HELP ASAP

[Ryan Rigdon]

I have worked this problem over and over and getting nowhere with it, its a a pain in the a#$. Please help me work this out.

the problem

If a point moves along a coordinate line so that its directed distance s (in meters) from 0 is given by

s = ((t+2)^3)/(t+3) at time t minutes, find its instantaneous velocity when t = 2.7 seconds.

 

[galactus]

Did you differentiate s(t)?. Velocity is the derivative of the position function.

\(\displaystyle s'(t)=\frac{(t+2)^{2}(2t+7)}{(t+3)^{2}}\)

Let t=27/10.

I like fractions over decimals. Use 2.7 if you wish.

You could also do it this way:

\(\displaystyle \lim_{t_{1}\to 2.7}\left[\frac{\frac{(t_{1}+2)^{3}}{t_{1}+3}-\frac{(2.7+2)^{3}}{2.7+3}}{t_{1}-2.7}\right]=\lim_{t_{1}\to 2.7}\left[\frac{570t_{1}^{2}+4959t_{1}+9847}{570(t_{1}+3)}\right]=\lim_{t_{1}\to 2.7}\left[\frac{10}{57(t_{1}+3)}+t_{1}+\frac{57}{10}\right]\)

Both methods should result in the same solution.

 

[Ryan Rigdon]

i found f(c+h) and f(c) first.

this is what i came up with

f(c+h) = f(2.7+h) = ((4.7+h)^3)/(t+3) = (103.823+66.27h+14.1h^(2)+h^(3))/(5.7+h)

f(c) = f(2.7) = ((4.7)^3)/(5.7) = 18.2145614035


i will try it again. with your suggestions.

 

[Ryan Rigdon]

i tried both ways and it yielded


its instantaneous velocity when t = 2.7 seconds is approx. 8.431 m/s

i rounded to the nearest thousandth place.

 

[galactus]

Yep, that's what I get. I assume that jives with what is in the book?. Assuming the solution is in the book

As a fraction it is \(\displaystyle \frac{136958}{16245} \;\ m/s\)