The parabola y=1-x^2, the lines x=1 and x =-1, and a line y = c for some c between 0 and 1. What should be the value of c so as to minimize the shaded

[Brenthenny]

The parabola y=1-x^2, the lines x=1 and x =-1, and a line y = c for some c between 0 and 1. What should be the value of c so as to minimize the shaded
The y axis is the bottom of the figure

 

[Harry_the_cat]

I think you mean the x-axis is at the bottom of the figure.

So, what have you tried? Where are you stuck?

 

[Steven G]

I have a question. How is the shaded region determined?

 

[MarkFL]

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?

 

[MarkFL]

Here's a live graph:

Desmos | Graphing Calculator

www.desmos.com

 

[Brenthenny]

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?

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 it

 

[Brenthenny]

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.

 

[Steven G]

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.

 

[MarkFL]

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.

Sorry, 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.

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.

Did you examine the live graph I posted? Looking at that should explain how the shading is determined.

 

[MarkFL]

I belong to a general chat forum, and there are some people there who enjoy math and I occasionally pose problems I find interesting on various math forums as challenges there. This was one such problem. I was simply asking for their best guesses, but one of the folks there stunned me by reasoning his way to a solution as follows:

"It's been many years since I did any calculus, but I consider the change in the area as \(c\) varies from 0 to 1. At first, since the majority of the line \(y=c\) is under the parabola, as \(c\) increases the area must be decreasing because we are losing more than we are gaining. Likewise as \(c\) approaches 1, the area must be increasing because we are gaining more than we are losing. And so it stands to reason that when the part of the line under the parabola is equal to the part not under the parabola, then we are gaining as much as we are losing, and the rate of change of the area is zero there, at \(x=\dfrac{1}{2}\) which implies:

[MATH]\sqrt{1-c}=\frac{1}{2}\implies c=\frac{3}{4}[/MATH]. "

I thought this was a very creative approach, worth sharing here.

 

[MarkFL]

Inspired by my buddy's reasoning, we could integrate with respect to \(y\) to get (still using only quadrant I):

[MATH]A(c)=\int_0^c 1-\sqrt{1-y}\,dy+\int_c^1 \sqrt{1-y}\,dy[/MATH]
Now we can simply use the derivative form of the FTOC to write:

[MATH]A'(c)=1-2\sqrt{1-c}=0\implies c=\frac{3}{4}[/MATH]

 

[Brenthenny]

Sorry, 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.

Thank you very much this was very helpful