# Almost got it, but not quite.

• Mar 3rd 2010, 10:21 AM
weekendclimber
Almost got it, but not quite.
So, I'm teaching myself calculus using the Open Course Ware at MIT and a few other online resources. I'm new to this forum and am hoping that if I get stuck anywhere I can bounce questions off someone here.

I think have most of this problem figure out, but I'm running into a dead-end. This shows up in the Problem Set 1, for course 18.01 'Single Variable Calculus'. Anyway here's the problem:

On the planet Quirk, a cell phone tower is a 100-foot pole on top of a green mound 1000 feet tall whose outline is described by the parabolic equation y = 1000 − x^2 (1000 minus x squared). An ant climbs up the mound starting from ground level (y = 0). At what height y does the ant begin to see the tower?
Now, I can derive the equation of the tangent line that meets at point (0, 1100) from the equation y = 1000 - x^2 as y = 1100 - 2x. Since there is only one point where these two equations meet I thought I should be able to use the quadradic equation to solve 0 = x^2 - 2x + 100 but I get an imaginary number ( +- sqrt( -396 )). Can anyone tell me where I'm off track? Thanks
• Mar 3rd 2010, 10:48 AM
Quote:

Originally Posted by weekendclimber
So, I'm teaching myself calculus using the Open Course Ware at MIT and a few other online resources. I'm new to this forum and am hoping that if I get stuck anywhere I can bounce questions off someone here.

I think have most of this problem figure out, but I'm running into a dead-end. This shows up in the Problem Set 1, for course 18.01 'Single Variable Calculus'. Anyway here's the problem:

On the planet Quirk, a cell phone tower is a 100-foot pole on top of a green mound 1000 feet tall whose outline is described by the parabolic equation y = 1000 − x^2 (1000 minus x squared). An ant climbs up the mound starting from ground level (y = 0). At what height y does the ant begin to see the tower?
Now, I can derive the equation of the tangent line that meets at point (0, 1100) from the equation y = 1000 - x^2 as y = 1100 - 2x. Since there is only one point where these two equations meet I thought I should be able to use the quadradic equation to solve 0 = x^2 - 2x + 100 but I get an imaginary number ( +- sqrt( -396 )). Can anyone tell me where I'm off track? Thanks

Your mistake is to use the slope of the line as 2.

the line equation you are using does not intersect the parabola,
hence the reason for your complex answers.

The derivative (instantaneous slope) of the parabola is -2x.

The slope of the line you are looking for is "m".

$\displaystyle 1100-mx=1000-x^2$

$\displaystyle x^2-mx+100=0$

$\displaystyle x=\frac{m\pm\sqrt{m^2-400}}{2}$

Therefore, to obtain a single positive solution for x

$\displaystyle m^2=400$

$\displaystyle m=\pm20$

The ant can can climb up either side, hence the positive and negative m.
• Mar 3rd 2010, 11:11 AM
Opalg
Quote:

Originally Posted by weekendclimber
On the planet Quirk, a cell phone tower is a 100-foot pole on top of a green mound 1000 feet tall whose outline is described by the parabolic equation y = 1000 − x^2 (1000 minus x squared). An ant climbs up the mound starting from ground level (y = 0). At what height y does the ant begin to see the tower?

At the point $\displaystyle (x_0,1000-x_0^2)$ on the parabola, the slope is $\displaystyle -2x_0$ and the equation of the tangent is $\displaystyle y - (1000-x_0^2) = -2x_0(x-x_0)$. The condition for this line to pass through the point (0,1100) is $\displaystyle 100-x_0^2 = -2x_0^2$, or $\displaystyle x_0^2 = 100$. So $\displaystyle x_0 = \pm10$, and the corresponding value of y is $\displaystyle 1000-100 = 900$.

Quote:

Originally Posted by weekendclimber
Now, I can derive the equation of the tangent line that meets at point (0, 1100) from the equation y = 1000 - x^2 as y = 1100 - 2x.

The mistake there is that you are using the same notation (x,y) to denote a fixed point on the parabola (the point where the tangent touches it), and a variable point on the tangent line. In my solution, I used $\displaystyle (x_0,y_0)$ for the point on the parabola, and (x,y) for a point on the tangent line. That way, you avoid confusing them.
• Mar 3rd 2010, 11:26 AM
weekendclimber
That makes sense. Thank you for your help.