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Calculus 3 : Calculus 3

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

Example Questions

Example Question #1283 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=x^{3}cos(y^{4})ln(z) - 4ze^{(y^{2})}\cdot tan(x)\\&\text{In the direction of }\overrightarrow{v}=(-14,0,-20).\end{align*}$

Possible Answers:

${- 195ye^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (293x^{2}y^{3}sin(y^{4}))}{z}}$

${3.28e^{(y^{2})}\cdot tan(x) + 2.28ze^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (0.82x^{3}cos(y^{4}))}{z}- 1.71x^{2}cos(y^{4})ln(z)}$

${\frac{(0.672x^{3}cos(y^{4}))}{z}- 1.3ze^{(y^{2})}\cdot (tan(x)^{2} + 1) - 2.69e^{(y^{2})}\cdot tan(x) + 0.975x^{2}cos(y^{4})ln(z)}$

${\frac{(24.4x^{3}cos(y^{4}))}{z}- 97.6ze^{(y^{2})}\cdot (tan(x)^{2} + 1) - 97.6e^{(y^{2})}\cdot tan(x) + 73.2x^{2}cos(y^{4})ln(z) - 97.6x^{3}y^{3}sin(y^{4})ln(z) - 195yze^{(y^{2})}\cdot tan(x)}$

Correct answer:

${3.28e^{(y^{2})}\cdot tan(x) + 2.28ze^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (0.82x^{3}cos(y^{4}))}{z}- 1.71x^{2}cos(y^{4})ln(z)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14)^2+(0)^2+(-20)^2}=24.41\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14}{24.41}=-0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{24.41}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-20}{24.41}=-0.82\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.57)(3x^{2}cos(y^{4})ln(z) - 4ze^{(y^{2})}\cdot (tan(x)^{2} + 1))+(0)(- 4x^{3}y^{3}sin(y^{4})ln(z) - 8yze^{(y^{2})}\cdot tan(x))+(-0.82)(\frac{(x^{3}cos(y^{4}))}{z}- 4e^{(y^{2})}\cdot tan(x))\\&D_{\overrightarrow{u}}(x,y,z)=3.28e^{(y^{2})}\cdot tan(x) + 2.28ze^{(y^{2})}\cdot (tan(x)^{2} + 1) -\frac{ (0.82x^{3}cos(y^{4}))}{z}- 1.71x^{2}cos(y^{4})ln(z)\end{align*}$

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Example Question #1284 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-y^{4}cos(x^{3})e^{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-1,-17,0).\end{align*}$

Possible Answers:

${68y^{3}cos(x^{3})e^{(z)} - 3x^{2}y^{4}sin(x^{3})e^{(z)}}$

${51.1x^{2}y^{4}sin(x^{3})e^{(z)} - 17y^{4}cos(x^{3})e^{(z)} - 68.1y^{3}cos(x^{3})e^{(z)}}$

${204x^{2}y^{3}sin(x^{3})e^{(z)}}$

${4y^{3}cos(x^{3})e^{(z)} - 0.18x^{2}y^{4}sin(x^{3})e^{(z)}}$

Correct answer:

${4y^{3}cos(x^{3})e^{(z)} - 0.18x^{2}y^{4}sin(x^{3})e^{(z)}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-1)^2+(-17)^2+(0)^2}=17.03\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-1}{17.03}=-0.06\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-17}{17.03}=-1\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{17.03}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.06)(3x^{2}y^{4}sin(x^{3})e^{(z)})+(-1)(-4y^{3}cos(x^{3})e^{(z)})+(0)(-1y^{4}cos(x^{3})e^{(z)})\\&D_{\overrightarrow{u}}(x,y,z)=4y^{3}cos(x^{3})e^{(z)} - 0.18x^{2}y^{4}sin(x^{3})e^{(z)}\end{align*}$

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Example Question #1285 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{4}sin(z^{3}) + y^{4}z^{3}cos(x)\\&\text{In the direction of }\overrightarrow{v}=(-12,-20,0).\end{align*}$

