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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
+
1/x
+
CHS
x<>y
STO 0
∫ B
RTN

Program B
Calculates the difference \(g(x)-f(x)\).
Code:
LBL B
RCL- 0
CHS
x<>y
RCL+ 0
RCL 0
RCL+ 0
1/x
+
*
RTN

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:

[Image: attachment.php?aid=2526]

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
+
1/x
+
ABS
CHS
x<>y
ABS
STO 0
∫ B
RTN

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


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RE: Brain Teaser - Area enclosed by a parabola and a line - Thomas Klemm - 09-13-2015 04:23 PM



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