- #1

skrat

- 748

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## 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.

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