AP Calculus BC - Calculus 3

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&1.26\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.685,0.729)\text{ and }\overrightarrow{u_2}=(-0.969,0.249)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3sin(\frac{(2x^{2})}{3}+\frac{ (2y^{2})}{3}))}{23}- 32e^{(- 2x^{2} - 2y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{23}- 32\cdot re^{(-2\cdot r^{2})})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.729}{0.685})=0.26\pi;\theta_2=arctan(\frac{0.249}{-0.969})=0.92\pi\&\text{Now, utilizing integral rules:}\&\int[rsin(ar^2)]=-\frac{cos(ar^2)}{2a}=-\frac{cos^2(\frac{a}{2}r^2)}{a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.26\pi}^{0.92\pi}\int_{0}^{1.26}(\frac{(3\cdot rsin(\frac{(2\cdot r^{2})}{3}))}{23}- 32\cdot re^{(-2\cdot r^{2})})drd\theta=(8e^{(-2\cdot r^{2})} -\frac{ (9cos(\frac{(2\cdot r^{2})}{3}))}{92})d\theta|_{0}^{1.26}\&\int_{0.26\pi}^{0.92\pi}(-7.616)d\theta=(-7.616\theta)|_{0.26\pi}^{0.92\pi}=-15.79\end{align*}$

Given the function $f(x,y)=e^{24xy+25x^2}$ , find the partial derivative

$f_y(x,y)=24xe^{24xy+25x^2}$

$f_y(x,y)=(24y+50x)e^{24xy+25x^2},$

$f_y(x,y)=e^{24y+50x}$

$f_y(x,y)=(24y+50x)e^{24y+50x}$

Explanation

To find the partial derivative of $f(x,y)=e^{24xy+25x^2}$ , we need to differentiate with respect to while holding constant. We can use the chain rule to get

$f_y(x,y)=\frac{\partial}{\partial y}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial y}(24xy+25x^2)$

$=24xe^{24xy+25x^2}$

$\begin{align*}&\text{Evaluate the following limit:}\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}\end{align*}$

$\frac{324}{7}$

$\infty$

$-\infty$

$-\frac{324}{7}$

Explanation

$\begin{align*}&\text{Examining this problem, we find that both}\&\text{numerator and denominator have zero values at the limit specified.}\&\text{Noting this, L'Hopital's method may be of use:}\&\text{If both numerator and denominator go to }0\text{ or }\pm\infty\&lim_{x\rightarrow n}\frac{f(x)}{g(x)}=lim_{x\rightarrow n}\frac{f'(x)}{g'(x)}\&\frac{9sin(6x)^{2}}{7x^{2}}\rightarrow\frac{0}{0}\&\frac{108cos(6x)sin(6x)}{14x}\rightarrow\frac{0}{0}\&\frac{648cos(6x)^{2} - 648sin(6x)^{2}}{14}\rightarrow\frac{648}{14}\&lim_{x\rightarrow0-}\frac{9sin(6x)^{2}}{7x^{2}}=\frac{324}{7}\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{16}-\frac{ (4cos(x^{2} + y^{2}))}{5})dA\text{, where D is}\&\text{the region of a circle centered on the origin with a radius of }\&0.97\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(-0.960,0.279)\text{ and }\overrightarrow{u_2}=(0.771,-0.637)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{(3e^{(-\frac{ (2x^{2})}{3}-\frac{ (2y^{2})}{3})})}{16}-\frac{ (4cos(x^{2} + y^{2}))}{5})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{16}-\frac{ (4\cdot rcos(r^{2}))}{5})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.279}{-0.960})=0.91\pi;\theta_2=arctan(\frac{-0.637}{0.771})=1.78\pi\&\text{Now, utilizing integral rules:}\&\int[rcos(ar^2)]=\frac{sin(ar^2)}{2a}\&\int[re^{ar^2}]=\frac{e^{ar^2}}{2a}\&\int_{0.91\pi}^{1.78\pi}\int_{0}^{0.97}(\frac{(3\cdot re^{(-\frac{(2\cdot r^{2})}{3})})}{16}-\frac{ (4\cdot rcos(r^{2}))}{5})drd\theta=(-\frac{ (2sin(r^{2}))}{5}-\frac{ (9e^{(-\frac{(2\cdot r^{2})}{3})})}{64})d\theta|_{0}^{0.97}\&\int_{0.91\pi}^{1.78\pi}(-0.2577)d\theta=(-0.2577\theta)|_{0.91\pi}^{1.78\pi}=-0.7\end{align*}$

