2.1.312 AP Calculus Exam Int of half circle

In summary, the function f(x) is defined by f(x)=√(25-x^2), with -5 ≤ x ≤ 5. The derivative f'(x) is found using the chain rule and is equal to -x/√(25-x^2). The equation for the tangent line at x = -3 is y = (-3/4)(x+3) + 4. The function g is defined as a piece-wise function, with f(x) as the first part and x+7 as the second part for the given intervals. The continuity at x = -3 is in question, as there is a corner at that point. The integral of x√(25-x^2)
  • #1
karush
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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.

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  • #2
karush said:
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
 
  • #3
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?
 
Last edited:
  • #4
karush said:
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
$\int_{25}^0 u^{1/2} (-\frac{1}{2}du)= \frac{1}{2}\int_0^{25} u^{1/2}du$ ($\int_a^b f(x)dx= -\int_b^a f(x)dx$)
 
  • #5
\(\displaystyle u^2=25-x^2\)

\(\displaystyle -u\,du=x\,dx\)

\(\displaystyle \int_0^5u^2\,du=\frac{125}{3}-0=\frac{125}{3}\)
 
  • #6
Greg said:
\(\displaystyle u^2=25-x^2\)

\(\displaystyle -u\,du=x\,dx\)

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

that was tricky!
 

FAQ: 2.1.312 AP Calculus Exam Int of half circle

1. What is the format of the 2.1.312 AP Calculus Exam Int of half circle?

The 2.1.312 AP Calculus Exam Int of half circle is a multiple-choice exam that consists of 40 questions. It is divided into two sections: Section I contains 30 questions and Section II contains 10 questions.

2. What topics are covered on the 2.1.312 AP Calculus Exam Int of half circle?

The exam covers topics related to the integration of half circles, including finding the area and arc length of a half circle, using the fundamental theorem of calculus, and applying integration techniques such as substitution and integration by parts.

3. How long is the 2.1.312 AP Calculus Exam Int of half circle?

The exam is 90 minutes long, with 50 minutes allotted for Section I and 40 minutes for Section II. There is also a 15-minute break between the two sections.

4. What is the passing score for the 2.1.312 AP Calculus Exam Int of half circle?

The passing score for the exam varies each year and is determined by the College Board. Generally, a score of 3 or higher is considered passing, with 5 being the highest score.

5. How can I prepare for the 2.1.312 AP Calculus Exam Int of half circle?

To prepare for the exam, it is recommended to review relevant course materials, practice with past AP Calculus exams, and use study guides and review books. It is also important to understand the format and content of the exam and to manage your time effectively during the exam.

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