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.