# Calculus - Related Rates Problem

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1. Mar 11, 2016

### RoyceB

• Originally posted in a non-homework section, so missing the template
Question: Two bikers leave a diner at the same time. Biker Slim rides at 85kmh [N] and Biker Haug rides at 120kmh [NE]. How fast is the distance between them changing 40 minutes after they left?

I suggest looking at my photos of the triangles and such, as explaining it over text can be a bit confusing.

Attempt:

http://imgur.com/a/lkyvL [Photos are out of order, sorry.]

So I drew out the triangle, and it is not a right angled triangle by nature. But I can break it apart into two right angle triangles.

From my photo, I solve for the speed of side X by using Sine Law, and also I found that length X = length (A-Z). From there and knowing (A) from the beginning, and having found (A-Z) I was able to determine Z using simple algebra. After that, and using pythagorean's theorem I was able to find side C with sides Z and X.

Once that was done, I took the lengths of X, Z, and C and put them into pythagorean's theorem and derived it, in search of dc/dt or the rate at which the distance between them.

Another guy in my class agrees with my answer but my teacher told me it's around 49. My teacher created and solved it himself so there is not really a verification we can do, he also makes lots of mistakes (as everyone does in math) so were are unsure who is right or if we made a flub somewhere. If someone could look over my work and point out/help me realize my mistake it would be appreciated.

2. Mar 11, 2016

### Staff: Mentor

Your triangle is incorrect. You have the long side as 120, which would be the distance (in km) that Haug went in one hour. Your triangle also doesn't include the rate that Slim is travelling.

At any time t (in hrs.), Slim will have covered 85t km and Haug will have covered 120t km. I used the Law of Cosines to get an expression for the distance between the two bikes at time t, and then differentiated it. In 40 minutes (2/3 hr), I get the distance between them changing at a rate of about 113 km/hr per hour.
Edit: I have a mistake in my work, due to an extra power of t in my distance.

Note that coursework problems such as this one should be posted in the Homework & Coursework sections, not in the technical math sections. I have moved your thread.

Last edited: Mar 11, 2016
3. Mar 11, 2016

### Ray Vickson

Please type out your actual answer here: every time I try to look at your photos I get sucked into an environment where I have to exit this Forum and sign in again. That is a common problem with attached images, and is one reason I rarely ever look at them; you should type out your work.

Anyway, as far as I can tell, your answer is correct.

I would do it using cartesian coordinates, rather than geometry: the coordinates of S and H are $(x_S,y_S) = (0,85 t)$ and $(x_H,y_H) = (120 \cos(\pi/4) t, 120 \sin(\pi/4) t)$ after $t$ hours of riding. Using $\sin(\pi/4) = \cos(\pi/4) = 1/\sqrt{2}$, the distance between them is
$$D(t) = \sqrt{(x_S-x_H)^2 + (y_S-y_H)^2} = ct,$$
where $c = \sqrt{21625 -10200 \,\sqrt{2}}$.

4. Mar 12, 2016

### epenguin

Er is the "40 minutes after they left" a red herring? Isn't it the same at all times?

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