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Alice and Bob go for a jog in the same direction along a straight path. For $0\leq t \leq20$, Alice’s velocity at time t is given by $A(t)=\frac{6010}{t^2-3t+50.5}$ meters per minute, and Bob’s velocity at time t is given by $B(t)=8.5t^3e^{-0.45t}$ meters per minute. Both of these velocities are always positive. Alice is 12 meters ahead of Bob at time $t=0$, and she remains ahead of Bob for $0\leq t\leq 20$.

What is the maximum distance between Alice and Bob over the time interval $0\leq t \leq 20$?

The answer is $d=1413.23$ when $t=4.58$.

I'm confused on how to get this answer...

I did $\int_0^{20} \frac{6010}{t^2-3t+50.5} - \int_0^{20} 8.5t^3e^{-0.45t}$

$\int_0^{20}\frac{6010}{t^2-3t+50.5} =1232.323$

$\int_0^{20} 8.5t^3e^{-0.45t}=1217.313$

and got $1232.323-1217.313=15.01=d$

I thought that the distance between Alice and Bob is $\int A(t)-\int(B(t)$ so I'm confused where I went wrong.

How can I solve this?

user130306
  • 1,890
  • You've got some elements correct, but how did you get from the difference in the integrals to $t \approx 15.05 $? The issue will be somewhere between those two statements, so if you show us your work, we can help pinpoint what went wrong. – Amaan M Mar 02 '21 at 01:06
  • thanks, i just updated it with what I did – user130306 Mar 02 '21 at 01:10

1 Answers1

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You've calculated the overall difference in the distances they traveled between $t=0$ and $t=20$, not the difference at a specific point in time. What we need is the difference as a function of $t$.

We're given $A(t)=\frac{6010}{t^2-3t+50.5}$ meters per minute, and $B(t)=8.5t^3e^{-0.45t}$. Since we know Alice is ahead of Bob at any given time, and we know that at $t = 0$ Alice is $12$ meters ahead of Bob, we can write our distance function as follows:

$$ d(t) = 12 + \int_0^t A(\tau) - B(\tau) d\tau $$

This is similar to what you wrote, except you integrated from $\tau=0$ to $\tau=20$. Now, the key to this problem is that we want to maximize that distance, so all we have to do is differentiate and set the derivative equal to $0$, which is very easy with the Fundamental Theorem of Calculus:

$$ \frac{d}{dt} d(t) = \frac{d}{dt} \left( 12 + \int_0^t A(\tau) - B(\tau) d\tau \right) = A(t) - B(t)$$ $$ \Rightarrow A(t) - B(t) = 0 $$ $$ \Rightarrow A(t) = B(t) $$ $$ \Rightarrow \frac{6010}{t^2 - 3t + 50.5} = 8.5t^3e^{-0.45t} . $$

We can use a calculator to solve numerically and find that $A(t) = B(t)$ at $t \approx 4.584$ or $t \approx 16.715$. These are the local extrema on the interval $t\in [0,20]$, so plug those values back into $d(t)$, plus check the endpoints at $t = 0$ and $t = 20$, and one of those four values will be your maximum distance.

Amaan M
  • 2,790
  • Thank you so much! Just one more question: when I do $d(4.584)=12+\int_{0}^{4.584} A(\tau)-B(\tau) d\tau$ I get 365.527. But this is wrong? – user130306 Mar 02 '21 at 01:44
  • Hmm, I'm seeing the same thing. The max still occurs at $t = 4.584$. Are you sure you've copied the functions exactly as they were in the problem? And that the answer source should be correct? – Amaan M Mar 02 '21 at 02:07
  • I'm 100% sure I copied the question correctly however the answer source might not be correct. So just to double check with you for $t=0,d=0,$, and $t=4.584, d=365.527$ and $t=16.715, d=13.344$ and $t=20, d=27.01$ – user130306 Mar 02 '21 at 02:18
  • if we both got the same thing then the answer source must be wrong – user130306 Mar 02 '21 at 02:19
  • At $t = 0$, you should have $d = 12$, don't forget your initial condition! I have the same answers for everything else. – Amaan M Mar 02 '21 at 02:22
  • oh right! yes thank you so much for all of your help. I understand now – user130306 Mar 02 '21 at 02:23
  • You're very welcome! – Amaan M Mar 02 '21 at 02:24