# Math Help - Coordinate type problem help

1. ## Coordinate type problem help

Dave is going to leave academia and go into business building grain silos. A grain silo is a cylinder with a hemispherical top, used to store grain for farm animals. Here is a 3D view, a cross-section, and the top view.

If Dave is standing next to a silo of cross-sectional radius r = 8
feet at the indicated position, his vision will be partially obstructed. Find the portion of the y-axis that Dave cannot see. (Hint: Let a be the x-coordinate of the point where line of sight #1 is tangent to the silo; compute the slope of the line using two points (the tangent point and (12, 0)). On the other hand, compute the slope of line of sight #1 by noting it is perpendicular to a radial line through the tangency point. Set these two calculations of the slope equal and solve for a. Enter your answer using interval notation. Round your answer to three decimal places.)

I have been stuck on this problem, any help on how to approach it? Thank you!

2. ## Re: Coordinate type problem help

cylinder is represented by a central circle with radius r=8. Write down the equation of that circle. Next find the equations for tangents passing through the point (12,0). You only need one of two tangents since you have symmetry about the abscissa. So if you can calculate the equation of one tangent line passing through the point (12,0) you can easily answer the question. If b is the point of intersection of one tangent line and y-axis then the portion that Dave cannot see is [-b,b]

3. ## Re: Coordinate type problem help

Well, so how do you calculate the line equation in this problem?

4. ## Re: Coordinate type problem help

Okay, I've actually just solved it, but it was a long circuitous route involving measuring height of a triangle (using sin, arcsin, tan etc.) as well as its many sides, and by that figuring out the m of the radial line. After that I deduced the m of our initial line by taking the negative reciprocal of radial line's m... And then it's easy.

So... I have really no idea what does that hint in the problem actually hint at . I think it actually only got me to spend few hours figuring out what it meant instead of really helping me, whereas I could've solved it my way.

So what does it hint at? Maybe I have missed something in my studies... something important.

5. ## Re: Coordinate type problem help

Let's say $(a,b)$ is the point on the upper semi-circle where the tangent line hits. Note that $b = \sqrt{64 - a^2}$. Now the tangent line has formula:

$y = \dfrac{\sqrt{64 - a^2}}{a - 12}(x - 12)$.

The radial line has slope $\dfrac{b}{a} = \dfrac{\sqrt{64 - a^2}}{a}$, and by perpendicularity, we know that:

$\dfrac{\sqrt{64-a^2}}{a-12}\cdot\dfrac{\sqrt{64-a^2}}{a} = -1$ that is:

$64 - a^2 = 12a - a^2 \implies 64 = 12a \implies a = \dfrac{16}{3}$, so $b = \dfrac{8\sqrt{5}}{3}$.

Now, what we WANT to know is the value of $y$ on the tangent line when $x = 0$, doubling this gives the answer.

And we have everything we need to calculate this:

$y = \dfrac{\frac{8\sqrt{5}}{3}}{\frac{-20}{3}}\cdot(-12) = \dfrac{24\sqrt{5}}{5}$ so the desired answer is $\dfrac{48\sqrt{5}}{5}$.

At least, I hope that's right...my old addled brain does make mistakes.

6. ## Re: Coordinate type problem help

Oh, wow, that's nice. Got quite a few insights, thanks.

The answer is surely correct (there are answers in the end of the book).

7. ## Re: Coordinate type problem help

Originally Posted by Deveno
Let's say $(a,b)$ is the point on the upper semi-circle where the tangent line hits. Note that $b = \sqrt{64 - a^2}$. Now the tangent line has formula:

$y = \dfrac{\sqrt{64 - a^2}}{a - 12}(x - 12)$.

The radial line has slope $\dfrac{b}{a} = \dfrac{\sqrt{64 - a^2}}{a}$, and by perpendicularity, we know that:

$\dfrac{\sqrt{64-a^2}}{a-12}\cdot\dfrac{\sqrt{64-a^2}}{a} = -1$ that is:

$64 - a^2 = 12a - a^2 \implies 64 = 12a \implies a = \dfrac{16}{3}$, so $b = \dfrac{8\sqrt{5}}{3}$.

Now, what we WANT to know is the value of $y$ on the tangent line when $x = 0$, doubling this gives the answer.

And we have everything we need to calculate this:

$y = \dfrac{\frac{8\sqrt{5}}{3}}{\frac{-20}{3}}\cdot(-12) = \dfrac{24\sqrt{5}}{5}$ so the desired answer is $\dfrac{48\sqrt{5}}{5}$.

At least, I hope that's right...my old addled brain does make mistakes.
Beautiful. Although I had a little trouble reconciling the following
Originally Posted by Deveno
so the desired answer is $\dfrac{48\sqrt{5}}{5}$.
It seems out of place somehow. Taking a cue from Sir maxpancho's last post, it seems that this is Problem 4.8 with a corresponding answer from Math 120 Materials Website - Dept of Math, Univ of Washington
Same place as OP.