Find constants of function with given conditions

[skrat]

Homework Statement

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

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.