Mechanical Design: Application of integration...please help

[roxstar1]

The surface of a machine part is the region between the graphs of y1 = l x l and y2 = 0.08x^2 + k

a) find k if the parabola is tangent to the graph of y1


b) find the surface area of the machine part


I am not sure how to find k in part a. I tried setting the two equations equal to each other and solving for k in terms of x, but this led to problems when trying to solve part b. any suggestions would be greatly appreciated.

 

[soroban]

Re: Mechanical Design: Application of integration...please h

Hello, roxstar1!

I'll do the first part . . .

The surface of a machine part is the region between the graphs of: $\displaystyle y_1\,=\,|x|$ and $\displaystyle y_2\,=\,0.08x^2\,+\,k$

a) find $\displaystyle k$ if the parabola is tangent to the graph of $\displaystyle y_1$

  \  *          |          *  /
    \           |           /
      \*        |        */
        o       |       o 
       Q  \ *   |   * /  P
            \  ***  /
              \ | /
     -----------+-----------

The parabola is tangent to $\displaystyle \,y_1\,=\,|x|\,$ at $\displaystyle P$ and $\displaystyle Q$.
$\displaystyle \;\;$Hence, they have equal slopes at $\displaystyle P$ and $\displaystyle Q$.

At $\displaystyle P$, the slope of $\displaystyle y_1\,=\,|x|\,$ is 1.
$\displaystyle \;\;$The slope of the parabola is: $\displaystyle \,y'\,=\,0.16x$
$\displaystyle \;\;$Then: $\displaystyle \,0.16x\,=\,1\;\;\Rightarrow\;\;x\,=\,6.25$
Hence, point $\displaystyle P$ is: $\displaystyle (6.25,\,6.25)$

This point satisfies the equation of the parabola,
$\displaystyle \;\;$so we have: $\displaystyle \;6.25 \:=\:0.08(6.25)^2\,+\,k\;\;\Rightarrow\;\;k\,=\,3.125$

The equation of the parabola is: $\displaystyle \:y_2\:=\:0.08x^2\,+\,3.125$

 

[Unco]

k will only move the parabola up and down, so setting 0.08x^2 + k = x, as you have, and setting the discriminant to zero will certainly work. (Using the derivative will work just as well.)

(a) gives you the x-ordinate of the intersection (provided we are sticking with the parabola being tangential to y1) and the equation for y2, allowing you to determine the area by integration -- again, just consider the area between y = x and y2 from x=0 to x=<from (a)>, and double it.