.66\cdot 4^{(x^{3})}x^{2}e^{(z^{4})})}{y^{2}}\end{align*}$

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Example Question #1324 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=cos(y^{2})e^{(z^{2})}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(-7,-13,0).\end{align*}$

Possible Answers:

${- 0.221cos(y^{2})e^{(z^{2})}sin(x) - 1.55ysin(y^{2})e^{(z^{2})}cos(x)}$

${0.47cos(y^{2})e^{(z^{2})}sin(x) + 1.76ysin(y^{2})e^{(z^{2})}cos(x)}$

${29.5zcos(y^{2})e^{(z^{2})}cos(x) - 14.8cos(y^{2})e^{(z^{2})}sin(x) - 29.5ysin(y^{2})e^{(z^{2})}cos(x)}$

${59yzsin(y^{2})e^{(z^{2})}sin(x)}$

Correct answer:

${0.47cos(y^{2})e^{(z^{2})}sin(x) + 1.76ysin(y^{2})e^{(z^{2})}cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-7)^2+(-13)^2+(0)^2}=14.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-7}{14.76}=-0.47\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{14.76}=-0.88\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{14.76}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.47)(-1cos(y^{2})e^{(z^{2})}sin(x))+(-0.88)(-2ysin(y^{2})e^{(z^{2})}cos(x))+(0)(2zcos(y^{2})e^{(z^{2})}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.47cos(y^{2})e^{(z^{2})}sin(x) + 1.76ysin(y^{2})e^{(z^{2})}cos(x)\end{align*}$

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Example Question #1325 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 3^ze^{(x^{2})}e^{(y)} - 2xy^{2}tan(z)\\&\text{In the direction of }\overrightarrow{v}=(0,-18,-2).\end{align*}$

Possible Answers:

${- 72.4y\cdot (tan(z)^{2} + 1) - 39.8\cdot 3^zxe^{(x^{2})}e^{(y)}}$

${- 36.2y^{2}tan(z) - 72.4xytan(z) - 36.2xy^{2}\cdot (tan(z)^{2} + 1) - 38\cdot 3^ze^{(x^{2})}e^{(y)} - 36.2\cdot 3^zxe^{(x^{2})}e^{(y)}}$

${- 3.92xytan(z) - 0.0242xy^{2}\cdot (tan(z)^{2} + 1) - 0.993\cdot 3^ze^{(x^{2})}e^{(y)}}$

${3.96xytan(z) + 0.22xy^{2}\cdot (tan(z)^{2} + 1) + 1.11\cdot 3^ze^{(x^{2})}e^{(y)}}$

Correct answer:

${3.96xytan(z) + 0.22xy^{2}\cdot (tan(z)^{2} + 1) + 1.11\cdot 3^ze^{(x^{2})}e^{(y)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-18)^2+(-2)^2}=18.11\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{18.11}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-18}{18.11}=-0.99\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-2}{18.11}=-0.11\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(- 2y^{2}tan(z) - 2\cdot 3^zxe^{(x^{2})}e^{(y)})+(-0.99)(- 4xytan(z) - 1\cdot 3^ze^{(x^{2})}e^{(y)})+(-0.11)(- 2xy^{2}\cdot (tan(z)^{2} + 1) - 1.1\cdot 3^ze^{(x^{2})}e^{(y)})\\&D_{\overrightarrow{u}}(x,y,z)=3.96xytan(z) + 0.22xy^{2}\cdot (tan(z)^{2} + 1) + 1.11\cdot 3^ze^{(x^{2})}e^{(y)}\end{align*}$

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Example Question #1326 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-cos(x^{2})ln(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,5,-17).\end{align*}$

Possible Answers:

${\frac{(3.84cos(x^{2}))}{z}}$

${-\frac{(3.69cos(x^{2}))}{z}}$

${35.4xln(z^{4})sin(x^{2}) -\frac{ (70.9cos(x^{2}))}{z}}$

Correct answer:

${\frac{(3.84cos(x^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(5)^2+(-17)^2}=17.72\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{17.72}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{5}{17.72}=0.28\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-17}{17.72}=-0.96\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(2xln(z^{4})sin(x^{2}))+(0.28)(0.0)+(-0.96)(-\frac{(4cos(x^{2}))}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(3.84cos(x^{2}))}{z}\end{align*}$

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Example Question #1327 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=sin(y^{3})e^{(x^{4})}sin(z)\\&\text{In the direction of }\overrightarrow{v}=(0,8,12).\end{align*}$

Possible Answers:

${14.4sin(y^{3})e^{(x^{4})}cos(z) + 43.3y^{2}cos(y^{3})e^{(x^{4})}sin(z) + 57.7x^{3}sin(y^{3})e^{(x^{4})}sin(z)}$

${173x^{3}y^{2}cos(y^{3})e^{(x^{4})}cos(z)}$

${0.689sin(y^{3})e^{(x^{4})}cos(z) + 0.908y^{2}cos(y^{3})e^{(x^{4})}sin(z)}$

${0.83sin(y^{3})e^{(x^{4})}cos(z) + 1.65y^{2}cos(y^{3})e^{(x^{4})}sin(z)}$

Correct answer:

${0.83sin(y^{3})e^{(x^{4})}cos(z) + 1.65y^{2}cos(y^{3})e^{(x^{4})}sin(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(8)^2+(12)^2}=14.42\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{14.42}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{8}{14.42}=0.55\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{12}{14.42}=0.83\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(4x^{3}sin(y^{3})e^{(x^{4})}sin(z))+(0.55)(3y^{2}cos(y^{3})e^{(x^{4})}sin(z))+(0.83)(sin(y^{3})e^{(x^{4})}cos(z))\\&D_{\overrightarrow{u}}(x,y,z)=0.83sin(y^{3})e^{(x^{4})}cos(z) + 1.65y^{2}cos(y^{3})e^{(x^{4})}sin(z)\end{align*}$

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Calculus 3 : Directional Derivatives

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