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A work crew is digging a pipeline. 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 lay on the bottom of the trench.

(a) What is the radius of the largest pipe that will lay 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 $a$ that will make the pipe lay in the bottom of the trench?

Any tips for starting points on how to solve this problem?

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  • 1
    $\begingroup$ Drawing a diagram is always a good start. $\endgroup$ – barrycarter Apr 2 '16 at 0:35
  • $\begingroup$ typo? largest value of $a$ that will lay in... $\endgroup$ – Narasimham Apr 2 '16 at 1:12
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(a) The trench cross-section has the equation $y=x^2$. By symmetry, the circular pipe cross-section that lies at the bottom of the trench will have its center on the $y$-axis. To reach the bottom of the trench, if the radius of the circle is $r$ then the center must be at the point $(0,r)$. You can figure out the equation for the circle from that information.

For the pipe to "lay at the bottom of the trench" there must be exactly one intersection between the two curves $y=x^2$ and the circle, namely at the origin. If there are more intersections, the pipe will not fall all the way to the bottom.

So find the intersections between the two curves, and find which values of $r$ will allow exactly one intersection. The largest such value is your desired pipe radius.

(b) Use the same idea. Find the equation for the circle of radius $3$ centered at $(0,3)$ and find the intersections with $y=ax^2$. Then find the largest value of $a$ that allows exactly one intersection.


There are other ways to solve this problem, such as using calculus to find the curvature of the parabola at its bottom. The method I gave above is suitable for pre-calculus, which is your tag for the problem.

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  • $\begingroup$ Analyzing curvature at the bottom of the parabola is not sufficient, by itself (see my comment to he answer from @Narasimham). Your argument about number of intersections is the correct one, I think. $\endgroup$ – bubba Apr 2 '16 at 1:58
  • $\begingroup$ @bubba: I left out one detail, which is to make sure that (for the found value of $r$) the circle, when translated straight up, has no intersections with the parabola. That is true for this problem, and I thought that was a little too picky for this level. $\endgroup$ – Rory Daulton Apr 2 '16 at 10:11
  • $\begingroup$ I think your answer is fine, except for your statement that the problem can also be solved by finding the curvature at the bottom of the ditch. That's true for a parabolic ditch, but it's not true in general, and it would take some work to figure out what special properties of the parabola are relevant. $\endgroup$ – bubba Apr 2 '16 at 10:19
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Standard equation of parabola

$$ y= \frac{x^2 }{2 R} \; ,\frac{y''}{(1+y'^2)^{3/2}} =1/R$$ is the curvature that should be made to match. ( The slope at bottom $y'$ vanishes), i.e., reciprocal of radius

First case

$$ 2 R = 1;\, R =\frac12 $$

Second case is general

comparing the coefficients, general formula $y = \dfrac{x^2}{1/a} $

$$ 1/a = 2\, R ; \, a = \frac{1}{2R} = \frac16. $$

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  • $\begingroup$ Why does equal curvature ensure that the pipe will fit? What if the curve describing the ditch had zero curvature at its bottom (like $y = x^4$)? By your argument, any pipe would fit, no matter how large it's radius, and this is obviously not true. $\endgroup$ – bubba Apr 2 '16 at 1:55
  • $\begingroup$ The OP defines trench as parabolically shaped. $\endgroup$ – Narasimham Apr 2 '16 at 10:19
  • $\begingroup$ Yes, and you must have made use of some special property of this particular parabolic shape somewhere in your argument, because you're argument doesn't work, in general. Do you know what property you used, and where? $\endgroup$ – bubba Apr 2 '16 at 10:58
  • $\begingroup$ I like the use of the second derivative in a pre-calculus problem :) $\endgroup$ – barrycarter Apr 2 '16 at 16:54
  • $\begingroup$ :), his question related to shapes reachable with elementary differentiation. When he tagged conic sections, circles i assumed he would have had some basic knowledge of calculus...Otherwise he should straight have asked for trench width, pipe radius relation. $\endgroup$ – Narasimham Apr 2 '16 at 18:49

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