Insertion of a circle into a parabola

[ryan_kalle]

Im clueless on both parts of this problem

"A motivated work crew of Marines is digging a pipeline through a frozen wilderness in Alaska. The cross section of the trench is in the shape of the parabola y = x^2. The pipe has a circular cross section. If the pipe is too large, then the pipe will not lie on the bottom of the trench.

A. What is the radius of the largest pipe that will lie on the bottom of the trench?


B. If the radius of the pipe is 3 and the trench is in the shape of y = ax^2, then what is the largest value of a for which the pipe will lie on the bottom of the trench? "

A) All I have been able to reason is

y=1X^2

X^2+(Y-h)^2=r^2

B) I have

X^2 + (Y-3)^2=9

y=aX^2

 

[galactus]

This ia a cool problem.

The parabola has equation given as \(\displaystyle y=x^{2}\)

This means the circle has equation of the form \(\displaystyle x^{2}+(y-r)^{2}=r^{2}\).............[1]

Sub \(\displaystyle y=x^{2}\) into [1] and solve for x and we get

\(\displaystyle x=\pm\sqrt{2r-1}\)

and since \(\displaystyle y=x^{2}\), \(\displaystyle y=2r-1\)

\(\displaystyle 2r-1= 0\)

\(\displaystyle r= \frac{1}{2}\)

The largest pipe has a radius of 1/2 units.

Using the same technique as above, the max size of pipe for a parabola \(\displaystyle y=ax^{2}\) is found to be \(\displaystyle r=\frac{1}{2a}\)

So, if r=3 what is a?.

 

[ryan_kalle]

I'm still not understanding. I can't figure from x^2+(y-r)^2=r^2 to x=+/- sqrt(2r-1). I try to sub like you say and I think my arithmetic is flawed.

 

[ryan_kalle]

This is my math and it is probably wrong

x^2 + (x^2-r)^2 = r^2

x^4 + x^2 - r^2 = r^2

x^4 + x^2 = 2r^2

x^2 + x = 2r

Then I get stuck.

 

[galactus]

Yeas, you got a little off kilter.

Anyway:

\(\displaystyle x^{2}+(y-r)^{2}=r^{2}\)

Expand and let \(\displaystyle y=x^{2}\)

\(\displaystyle x^{4}+x^{2}-2rx^{2}+r^{2}=r^{2}\)

\(\displaystyle x^{4}+x^{2}-2rx^{2}=0\)

\(\displaystyle x^{4}+x^{2}(1-2r)=0\)

\(\displaystyle x^{4}-x^{2}(2r-1)=0\)

\(\displaystyle x^{4}=x^{2}(2r-1)\)

Divide by x^2:

\(\displaystyle x^{2}=2r-1\)

Since \(\displaystyle y=x^{2}\):

\(\displaystyle y=2r-1\)

Then, let y=0 and solve for r.

 

[Denis]

This is my math and it is probably wrong
x^2 + (x^2-r)^2 = r^2
x^4 + x^2 - r^2 = r^2

Ya seem to have forgotten the basics...
(x^2 - r)^2 = x^4 - 2rx^2 + r^2

 

[ryan_kalle]

Yeah I figured it out. I went this direction with the first.

y=x^2
x^2 + (y-r)^2 = r^2

y + (y-r)(y-r) = r^2

y + y^2 - 2ry + r^2 = r^2

Subtract r^2 from both sides and factor out a y

y(1-2r)=0

y=0 1-2r=0

r=1/2

Likewise with the second part

y=ax^2
x^2 + (y-3)^2 = 3^2

x^2 + y^2 - 6y + 9 = 9

y=ax^2
x^2=y/a

y/a + y^2 - 6y = 0

y(1/a + y - 6) = 0

y=0

1/a -6 = 0

1/a=6

a=1/6