# Math Help - Maximum Difference

1. ## Maximum Difference

I have a problem that I am working on called "Calvin and Phoebe's Velocity Problem."

The two characters distance equations are given as:
Calvin: s(t)=t^3
Phoebe: s(t)=e^(t) - 1
s is feet and t is seconds.

Part one was to decided which one had the greatest initial acceleration. I figured it was Phoebe by entering 1 into each equation
Phoebe: 1.71828
Calvin: 1

Next was to find when Calvin catches up with phoebe. This was there I had problems. I was not able to do this algebraically, but finding the intersection graphically I got t=1.545. But I need to know how to do it algebraically.

I also have to find the maximum distance between the two cars on the interval (1,1.545). I started by setting D=(e^(t) - 1) - (t^3)
I thought I had to then take the derivatives and set that equal to zero, then solve for t, but when I did that i get stuck when i reach t=ln(3t^2). Where would I need to go from there? Solving graphically I got .91 ft

I also have to find maximum difference in velocity, which I am guessing is done similarly to the one previously. Again I get stuck when setting the derivatives equal to 0. Graphically I got .2157 ft/sec.

It would be greatly appreciated if anybody could help me with what I am doing wrong, how to find these answers algebraically. Thank you.

2. Originally Posted by TheEdge
I have a problem that I am working on called "Calvin and Phoebe's Velocity Problem."

The two characters distance equations are given as:
Calvin: s(t)=t^3
Phoebe: s(t)=e^(t) - 1
s is feet and t is seconds.

Part one was to decided which one had the greatest initial acceleration. I figured it was Phoebe by entering 1 into each equation
Phoebe: 1.71828
Calvin: 1

Next was to find when Calvin catches up with phoebe. This was there I had problems. I was not able to do this algebraically, but finding the intersection graphically I got t=1.545. But I need to know how to do it algebraically.

I also have to find the maximum distance between the two cars on the interval (1,1.545). I started by setting D=(e^(t) - 1) - (t^3)
I thought I had to then take the derivatives and set that equal to zero, then solve for t, but when I did that i get stuck when i reach t=ln(3t^2). Where would I need to go from there? Solving graphically I got .91 ft

I also have to find maximum difference in velocity, which I am guessing is done similarly to the one previously. Again I get stuck when setting the derivatives equal to 0. Graphically I got .2157 ft/sec.

It would be greatly appreciated if anybody could help me with what I am doing wrong, how to find these answers algebraically. Thank you.
you got part one correct, but your method is wrong.

initial acceleration is measured at time t = 0

for calvin ...
$s(t) = t^3$
$v(t) = 3t^2$
$a(t) = 6t$

for phoebe ...
$s(t) = e^t - 1$
$v(t) = e^t$
$a(t) = e^t$

note that at t = 1, calvin's acceleration is greater ... $6 > e$

at t = 0, phoebe's acceleration is greater ... $e^0 = 1 > 6(0) = 0$

part 2 is correct ... calvin catches up at t = 1.545
you cannot solve this equation using methods of elementary algebra.

distance between the two is the absolute value of the difference in their position functions ...

D = |t^3 - (e^t - 1)|

graph this on the given interval to find the max difference.

difference in velocity, V = |3t^2 - e^t|

once again, note that this is strictly a calculator driven problem ... you will not be able to solve these equations algebraically.