# Find constants of function with given conditions

• skrat
In summary: The second integral is ##\int_{-a}^a \left(1 - \left(\frac{x}{a}\right)^2\right)\,dx = \int_{-a}^a \left(1 - \frac{x^2}{a^2}\right)\,dx = \int_{-a}^a \frac{a^2 - x^2}{a^2}\,dx##. Now use the fact that ##\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx## when ##f## is an even (or odd) function.
skrat

## Homework Statement

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

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})##?

## 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.

Last edited:
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.

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

Thank you!

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

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

haruspex said:
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##.

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

Agreed.

## 1. How do I find the constants of a function with given conditions?

To find the constants of a function with given conditions, you need to have a clear understanding of the conditions and the function itself. Then, you can use algebraic manipulation and solving techniques to determine the values of the constants. It may also be helpful to draw a graph of the function to visualize the conditions and the possible values of the constants.

## 2. What are the common conditions given when finding constants of a function?

The conditions given when finding constants of a function can vary, but some common ones include the function's domain and range, the function's behavior at certain points or intervals, and the function's relationship to other functions or equations.

## 3. Can I use calculus to find the constants of a function?

Yes, calculus can be used to find the constants of a function with given conditions. Techniques such as differentiation and integration can help determine the values of the constants. However, it is important to have a solid understanding of calculus concepts before using them to find constants.

## 4. Is there a specific method or formula for finding constants of a function?

There is no one specific method or formula for finding constants of a function with given conditions. The approach will depend on the specific conditions and the function itself. In some cases, a trial and error method may be necessary, while in others, algebraic manipulation or calculus techniques may be more effective.

## 5. Can I use technology to find the constants of a function?

Yes, technology such as graphing calculators or online graphing tools can be helpful in finding the constants of a function. These tools can graph the function and its given conditions, making it easier to visualize and determine the values of the constants. However, it is important to understand the underlying concepts and not solely rely on technology to solve the problem.

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