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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 #341 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-\frac{(4^{(y^{2})}ln(x^{2}))}{z}\\&\text{In the direction of }\overrightarrow{v}=(-12.26,-16.17,-7.03).\end{align*}$

Possible Answers:

${\frac{(24.5\cdot 4^{(y^{2})})}{(xz)}-\frac{ (7.03\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (44.8\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

${\frac{(119\cdot 4^{(y^{2})}y)}{(xz^{2})}}$

${\frac{(1.14\cdot 4^{(y^{2})})}{(xz)}-\frac{ (0.33\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (2.08\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

${\frac{(21.5\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}-\frac{ (43\cdot 4^{(y^{2})})}{(xz)}-\frac{ (59.6\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

Correct answer:

${\frac{(1.14\cdot 4^{(y^{2})})}{(xz)}-\frac{ (0.33\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (2.08\cdot 4^{(y^{2})}yln(x^{2}))}{z}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-12.26)^2+(-16.17)^2+(-7.03)^2}=21.48\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-12.26}{21.48}=-0.57\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-16.17}{21.48}=-0.75\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-7.03}{21.48}=-0.33\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.57)(-\frac{(2\cdot 4^{(y^{2})})}{(xz)})+(-0.75)(-\frac{(2.77\cdot 4^{(y^{2})}yln(x^{2}))}{z})+(-0.33)(\frac{(4^{(y^{2})}ln(x^{2}))}{z^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(1.14\cdot 4^{(y^{2})})}{(xz)}-\frac{ (0.33\cdot 4^{(y^{2})}ln(x^{2}))}{z^{2}}+\frac{ (2.08\cdot 4^{(y^{2})}yln(x^{2}))}{z}\end{align*}$

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Example Question #342 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=-3^{(y^{2})}ln(x^{4})sin(z^{4})\\&\text{In the direction of }\overrightarrow{v}=(8.74,-13.71,-14.88).\end{align*}$

Possible Answers:

${-\frac{(775\cdot 3^{(y^{2})}yz^{3}cos(z^{4}))}{x}}$

${1.36\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (1.6\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 2.72\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

${30.1\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (35\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 59.5\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

${-\frac{ (88.2\cdot 3^{(y^{2})}sin(z^{4}))}{x}- 48.4\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) - 88.2\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

Correct answer:

${1.36\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (1.6\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 2.72\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(8.74)^2+(-13.71)^2+(-14.88)^2}=22.04\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{8.74}{22.04}=0.4\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-13.71}{22.04}=-0.62\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14.88}{22.04}=-0.68\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.4)(-\frac{(4\cdot 3^{(y^{2})}sin(z^{4}))}{x})+(-0.62)(-2.2\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}))+(-0.68)(-4\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4}))\\&D_{\overrightarrow{u}}(x,y,z)=1.36\cdot 3^{(y^{2})}yln(x^{4})sin(z^{4}) -\frac{ (1.6\cdot 3^{(y^{2})}sin(z^{4}))}{x}+ 2.72\cdot 3^{(y^{2})}z^{3}cos(z^{4})ln(x^{4})\end{align*}$

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Example Question #343 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=- 3zsin(y^{3}) - 4^{(y^{4})}xsin(z^{3})\\&\text{In the direction of }\overrightarrow{v}=(-5.63,-7.28,19.88).\end{align*}$

Possible Answers:

${-364\cdot 4^{(y^{4})}y^{3}z^{2}cos(z^{3})}$

${0.26\cdot 4^{(y^{4})}sin(z^{3}) - 2.73sin(y^{3}) + 2.97y^{2}zcos(y^{3}) - 2.73\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) + 1.83\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

${- 65.7sin(y^{3}) - 21.9\cdot 4^{(y^{4})}sin(z^{3}) - 197y^{2}zcos(y^{3}) - 65.7\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) - 121\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

${- 2.48sin(y^{3}) - 0.0676\cdot 4^{(y^{4})}sin(z^{3}) - 0.98y^{2}zcos(y^{3}) - 2.48\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) - 0.604\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

Correct answer:

