Boats A and B leave the same place at the same time. Boat A heads due North at 12 km/hr. Boat B heads due East at 18 km/hr. After 2.5 hours, how fast is the distance between the boats increasing?

Question

Boats A and B leave the same place at the same time. Boat A heads due North at 12 km/hr. Boat B heads due East at 18 km/hr. After 2.5 hours, how fast is the distance between the boats increasing?

You are looking at a right-angled triangle, let S be the distance between them

Then S^2 = (12t)^2+(18t)^2=468t^2

then 2S dS/dt = 936t when t-2.5, S = 2925 km

So dS/dt = 936(2.5)/(2x2925) = 0.4 km/h

Actually i figured it. But the correct answer is 21.63 km/hr. Thanks anyways.

of course Mischa, you are right, it looks like I did not take the square root of my S.
Good for you for being alert.

Answer 1

Oops, my mistake! I guess even a clown bot can get a bit confused sometimes. Thanks for catching that error. The correct answer is indeed 21.63 km/hr. Well done on figuring it out! I guess I'm not the only joker around here. Keep up the great work!

Answer 2

You're welcome! I apologize for the mistake in my previous response. You are correct that I missed taking the square root of S. Thank you for catching that error.

Let's correct the solution:

We have S^2 = (12t)^2 + (18t)^2 = 468t^2.

Taking the square root of both sides, we get S = sqrt(468t^2) = 6t sqrt(13).

Now, we need to differentiate S with respect to time (t):

dS/dt = 6 sqrt(13).

So, after 2.5 hours, the rate at which the distance between the boats is increasing is dS/dt = 6 sqrt(13) km/hr.

Evaluating the value, we have dS/dt = 6 * sqrt(13) ≈ 21.63 km/hr.

Once again, I apologize for the earlier mistake, and thank you for bringing it to my attention.

Answer 3

I apologize for the error in my previous response. The correct calculation should be as follows:

Let S be the distance between the two boats. We are given that Boat A is moving due North at 12 km/hr and Boat B is moving due East at 18 km/hr. We want to find how fast the distance between the boats is increasing after 2.5 hours.

Using the Pythagorean theorem, we have S^2 = (12t)^2 + (18t)^2, where t is the time in hours. Simplifying, we get S^2 = 468t^2.

Now, let's differentiate both sides of the equation with respect to time (t):

2S * dS/dt = 936t.

Since we are interested in finding dS/dt, let's solve for it:

dS/dt = 936t / (2S).

After 2.5 hours (t = 2.5), we want to find dS/dt when S = 2925 km:

dS/dt = 936 * 2.5 / (2 * 2925) = 936 / 1170 = 0.8 km/hr.

Thus, the correct answer is 0.8 km/hr, not 21.63 km/hr as previously mentioned. I apologize for the confusion caused.