Possible Answers:

${-280y^{3}z^{2}sin(x)}$

${93.3y^{3}sin(z^{3}) + 70y^{4}z^{2}cos(z^{3}) + 93.3y^{3}z^{3}cos(x) + 70y^{4}z^{2}cos(x) - 23.3y^{4}z^{3}sin(x)}$

${2.96y^{3}sin(z^{3}) + 2.96y^{3}z^{3}cos(x) - 0.26y^{4}z^{3}sin(x)}$

${0.51y^{4}z^{3}sin(x) - 3.44y^{3}z^{3}cos(x) - 3.44y^{3}sin(z^{3})}$

Correct answer:

${0.51y^{4}z^{3}sin(x) - 3.44y^{3}z^{3}cos(x) - 3.44y^{3}sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12)^2+(-20)^2+(0)^2}=23.32\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12}{23.32}=-0.51\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-20}{23.32}=-0.86\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{23.32}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.51)(-1y^{4}z^{3}sin(x))+(-0.86)(4y^{3}sin(z^{3}) + 4y^{3}z^{3}cos(x))+(0)(3y^{4}z^{2}cos(z^{3}) + 3y^{4}z^{2}cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=0.51y^{4}z^{3}sin(x) - 3.44y^{3}z^{3}cos(x) - 3.44y^{3}sin(z^{3})\end{align*}$

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Example Question #1286 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=5ln(x^{2})ln(y^{2}) + 2^{(z^{2})}xy^{2}\\&\text{In the direction of }\overrightarrow{v}=(18,20,0).\end{align*}$

Possible Answers:

${74.6\cdot 2^{(z^{2})}yz}$

${\frac{(269ln(y^{2}))}{x}+\frac{ (269ln(x^{2}))}{y}+ 26.9\cdot 2^{(z^{2})}y^{2} + 53.8\cdot 2^{(z^{2})}xy + 37.3\cdot 2^{(z^{2})}xy^{2}z}$

${\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy}$

${\frac{(4.49ln(y^{2}))}{x}+\frac{ (5.48ln(x^{2}))}{y}+ 0.449\cdot 2^{(z^{2})}y^{2} + 1.1\cdot 2^{(z^{2})}xy}$

Correct answer:

${\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(18)^2+(20)^2+(0)^2}=26.91\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{18}{26.91}=0.67\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{26.91}=0.74\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{0}{26.91}=0\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.67)(\frac{(10ln(y^{2}))}{x}+ 2^{(z^{2})}y^{2})+(0.74)(\frac{(10ln(x^{2}))}{y}+ 2\cdot 2^{(z^{2})}xy)+(0)(1.39\cdot 2^{(z^{2})}xy^{2}z)\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(6.7ln(y^{2}))}{x}+\frac{ (7.4ln(x^{2}))}{y}+ 0.67\cdot 2^{(z^{2})}y^{2} + 1.48\cdot 2^{(z^{2})}xy\end{align*}$

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Example Question #1287 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(2cos(y^{3})sin(x^{3}))}{z}-\frac{ (ln(y^{3})sin(z^{2}))}{x^{2}}\\&\text{In the direction of }\overrightarrow{v}=(2,0,-6).\end{align*}$

Possible Answers:

${\frac{(12.6ln(y^{3})sin(z^{2}))}{x^{3}}-\frac{ (12.6cos(y^{3})sin(x^{3}))}{z^{2}}-\frac{ (19sin(z^{2}))}{(x^{2}y)}-\frac{ (12.6zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (37.9x^{2}cos(x^{3})cos(y^{3}))}{z}-\frac{ (37.9y^{2}sin(x^{3})sin(y^{3}))}{z}}$

${\frac{(12cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (4ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (12zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (12x^{2}cos(x^{3})cos(y^{3}))}{z}}$

${\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}}$

${\frac{(75.8zcos(z^{2}))}{(x^{3}y)}+\frac{ (114x^{2}y^{2}cos(x^{3})sin(y^{3}))}{z^{2}}}$

Correct answer:

${\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}}$

Explanation:

$\begin{align*}\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{2}{6.32}=0.32\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{0}{6.32}=0\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-6}{6.32}=-0.95\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.32)(\frac{(2ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (6x^{2}cos(x^{3})cos(y^{3}))}{z})+(0)(-\frac{ (3sin(z^{2}))}{(x^{2}y)}-\frac{ (6y^{2}sin(x^{3})sin(y^{3}))}{z})+(-0.95)(-\frac{ (2cos(y^{3})sin(x^{3}))}{z^{2}}-\frac{ (2zcos(z^{2})ln(y^{3}))}{x^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.9cos(y^{3})sin(x^{3}))}{z^{2}}+\frac{ (0.64ln(y^{3})sin(z^{2}))}{x^{3}}+\frac{ (1.9zcos(z^{2})ln(y^{3}))}{x^{2}}+\frac{ (1.92x^{2}cos(x^{3})cos(y^{3}))}{z}\end{align*}$

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Example Question #1288 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=yln(x^{2})ln(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(0,13,-16).\end{align*}$

Possible Answers:

${13ln(x^{2})ln(z^{4}) -\frac{ (64yln(x^{2}))}{z}}$

${0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}}$

${20.6ln(x^{2})ln(z^{4}) +\frac{ (82.5yln(x^{2}))}{z}+\frac{ (41.2yln(z^{4}))}{x}}$

${\frac{165.0}{(xz)}}$

Correct answer:

${0.63ln(x^{2})ln(z^{4}) -\frac{ (3.12yln(x^{2}))}{z}}$

Explanation:

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

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Example Question #1289 : Partial Derivatives

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

Possible Answers:

${\frac{(163xy^{2})}{z}}$

${\frac{(24.0tan(x))}{z}- 39x^{2}y^{2}ln(z^{2}) +\frac{ (8x^{2}y^{3})}{z}}$

${\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}}$

${\frac{(81.6tan(x))}{z}+ 40.8ln(z^{2})\cdot (tan(x)^{2} + 1) + 40.8x^{2}y^{2}ln(z^{2}) +\frac{ (27.2x^{2}y^{3})}{z}+ 27.2xy^{3}ln(z^{2})}$

Correct answer:

${\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(-13)^2+(4)^2}=13.6\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{13.6}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13}{13.6}=-0.96\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{4}{13.6}=0.29\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(3ln(z^{2})\cdot (tan(x)^{2} + 1) + 2xy^{3}ln(z^{2}))+(-0.96)(3x^{2}y^{2}ln(z^{2}))+(0.29)(\frac{(6.0tan(x))}{z}+\frac{ (2x^{2}y^{3})}{z})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.74tan(x))}{z}- 2.88x^{2}y^{2}ln(z^{2}) +\frac{ (0.58x^{2}y^{3})}{z}\end{align*}$

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Example Question #1290 : Partial Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(x^{4}cos(z))}{y}\\&\text{In the direction of }\overrightarrow{v}=(0,20,-7).\end{align*}$

Possible Answers:

${\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}}$

${-\frac{(84.8x^{3}sin(z))}{y^{2}}}$

${\frac{(20x^{4}cos(z))}{y^{2}}-\frac{ (7x^{4}sin(z))}{y}}$

${\frac{(21.2x^{4}cos(z))}{y^{2}}-\frac{ (84.8x^{3}cos(z))}{y}+\frac{ (21.2x^{4}sin(z))}{y}}$

Correct answer:

${\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(0)^2+(20)^2+(-7)^2}=21.19\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{0}{21.19}=0\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{21.19}=0.94\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7}{21.19}=-0.33\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0)(-\frac{(4x^{3}cos(z))}{y})+(0.94)(\frac{(x^{4}cos(z))}{y^{2}})+(-0.33)(\frac{(x^{4}sin(z))}{y})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.94x^{4}cos(z))}{y^{2}}-\frac{ (0.33x^{4}sin(z))}{y}\end{align*}$

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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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Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

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A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

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Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

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I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

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Calculus 3 : Calculus 3

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