What is another, perhaps quicker and nicer way of solving this question?

Question

In a calculus book, a question reads:

A car is traveling at night along a highway shaped like a parabola with its vertex at the origin. The car starts at a point 100 m west and 100 m north of the origin and travels in an easterly direction. There is a statue located 100 m east and 50 m north of the origin. At what point on the highway will the car's headlights illuminate the statue?

I want to know how you would solve this problem, because the method I am using is very round-about. There must be a much more logical way of solving this problem, and I hope you will share it with me. Thank you.

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This is my solution..

The function appears to be of the form $$y = f(x) = ax^2$$

$$100 = a(100)^2$$ $$a = 1/100$$ $$y = \frac{x^2}{100}$$

Then I decided (by solving a few tangent equations and generalizing) that the tangent line expression at a point $(p, f(x))$ is given by the function:

$$T(p) = T_m(p)*x + T_b(p)$$

Where $T_m(p)$ represents the slope for a point, $p$ on $f(x)$ and $T_b(p)$ represents the y-intercept value.

$$T_m(p) = \frac{df(p)}{dx} =\frac{p}{50}$$

$$T_b(p) = f(p) - \frac{df(p)}{dx} * p = -\frac{p^2}{100}$$

So given that the statue rests on $(100, 50)$ we know the equation of this tangent line outputs $50$ when $100$ is input.

$$T(p) = \frac{p}{50}*x - \frac{p^2}{100}$$

$$50 = \frac{p}{50}*100 - \frac{p^2}{100}$$

the $x$ position of the car is at a point $p$ such that $p$ satisfies the equation.

I end up getting the positive result $x = 100 - 50\sqrt{2}$

Answer 1

Your solution is great. Here's my approach.

We know that the parabola has equation $f(x)=\dfrac{x^2}{100}$. Now suppose that the car illuminates the statue when $x=p$. Then the car's location must be at $\left(p,\frac{p^2}{100}\right)$, and the line connecting this point to $(100,50)$ must have a slope of $f'(p)=\dfrac{p}{50}$. Hence, using the slope formula, we obtain: $$ \begin{align*} \dfrac{\frac{p^2}{100}-50}{p-100} &= \dfrac{p}{50} \\ \dfrac{p^2-5000}{100(p-100)} &= \dfrac{p}{50} \\ \dfrac{p^2-5000}{2(p-100)} &= p \\ p^2-5000 &= 2p(p-100) \\ p^2-5000 &= 2p^2-200p \\ 0 &= p^2-200p+5000 \\ p &= 100 \pm 50\sqrt2 \\ \end{align*} $$

Hence, since we know that the car travels from left to right, we reject the larger solution and conclude that the car illuminates the statue at $x=100-50\sqrt2$.

Answer 2

A coworker came to me with this problem today, and I found the solution a slightly different way. My approach wasn’t any simpler, but I still thought I would share:

As both of your approaches demonstrated, $f(x)=\frac{x^2}{100}$ and $f'(x)=\frac{x}{50}$. We also know that the tangent line intersects $(100,50)$. So the tangent line can be defined as: $$y=mx+b$$ $$50=\frac{x}{50}100+b$$ $$b=50-2x$$ $$y=\frac{x}{50}x+50-2x$$ $$y=\frac{x^2}{50}-2x+50$$ Solve the system: $y=\frac{x^2}{100}$ and $y=\frac{x^2}{50}-2x+50$. $$\frac{x^2}{100}=\frac{x^2}{50}-2x+50$$ $$\frac{x^2}{100}-2x+50=0$$ $$x^2-200x+5000=0$$ $$x=100±50\sqrt{2}$$ I ignored that value where $x>100$, so the car's location at which its headlight's illuminate the statue is:$$(100-50\sqrt{2}, 150-100\sqrt{2})$$