Hi there, could you please help me with this confusing question:

After (t) seconds, a projectile is at height (h) metres where h(t)=300+24t-3t^(2). Find the rate of change of height at t=2,t=3 and t=4 seconds (use a time interval of 0.01) seconds. Then describe what is happening to the projectile. Sketch the path.

I wouldn't know find the rates of change using the time interval of 0.01 seconds and then I wouldn't know ow to sketch it.

Thanyou

I guess it wants you to, for t=2, find the height at t = 1.995 and at t = 2.005 and find the change in height, then you can divide this by 0.01 to find the change in height per second, which is th rate of change at t=2 in a time interval of 0.01.

Does that make sense?

No sorry it doesn't really make sense to me. -_-

That's okay, I hate the word "rate" because it can mean so many things in different contexts.

You must have example questions that you have worked through in class or something? What I wrote above is what I would do if I was given the question, but there are certainly something's your teacher might do differently.

Thanks anyway
Could you please do one for an example
That would be much appreciated

Okay let's see, we want the rate of change at t=2 in an interval of 0.01, so we need to take the height at t = 2-0.005 and t = 2+0.005

so this gives us h(1.995)=335.939925m and h(2.005)=336.059925m

so the change is 336.059925-335.939925 = 0.12m in 0.01 seconds.

so the rate of change is 0.12/0.01 = 12 meters per second

Oh cool
Thanyou so much for your help, much appreciated

We could also use the instantaneous rate of change of y w.r.t. x

m_{tan}=\lim_{x_{1}\rightarrow{x_{0}}}\frac{f(x_{1 })-f(x_{0})}{x_{1}-x_{0}}


\frac{-3t^{2}+24t+300-336}{t-2}=-3(t-6)

\lim_{t\to\2}-3(t-6)=\fbox{12}


Just a thought.