${0.26\cdot 4^{(y^{4})}sin(z^{3}) - 2.73sin(y^{3}) + 2.97y^{2}zcos(y^{3}) - 2.73\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) + 1.83\cdot 4^{(y^{4})}xy^{3}sin(z^{3})}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-5.63)^2+(-7.28)^2+(19.88)^2}=21.91\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-5.63}{21.91}=-0.26\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-7.28}{21.91}=-0.33\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{19.88}{21.91}=0.91\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.26)(-1\cdot 4^{(y^{4})}sin(z^{3}))+(-0.33)(- 9y^{2}zcos(y^{3}) - 5.55\cdot 4^{(y^{4})}xy^{3}sin(z^{3}))+(0.91)(- 3sin(y^{3}) - 3\cdot 4^{(y^{4})}xz^{2}cos(z^{3}))\\&D_{\overrightarrow{u}}(x,y,z)=0.26\cdot 4^{(y^{4})}sin(z^{3}) - 2.73sin(y^{3}) + 2.97y^{2}zcos(y^{3}) - 2.73\cdot 4^{(y^{4})}xz^{2}cos(z^{3}) + 1.83\cdot 4^{(y^{4})}xy^{3}sin(z^{3})\end{align*}$

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Example Question #344 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=z^{3}sin(y^{4})cos(x)\\&\text{In the direction of }\overrightarrow{v}=(-14.68,12.52,7.66).\end{align*}$

Possible Answers:

${0.411z^{2}sin(y^{4})cos(x) - 0.504z^{3}sin(y^{4})sin(x) + 1.44y^{3}z^{3}cos(y^{4})cos(x)}$

${1.11z^{2}sin(y^{4})cos(x) + 0.71z^{3}sin(y^{4})sin(x) + 2.4y^{3}z^{3}cos(y^{4})cos(x)}$

${62.3z^{2}sin(y^{4})cos(x) - 20.8z^{3}sin(y^{4})sin(x) + 83y^{3}z^{3}cos(y^{4})cos(x)}$

${-249y^{3}z^{2}cos(y^{4})sin(x)}$

Correct answer:

${1.11z^{2}sin(y^{4})cos(x) + 0.71z^{3}sin(y^{4})sin(x) + 2.4y^{3}z^{3}cos(y^{4})cos(x)}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14.68)^2+(12.52)^2+(7.66)^2}=20.76\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14.68}{20.76}=-0.71\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{12.52}{20.76}=0.6\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{7.66}{20.76}=0.37\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.71)(-1z^{3}sin(y^{4})sin(x))+(0.6)(4y^{3}z^{3}cos(y^{4})cos(x))+(0.37)(3z^{2}sin(y^{4})cos(x))\\&D_{\overrightarrow{u}}(x,y,z)=1.11z^{2}sin(y^{4})cos(x) + 0.71z^{3}sin(y^{4})sin(x) + 2.4y^{3}z^{3}cos(y^{4})cos(x)\end{align*}$

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Example Question #345 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=\frac{(z^{2}sin(x^{4}))}{y^{2}}\\&\text{In the direction of }\overrightarrow{v}=(4.93,2.88,2.11).\end{align*}$

Possible Answers:

${\frac{(0.7zsin(x^{4}))}{y^{2}}-\frac{ (0.94z^{2}sin(x^{4}))}{y^{3}}+\frac{ (3.24x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

${\frac{(12.2zsin(x^{4}))}{y^{2}}-\frac{ (12.2z^{2}sin(x^{4}))}{y^{3}}+\frac{ (24.4x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

${\frac{(4.22zsin(x^{4}))}{y^{2}}-\frac{ (5.76z^{2}sin(x^{4}))}{y^{3}}+\frac{ (19.7x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

${-\frac{(97.4x^{3}zcos(x^{4}))}{y^{3}}}$

Correct answer:

${\frac{(0.7zsin(x^{4}))}{y^{2}}-\frac{ (0.94z^{2}sin(x^{4}))}{y^{3}}+\frac{ (3.24x^{3}z^{2}cos(x^{4}))}{y^{2}}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(4.93)^2+(2.88)^2+(2.11)^2}=6.09\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{4.93}{6.09}=0.81\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{2.88}{6.09}=0.47\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{2.11}{6.09}=0.35\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(0.81)(\frac{(4x^{3}z^{2}cos(x^{4}))}{y^{2}})+(0.47)(-\frac{(2z^{2}sin(x^{4}))}{y^{3}})+(0.35)(\frac{(2zsin(x^{4}))}{y^{2}})\\&D_{\overrightarrow{u}}(x,y,z)=\frac{(0.7zsin(x^{4}))}{y^{2}}-\frac{ (0.94z^{2}sin(x^{4}))}{y^{3}}+\frac{ (3.24x^{3}z^{2}cos(x^{4}))}{y^{2}}\end{align*}$

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Example Question #351 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(x,y,z)\text{ where }f(x,y,z)=y^{4}sin(x^{4})e^{(z^{2})}\\&\text{In the direction of }\overrightarrow{v}=(-14.93,-4.93,-3.31).\end{align*}$

Possible Answers:

${514x^{3}y^{3}zcos(x^{4})e^{(z^{2})}}$

${64.3y^{3}sin(x^{4})e^{(z^{2})} + 32.1y^{4}zsin(x^{4})e^{(z^{2})} + 64.3x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

${- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

${- 19.7y^{3}sin(x^{4})e^{(z^{2})} - 6.62y^{4}zsin(x^{4})e^{(z^{2})} - 59.7x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

Correct answer:

${- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}}$

Explanation:

$\begin{align*}&\text{To find }D_{\overrightarrow{u}}(x,y,z)\\&\text{First make sure that }\overrightarrow{v}\\&\text{is in unit vector form. Find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-14.93)^2+(-4.93)^2+(-3.31)^2}=16.07\\&\text{Unit vector components are:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-14.93}{16.07}=-0.93\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{-4.93}{16.07}=-0.31\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-3.31}{16.07}=-0.21\\&D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(x,y,z)=(-0.93)(4x^{3}y^{4}cos(x^{4})e^{(z^{2})})+(-0.31)(4y^{3}sin(x^{4})e^{(z^{2})})+(-0.21)(2y^{4}zsin(x^{4})e^{(z^{2})})\\&D_{\overrightarrow{u}}(x,y,z)=- 1.24y^{3}sin(x^{4})e^{(z^{2})} - 0.42y^{4}zsin(x^{4})e^{(z^{2})} - 3.72x^{3}y^{4}cos(x^{4})e^{(z^{2})}\end{align*}$

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Example Question #352 : Directional Derivatives

Find the directional derivative D_uf(x,y) for the function:

f(x,y)=ln(xy)+ye^x

in the direction of the vector $\vec{u}=2\hat{i}+\hat{j}$

Possible Answers:

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[\frac{xe^x(x+1)+2x+y}{x} \right]$

$D_uf(x,y) =2\left(\frac{1}{y}+e^x \right)\hat{i}+\left(\frac{1}{x}+ye^x\right)\hat{j}$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]$

$D_uf(x,y) =\frac{\sqrt{7}}{3}\left[ 7e^x(x+1)-\frac{2}{y} \right]$

$D_uf(x,y) =2\left(\frac{1}{x}+ye^x \right)\hat{i}+\left(\frac{1}{y}+e^x\right)\hat{j}$

Correct answer:

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]$

Explanation:

Find the directional derivative for the function:

f(x,y)=ln(xy)+ye^x

in the direction of the vector $\vec{u}=2\hat{i}+\hat{j}$

Find the gradient of

The gradient by definition is the vector:

$\vec{\bigtriangledown} f(x,y)=\frac{\partial}{\partial x}f(x,y)\hat{i}+\frac{\partial}{\partial y}f(x,y)\hat{j}$ (1)

______________________________________________________________

Differentiate with respect to while holding constant:

$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left[\ln(xy)+ye^x\right ]=\frac{y}{xy}+ye^x$

$\frac{\partial}{\partial x}f(x,y)=\frac{1}{x}+ye^x$

______________________________________________________________

Differentiate with respect to while holding constant:

$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left[\ln(xy)+ye^x\right ]=\frac{x}{xy}+e^x$

$\frac{\partial}{\partial y}f(x, y)=\frac{1}{y}+e^x$

______________________________________________________________

Enter the components of the gradient, writing in the form of Equation (1)

$\vec{\bigtriangledown}f(x,y)=\left(\frac{1}{x}+ye^x\right)\hat{i}+ \left(\frac{1}{y}+e^x\right)\hat{j}$

_______________________________________________________________

Find the unit vector in the direction of $\vec{u}$

$\vec{u}=2\hat{i}+\hat{j}$

Divide the vector by its' magnitude $\left \| \vec{u}\right \|$ to obtain the unit vector $\hat{u}$ ,

$\hat{u}=\frac{\vec{u}}{\left \| \vec{u}\right \|} =\frac{\vec{u}}{\sqrt{2^2+1^2}}=\frac{2\hat{i}+\hat{j}}{\sqrt{5}}$

$\hat{u}=\frac{\sqrt{5}}{5}(2\hat{i}+\hat{j})$

You can check the magnitude of the unit vector and see that it is equal to 1 as required. It also has the same direction as the original vector $\vec{u}$ .

