# Thread: rate of change and extrema

1. ## rate of change and extrema

Hi, I have a strangely worded problem that I'm having problems interpreting/solving

(calculator active) The rate of change R, in km/hr, of the altitude of a hot air balloon is given by $R(t) = t^3 - 4t^2 + 6$ for time 0<(or equal to) t <(or equal to) 4, where t is measured in hours. Assume the balloon is initially at ground level.

a. what is the max altitude of the balloon during the interval given above?

Well, since the equation given above is the rate of change I plugged it into my calculator to find the critical points, I got t=1.572 and t=3.514. I then took the derivative of that equation and plugged in my critical points to determine the max. I got a max from t=1.572

I know there is a max altitude at time = 1.572 but how do I figure out what the altitude is at this time? I know I have to come up with a position equation but I'm a little lost on that.

All help is appreciated, thanks.

2. ## Re: rate of change and extrema

The rate of change of a function is its derivative. Saying that "The rate of change R, in km/hr, of the altitude of a hot air balloon is given by $R(t)= t^3= 4t^2+ 6$" means that, taking h(t) to be the altitude (in km) at time t (in hours), $\frac{dh}{dt}= t^3- 4t^2+ 6$. Take the "anti-derivative" to find h(t). Of course, that will involve a arbitrary "constant of integration". Use "Assume the balloon is initially at ground level" to find that constant.

3. ## Re: rate of change and extrema

Originally Posted by HallsofIvy
The rate of change of a function is its derivative. Saying that "The rate of change R, in km/hr, of the altitude of a hot air balloon is given by $R(t)= t^3= 4t^2+ 6$" means that, taking h(t) to be the altitude (in km) at time t (in hours), $\frac{dh}{dt}= t^3- 4t^2+ 6$. Take the "anti-derivative" to find h(t). Of course, that will involve a arbitrary "constant of integration". Use "Assume the balloon is initially at ground level" to find that constant.
Oh ok, I know about anti differntiation, just didn't know what to use as a constant. Using your help, I came up with an altitude equation:

$h(t) = (1/4)t^4 - (4/3)t^3 + 6t + 0$

Now to find the max altitude I plugged in the critical point t = 1.572 which is a "max" into my altitude equation and I get:

5.779 km, does this seem correct?

4. ## Re: rate of change and extrema

also, there is another question for this equation:

b. at what time is the altitude of the balloon increasing most rapidly?

I used the derivative of the rate of change to find this, for wherever the value of the derivative of the rate of change is greatest should be where the altitude is increasing most rapidly? Am i correct on this?

If so, then the answer should be t = 4

5. ## Re: rate of change and extrema

Originally Posted by TacticalPro
5.779 km, does this seem correct?
Yes.

Originally Posted by TacticalPro
b. at what time is the altitude of the balloon increasing most rapidly?

I used the derivative of the rate of change to find this, for wherever the value of the derivative of the rate of change is greatest should be where the altitude is increasing most rapidly?
No, when the range of change (the velocity) is the greatest, the altitude is increasing most rapidly.

6. ## Re: rate of change and extrema

Originally Posted by emakarov
Yes.

No, when the range of change (the velocity) is the greatest, the altitude is increasing most rapidly.
Oh, well then wouldn't it still be t = 4? I plugged the "velocity" equation into my calc and at t=4 it is increasing the most

7. ## Re: rate of change and extrema

Originally Posted by TacticalPro
Oh, well then wouldn't it still be t = 4? I plugged the "velocity" equation into my calc and at t=4 it is increasing the most
When the rate of change of altitude is at a maximum either dR/dt=0 or t is an end point of the interval of interest (in this case 0 or 4 hours). To answer this you need to find the roots of dR/dt=0 and evaluate R there and at t=0 and t=4, the maximum you seek is the largest of these values of R, and the time/s is the corresponding time/s.

CB