Brain Teaser - Area enclosed by a parabola and a line
|
09-13-2015, 04:23 PM
(This post was last modified: 09-13-2015 04:26 PM by Thomas Klemm.)
Post: #6
|
|||
|
|||
RE: Brain Teaser - Area enclosed by a parabola and a line
Let's assume that \(P=(u, u^2)\) is the intersection of the function \(f(x)=x^2\) and its normal line \(g(x)\).
Then from \(f'(x)=2x\) we can conclude that: \(\begin{bmatrix} x-u \\ y-u^2 \end{bmatrix}\cdot \begin{bmatrix} 1 \\ 2u \end{bmatrix}=0 \) Thus: \(x-u+2u(y-u^2)=0\) From the equation above we can find the formula for \(g(x)\): \(y=\frac{u-x}{2u}+u^2\) Thus we get for the difference \(g(x)-f(x)\): \(y=\frac{u-x}{2u}+u^2-x^2=(u-x)(\frac{1}{2u}+u+x)\) When we set both factors to 0 we can find the lower and upper limit of the integral. We already know that the upper limit is: \(x=u\). The lower limit is: \(x=-\left (u+\frac{1}{2u}\right ) \) Now let's pick the HP-15C and write two small programs. Program A Calculates the limits to integrate program B. Code: LBL A Program B Calculates the difference \(g(x)-f(x)\). Code: LBL B We can test this for \(x = 1\): 1 ENTER ENTER ENTER f A 2.6042 From the plot we can conclude that the minimum is somewhere between 0.1 and 1: Here comes the tricky part. We have to force the solver to only search for positive values. The easiest way to do this is to add the ABS function to the program A: Code: LBL A Now we can try to solve for a solution: 0.1 ENTER 1 f SOLVE A Error 8 This means that no solution was found. However we can still find the desired result in the stack: z: 1.3339 y: 0.5354 x: 0.5085 Cheers Thomas |
|||
« Next Oldest | Next Newest »
|
User(s) browsing this thread: 1 Guest(s)