Brenthenny
New member
- Joined
- Mar 10, 2020
- Messages
- 4
Perfect thank you very much that is how I attempted the problems I just wasn't certain if I was on the right track. I appreciate itHello, and welcome to FMH!
Because of even symmetry, we need only look at the shaded area in quadrant I. I would begin by locating the point of intersection between the parabola and the horizontal line:
[MATH]1-x^2=c[/MATH]
[MATH]x=\sqrt{1-c}[/MATH]
And so the area \(A\) is:
[MATH]A(c)=\int_0^{\sqrt{1-c}} 1-x^2-c\,dx+\int_{\sqrt{1-c}}^1 c-(1-x^2)\,dx[/MATH]
Differentiation under the integral sign and equating the result to zero yields:
[MATH]A'(c)=\left(-\sqrt{1-c}\right)+\left(1-\sqrt{1-c}\right)=0[/MATH]
Can you proceed?
We haven't learned differentiating under the integral other than using the Fundamental Theory of calculus am I correct to assume that would work here.Hello, and welcome to FMH!
Because of even symmetry, we need only look at the shaded area in quadrant I. I would begin by locating the point of intersection between the parabola and the horizontal line:
[MATH]1-x^2=c[/MATH]
[MATH]x=\sqrt{1-c}[/MATH]
And so the area \(A\) is:
[MATH]A(c)=\int_0^{\sqrt{1-c}} 1-x^2-c\,dx+\int_{\sqrt{1-c}}^1 c-(1-x^2)\,dx[/MATH]
Differentiation under the integral sign and equating the result to zero yields:
[MATH]A'(c)=\left(-\sqrt{1-c}\right)+\left(1-\sqrt{1-c}\right)=0[/MATH]
Can you proceed?
We haven't learned differentiating under the integral other than using the Fundamental Theory of calculus am I correct to assume that would work here.
Mark, can you please explain to me how the area becomes shaded and how y=c would change that shaded area?
Thank you for your time.
Thank you very much this was very helpfulSorry, I took an educated guess that the goal of this problem was to use the method I did, however, the problem certainly can be worked by using the antiderivative form of the FTOC to get the area function, and then differentiating the result.
Let's go back to:
[MATH]A(c)=\int_0^{\sqrt{1-c}} 1-x^2-c\,dx+\int_{\sqrt{1-c}}^1 c-(1-x^2)\,dx[/MATH]
Let's evaluate the definite integrals separately.
[MATH]\int_0^{\sqrt{1-c}} 1-x^2-c\,dx=\left[-\frac{1}{3}x^3+(1-c)x\right]_0^{\sqrt{1-c}}=\frac{2}{3}(1-c)^{\frac{3}{2}}[/MATH]
[MATH]\int_{\sqrt{1-c}}^1 c-(1-x^2)\,dx=\left[\frac{1}{3}x^3-(1-c)x\right]_{\sqrt{1-c}}^1=-\frac{2}{3}+c+\frac{2}{3}(1-c)^{\frac{3}{2}}[/MATH]
And so we have:
[MATH]A(c)=\frac{2}{3}(1-c)^{\frac{3}{2}}-\frac{2}{3}+c+\frac{2}{3}(1-c)^{\frac{3}{2}}=\frac{4}{3}(1-c)^{\frac{3}{2}}+c-\frac{2}{3}[/MATH]
And then, differentiating with respect to \(c\), we find:
[MATH]A'(c)=2(1-c)^{\frac{1}{2}}(-1)+1=1-2\sqrt{1-c}=0[/MATH]
This is equivalent to what I posted before. This implies:
[MATH]c=\frac{3}{4}[/MATH]
I will leave it to you to show this critical value is at a minimum, using either the first or second derivative tests.
Did you examine the live graph I posted? Looking at that should explain how the shading is determined.