_______________________________________________________________

The directional derivative of f(x,y) in the direction of the vector $\hat{u}$ is the dot product of the gradient $\vec{\bigtriangledown} f(x,y)$ and $\hat{u}$ . Dot products are also referred to as "projections," since the give the component of some vector in the direction of another.

$D_uf(x,y)= \vec{\bigtriangledown} f(x,y)\cdot \hat{u}$

$D_uf(x,y)= \left[ \left(\frac{1}{x}+ye^x\right)\hat{i}+ \left(\frac{1}{y}+e^x\right)\hat{j}\right]\cdot\left[\frac{2\sqrt{5}}{5}\hat{i}+\frac{\sqrt{5}}{5}\hat{j}\right]$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[\left(\frac{2}{x}+2ye^x\right)+ \left(\frac{1}{y}+e^x\right) \right ]$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ e^x(2y+1)+\frac{1}{y}+\frac{2}{x} \right]$

$D_uf(x,y) =\frac{\sqrt{5}}{5}\left[ \frac{xye^x(2y+1)+x+2y}{xy}\right]$

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Example Question #1 : Differentials

Compute the differentials for the following function.

$z=e^{x^2+4y^2}\sin(y)$

Possible Answers:

$dz=2xe^{x^2+4y^2}\sin(y)+8y\cos(y)e^{x^2+4y^2}$

$dz=2xe^{x^2+4y^2}\sin(y)dx$

$dz=8y\cos(y)e^{x^2+4y^2} dy$

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

$dz=2xe^{x^2+4y^2}\sin(y)$

Correct answer:

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

Explanation:

What we need to do is take derivatives, and remember the general equation.

$z=e^{x^2+4y^2}\sin(y)$

w=g(x,y,z)

dw=g_xdx+g_ydy+g_zdz

When taking the derivative with respect to y recall that the product rule needs to be used.

$dz=2xe^{x^2+4y^2}\sin(y)dx+(8y\sin(y)e^{x^2+4y^2}+cos(y)e^{x^2+4y^2}) dy$

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Example Question #1 : Differentials

Find the total differential , , of the function

z=e^x + cos (y)

Possible Answers:

dz=xe^x dx+sin(y)dy

dz=e^x dx-sin(y)dy

dz=cos(y) dx-e^xdy

dz=e^x dx+sin(y)dy

Correct answer:

dz=e^x dx-sin(y)dy

Explanation:

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

We first find

$\frac{\partial z}{\partial x}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=e^x$

We then find

$\frac{\partial z}{\partial y}$

by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=-sin(y)$

We then substitute these partial derivatives into the first equation to get the total differential

dz=e^x dx-sin(y)dy

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Example Question #3 : Differentials

Find the total differential, , of the function

$z=e^{x^2+y^2}+sec(2x)$

Possible Answers:

$dz=[2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

$dz=[2xe^{x^2+y^2}]dx+2ye^{x^2+y^2}dy$

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

Correct answer:

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

Explanation:

The total differential is defined as

$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$

We first find $\frac{\partial z}{\partial x}$ by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}[e^{x^2+y^2}+sec(2x)]=2xe^{x^2+y^2}+2sec(2x)tan(2x)$

We then find $\frac{\partial z}{\partial y}$ by taking the derivative with respect to and treating as a constant.

$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}[e^{x^2+y^2}+sec(2x)]=2ye^{x^2+y^2}$

We then substitute these partial derivatives into the first equation to get the total differential

$dz=[2xe^{x^2+y^2}+2sec(2x)tan(2x)]dx+2ye^{x^2+y^2}dy$

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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

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* 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.

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Calculus 3 : Calculus 3

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Example Question #341 : Directional Derivatives

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