Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

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

Example Questions

Example Question #1361 : 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:

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

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

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

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

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:

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

Report an Error

Example Question #1362 : 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:

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

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

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

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

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:

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

Report an Error

Example Question #291 : Directional 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:

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

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #1364 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #292 : Directional 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:

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

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #1366 : 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:

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #1367 : 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:

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

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #301 : Directional 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:

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

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

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

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

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:

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

Report an Error

Example Question #302 : Directional 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:

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

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

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

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

Correct answer:

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

Explanation:

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

Report an Error

Example Question #303 : Directional 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:

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

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

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

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

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:

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

Report an Error

1723729021 1e157913 b243 4ade a7a5 12f8f50d719f

View Calculus 3 Tutors

James
Certified Tutor

Universidad de Los Andes, Bachelor, B.Eng. Industrial Engineering. Sheridan College, Master's/Graduate, Honours Ineractive Me...

View Calculus 3 Tutors

Robert
Certified Tutor

Rensselaer Polytechnic Institute, Bachelors, Electrical Engineering. Brooklyn Polytechnic, Masters, Computer Science.

View Calculus 3 Tutors

Benjamin
Certified Tutor

Portland State University, Bachelor of Science, Computer Science.

All Calculus 3 Resources

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

Popular Subjects

ISEE Tutors in New York City, Physics Tutors in Seattle, Physics Tutors in Philadelphia, Chemistry Tutors in Seattle, French Tutors in Washington DC, Reading Tutors in San Francisco-Bay Area, Algebra Tutors in Los Angeles, Statistics Tutors in Boston, GMAT Tutors in San Francisco-Bay Area, LSAT Tutors in Houston

Popular Courses & Classes

SSAT Courses & Classes in Phoenix, SAT Courses & Classes in Washington DC, Spanish Courses & Classes in Los Angeles, Spanish Courses & Classes in Boston, ACT Courses & Classes in Boston, SAT Courses & Classes in Boston, GMAT Courses & Classes in Washington DC, MCAT Courses & Classes in San Diego, LSAT Courses & Classes in Washington DC, GRE Courses & Classes in Philadelphia

Popular Test Prep

GMAT Test Prep in Phoenix, ISEE Test Prep in Dallas Fort Worth, SAT Test Prep in Miami, SAT Test Prep in Atlanta, GRE Test Prep in Chicago, LSAT Test Prep in Chicago, SAT Test Prep in Phoenix, ISEE Test Prep in Houston, GRE Test Prep in San Diego, MCAT Test Prep in Phoenix

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

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.

Please follow these steps to file a notice:

You must include the following:

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.

Send your complaint to our designated agent at:

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

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

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.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #1361 : Partial Derivatives

Example Question #1362 : Partial Derivatives

Example Question #291 : Directional Derivatives

Example Question #1364 : Partial Derivatives

Example Question #292 : Directional Derivatives

Example Question #1366 : Partial Derivatives

Example Question #1367 : Partial Derivatives

Example Question #301 : Directional Derivatives

Example Question #302 : Directional Derivatives

Example Question #303 : Directional Derivatives

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: