Find constants of function with given conditions

[skrat]

Homework Statement

Let ##a>0## and [itex]y(x)=\left\{\begin{matrix}
-x ;& x<-a\\
Cx^2+D;& -a<x<a\\
x;& x>a
\end{matrix}\right.[/itex]

a) Find ##C## and ##D## so that ##y\in C^1(\mathbb{R})##
b) For A>a calculate ##\int_{-A}^{A}(1-({y}')^2)dx##
c) Is it possible to find ##C## and ##D## so that ##y\in C^2(\mathbb{R})##?

Homework Equations

The Attempt at a Solution

Could somebody please check if there is anything ok?

a)
##{y}'(x)=\left\{\begin{matrix}
-1 ;& x<-a\\
2Cx;& -a<x<a\\
1;& x>a
\end{matrix}\right.##

Than ##{y}'(a)=-1=2Ca##, therefore ##C=\frac{1}{2a}##.

We also know that ##y(a)=Ca^2+D=a## therefore ##D=\frac{a}{2}##.

b)
For ##A>a## and ##y(x)=\left\{\begin{matrix}
-x ;& x<-a\\
\frac{1}{2a}x^2+\frac{a}{2};& -a<x<a\\
x;& x>a
\end{matrix}\right.## the integral is

##\int_{-A}^{A}(1-({y}')^2)dx=\int_{-A}^{-a}(1-({y}')^2)dx+\int_{-a}^{a}(1-({y}')^2)dx+\int_{a}^{A}(1-({y}')^2)dx##

First and last integral are both 0 ahile the second is ##\int_{-a}^{a}(1-({y}')^2)dx=\int_{-a}^{a}(1-(\frac{x}{a})^2)dx=\frac{16}{15}a##

That's IF I didn't make a mistake...

c)
##{y}''(x)=\left\{\begin{matrix}
0;& x<-a\\
2C;& -a<x<a\\
0;& x>a
\end{matrix}\right.##

Everything suggests that ##C=0##, therefore the answer is NO.

 

[haruspex]

I think you meant y'(a) = 1, not -1.
I don't see how you get 16/15 in the final step in part b.
Other than that, all looks good.

 

[skrat]

Yes, I meant y'(a)=1.

Thank you!

 

[haruspex]

Yes, I meant y'(a)=1.

Thank you!

But what about the 16/15? That looks wrong to me.

 

[skrat]

But what about the 16/15? That looks wrong to me.

=) It is also wrong. The right result should be ##2a-\frac{2}{3}a=\frac{4}{3}a##.

 

[haruspex]

=) It is also wrong. The right result should be ##2a-\frac{2}{3}a=\frac{4}{3}a##.

Agreed.