MHB 2.1.312 AP Calculus Exam Int of half circle

[karush]

The function f is defined by
$$f(x)=\sqrt{25-x^2},\quad -5\le x \le 5$$
(a) Find $f'(x)$ apply chain rule
$$
\dfrac{d}{dx}(25-x^2)^{1/2}
=\dfrac{1}{2}(25-x^2)^{-1/2}2x
=-\frac{x}{\sqrt{25-x^2}}$$
(b) Write an equation for the tangent line to the graph of f at $x=-3$
$$f'(-3)=-\frac{3}{\sqrt{25-(3)^2}}
=-\dfrac{3}{4}=m$$
then f(-3)=4 $y=mx+b$ so $y=-\dfrac{3}{4}(x+3)+4$
(c) Let g be the function defined by $\biggr\{\begin{array}{ll}
f(x) &\textit{for } -5\le x \ge 5\\
x+7 &\textit{for } 3\le x\le 5
\end{array}$
Is g continuous at $x=-3$ (d) Find the value of
$\displaystyle
\int_0^5 x \sqrt{25-x^2}\, dx$

---------------------------------------------------------------
ok i think a and b are okbut (c) x+7 is not a tangent line but looks continuous by the inequalities(d) I assume they tossed in the x for a u substitution method.

View attachment 9331

 

[skeeter]

The function f is defined by
$$f(x)=\sqrt{25-x^2},\quad -5\le x \le 5$$
(a) Find $f'(x)$ apply chain rule
$$
\dfrac{d}{dx}(25-x^2)^{1/2}
=\dfrac{1}{2}(25-x^2)^{-1/2}{\color{red}(-2x)}
=-\frac{x}{\sqrt{25-x^2}}$$
(b) Write an equation for the tangent line to the graph of f at $x=-3$
$$f'(-3)=-\frac{{\color{red}(-3)}}{\sqrt{25-{\color{red}{(-3)}}^2}}
={\color{red}{+}}\dfrac{3}{4}=m$$
then f(-3)=4 $y=mx+b$ so $y={\color{red}{+}}\dfrac{3}{4}(x+3)+4$
(c) Let g be the function defined by $\biggr\{\begin{array}{ll}
f(x) &\textit{for } -5\le x \ge 5 {\color{red}{\text{ ?}}}\\
x+7 &\textit{for } 3\le x\le 5
\end{array}$

Is g continuous at $x=-3 {\color{red}{\text{ ?}}}$ (d) Find the value of
$\displaystyle
\int_0^5 x \sqrt{25-x^2}\, dx$

---------------------------------------------------------------
ok i think a and b are okbut (c) x+7 is not a tangent line but looks continuous by the inequalities(d) I assume they tossed in the x for a u substitution method.

(c) check the intervals of the piece-wise function again and the point where continuity is in question

(d) use substitution and evaluate

 

[karush]

Let g be the function defined by $$\biggr\{\begin{array}{ll}
f(x) &\textit{for } -5\le x < -3\\
x+7 &\textit{for } -3\le x\le 5
\end{array}$$
Is g continuous at $x=-3$
there is no break by means of the inequality but it is a corner (d) Find the value of
$\displaystyle
\int_0^5 x \sqrt{25-x^2}\, dx$$u=25-x^2 $ and $x=\sqrt{25-u}$ then $\dfrac{du}{2\sqrt{25-u}}=dx$
then
$\displaystyle\int_0^5 \sqrt{u}\sqrt{25-u}\dfrac{du}{2\sqrt{25-u}}
=\dfrac{1}{2}\int_0^5 \sqrt{u} \, du=\dfrac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}\biggr|_0^5\right]$

got this far ... different u maybe or change in limits?

 

[HallsofIvy]

Let g be the function defined by $$\biggr\{\begin{array}{ll}
f(x) &\textit{for } -5\le x < -3\\
x+7 &\textit{for } -3\le x\le 5
\end{array}$$
Is g continuous at $x=-3$
there is no break by means of the inequality but it is a corner

.
So what is your answer to this question? Is g continuous at x= -3?

(d) Find the value of
$\displaystyle
\int_0^5 x \sqrt{25-x^2}\, dx$$u=25-x^2 $ and $x=\sqrt{25-u}$ then $\dfrac{du}{2\sqrt{25-u}}=dx$
then

Simpler is $u= 25- x^2$ so $du= -2x dx$ and then $-\frac{1}{2}du= xdx$
since you already have "x" in the integral.

$\displaystyle\int_0^5 \sqrt{u}\sqrt{25-u}\dfrac{du}{2\sqrt{25-u}}
=\dfrac{1}{2}\int_0^5 \sqrt{u} \, du=\dfrac{1}{2}\left[\frac{2}{3}u^{\frac{3}{2}}\biggr|_0^5\right]$

got this far ... different u maybe or change in limits?

With $u= 25- x^2$, $-\frac{1}{2}du= xdx$, when x= 0, u= 25- 0= 25 and when x= 5, u= 25- 25= 0. The integral becomes
$\int_{25}^0 u^{1/2} (-\frac{1}{2}du)= \frac{1}{2}\int_0^{25} u^{1/2}du$ ($\int_a^b f(x)dx= -\int_b^a f(x)dx$)

 

[Greg]

$$u^2=25-x^2$$

$$-u\,du=x\,dx$$

$$\int_0^5u^2\,du=\frac{125}{3}-0=\frac{125}{3}$$

 

[karush]

$$u^2=25-x^2$$

$$-u\,du=x\,dx$$

$$\int_0^5u^2\,du=\frac{125}{3}-0=\frac{125}{3}$$

that was tricky!