Find $f_{xy}$ of the function $f(x,y,z)=3y\sin(x)+8x^2yz^3$

$3\cos(x)+16xz^3$

$5\cos(x)+16x^2z^4$

$3\cos(x)+14xz^3$

$3\sin(x)+16xz^3$

Explanation

To find $f_{xy}$ of the function, you take two consecutive partial derivatives:

$\frac{\partial }{\partial x}(3y\sin(x)+8x^2yz^3)=3y\cos(x)+16xyz^3$

$\frac{\partial }{\partial y}(3y\cos(x)+16xyz^3)=3\cos(x)+16xz^3$

Given the function $f(x,y,z)=5e^{x+y+xyz}$ , find the partial derivative

$f_x(x,y,z)=5(1+yz)e^{x+y+xyz}$

$f_x(x,y,z)=5yze^{x+y+xyz}$

$f_x(x,y,z)=5(1+xz)e^{x+y+xyz}$

Explanation

We can find the partial derivative of the function $f(x,y,z)=5e^{x+y+xyz}$ by taking its derivative with respect to while holding and constant. We will also use the chain rule:

$f_x(x,y,z)=\frac{\partial}{\partial x}5e^{x+y+xyz}=5e^{x+y+xyz}\frac{\partial}{\partial x}(x+y+xyz)$

$=5(1+yz)e^{x+y+xyz}$

Screen shot 2015 07 07 at 3.50.30 pm

Given the graph of above, what is $\lim_{x\rightarrow 0}f(x)$ ?

$\infty$

$-\infty$

Explanation

Examining the graph above, we need to look at three things:

What is the limit of the function as approaches zero from the left?
What is the limit of the function as approaches zero from the right?
What is the function value as and is it the same as the result from statement one and two?

Looking at the graph we can determine that $\lim_{x\rightarrow 0}f(x)=\infty$ as approaches because from both the left and right sides of zero, the function is approaching infinity.

$\begin{align*}&\text{Calculate the limits of the function}\&\lim_{(x,y)\rightarrow(0+,0+)}\frac{(17x^{2}y^{2} + 14xy^{3} - 5y^{4})}{(8x^{3}y - 15x^{2}y^{2} + 3x^{4} + 2y^{4})}\&\text{along the y-axis and the line, }y=x\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{5}{2}\&y=x:-13\end{align*}$

$\begin{align*}&\text{y-axis: }-156\&y=x:-\frac{95}{2}\end{align*}$

$\begin{align*}&\text{y-axis: }117\&y=x:45\end{align*}$

$\begin{align*}&\text{y-axis: }-\frac{5}{34}\&y=x:-\frac{13}{4}\end{align*}$

Explanation

$\begin{align*}&\text{To approach this problem, it is important to understand what}\ &\text{it is asking. Along the y-axis, }x=0\text{. Let us make that substitution:}\&\frac{(17(0)^{2}y^{2} + 14(0)y^{3} - 5y^{4})}{(8(0)^{3}y - 15(0)^{2}y^{2} + 3(0)^{4} + 2y^{4})}=\frac{-5y^{4}}{2y^{4}}\&lim_{(0,y)\rightarrow(0+,0+)}\frac{-5y^{4}}{2y^{4}}=-\frac{5}{2}\&\text{We can make a similar substitution for }y=x:\&\lim_{(x,x)\rightarrow(0+,0+)}\frac{(17x^{2}(x)^{2} + 14x(x)^{3} - 5(x)^{4})}{(8x^{3}(x) - 15x^{2}(x)^{2} + 3x^{4} + 2(x)^{4})}=-13\&\text{As these two limits are different, the limit does not exist.}\end{align*}$

$\begin{align*}&\text{Evaluate the triple integral}\int_{3}^{5.5}\int_{4}^{5}\int_{-6}^{-1.5}(\frac{(3\cdot 3^{(\frac{z}{2})}\cdot 3^y)}{(74x^{2})})dxdydz\end{align*}$

Explanation

$\begin{align*}&\text{Performing a triple integral, the order of integration does}\&\text{not entirely matter.}\&\text{For example:}\&\int_{s}^{t} \int_{c}^{d} \int_{a}^b f(x,y,z)dxdydz=\int_{a}^{b} \int_{c}^{d} \int_{s}^t f(x,y,z)dzdydx\&\text{Considering our integral, and keeping savvy about }\&\text{utilizing integral rules:}\&\int[b^{ax}]=\frac{b^{ax}}{aln(b)}\end{align*}$

$\begin{align*}&\int_{3}^{5.5}\int_{4}^{5}\int_{-6}^{-1.5}(\frac{(3\cdot 3^{(\frac{z}{2})}\cdot 3^y)}{(74x^{2})})dxdydz\&\text{The approach is simply to take it step by step:}\&\int_{3}^{5.5}\int_{4}^{5}\int_{-6}^{-1.5}(\frac{(3\cdot 3^{(\frac{z}{2})}\cdot 3^y)}{(74x^{2})})dxdydz=\int_{3}^{5.5}\int_{4}^{5}(-\frac{(3\cdot 3^{(y +\frac{ z}{2})})}{(74x)})dydz|_{-6}^{-1.5}\&\int_{3}^{5.5}\int_{4}^{5}(\frac{(3\cdot 3^{(y +\frac{ z}{2})})}{148})dydz=\int_{3}^{5.5}(\frac{(3\cdot 3^{(y +\frac{ z}{2})})}{(148ln(3))})dz|_{4}^{5}\&\int_{3}^{5.5}(\frac{(243\cdot (3^{(\frac{1}{2})})^z)}{(74ln(3))})dz=\frac{(243\cdot 3^{(\frac{z}{2})})}{(37ln(3)^{2})}|_{3}^{5.5}=83.36\end{align*}$

$\begin{align*}&\text{Evaluate the integral, }\iint_{D}(\frac{5}{(x^{2} + y^{2})}+ 2\cdot (\frac{2}{5})^{(x^{2} + y^{2})})dA\text{, where D is}\&\text{the region defined between two circles centered on the origin with radii }\&0.48\text{ and }1.84\&\text{and encompassed by vectors from the origin }\&\overrightarrow{u_1}=(0.891,0.454)\text{ and }\overrightarrow{u_2}=(-0.661,-0.750)\&\text{counterclockwise from }\overrightarrow{u_1}\end{align*}$

Explanation

$\begin{align*}&\text{The current form of the integral is rather unwieldy, due to the }x^2\text{ and }y^2\text{ terms.}\&\text{An approach that would be beneficial is a conversion to polar form:}\&r=cos(\theta);r=sin(\theta)\&r^2=x^2+y^2\&dA=rdrd\theta\&\text{With this we can find: }\iint_{D}(\frac{5}{(x^{2} + y^{2})}+ 2\cdot (\frac{2}{5})^{(x^{2} + y^{2})})dA\rightarrow\int_{\theta_1}^{\theta_2}\int_{r_1}^{r_2}(2\cdot (\frac{2}{5})^{(r^{2})}\cdot r +\frac{ 5}{r})drd\theta\&\text{The bounds of our radius we can readily take from the problem. To find the angle ranges:}\end{align*}$

$\begin{align*}&\theta_1=arctan(\frac{0.454}{0.891})=0.15\pi;\theta_2=arctan(\frac{-0.750}{-0.661})=1.27\pi\&\text{Now, utilizing integral rules:}\&\int[\frac{a}{r}]=aln(r)=ln(r^a)\&\int[rb^{ar^2}]=\frac{b^{ar^2}}{2aln(b)}\&\int_{0.15\pi}^{1.27\pi}\int_{0.48}^{1.84}(2\cdot (\frac{2}{5})^{(r^{2})}\cdot r +\frac{ 5}{r})drd\theta=(5ln(r) + (\frac{\frac{2}{5})^{(r^{2})}}{ln(\frac{2}{5})})d\theta|_{0.48}^{1.84}\&\int_{0.15\pi}^{1.27\pi}(7.553)d\theta=(7.553\theta)|_{0.15\pi}^{1.27\pi}=26.58\end{align*}$

Calculus 3

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