Calculating Instantaneous Velocity in Calculus

[silverbell]

Homework Statement

A ball is thrown into the air a velocity of 49 ft/s. Its height in feet after t seconds is given by y=49t-10t^2.

A. Find the average velocity for the time period beginning when t=3 and lasting
0.01 s:
0.005 s:

B. Estimate the instantaneous velocity when t=3.

Homework Equations

Integrals

The Attempt at a Solution

I'm not exactly sure if I'm approaching this problem at the right angle...

A) 0.01s

1/ [(3.01) -3] ∫ from 3 to 3.01 49t-10t^2 dx

[1/0.01] ∫ from 3 to 3.01 49t-10t^2 dx

[1/0.01] ∫ from 3 to 3.01 (49t^2)/2 - (10t^3)/3 <------ integrated

[1/0.01] [(49(3.01)^2)/2 - (10(3.01)^3)/3] - [1/0.01] [ 49(3)^2)/2 - (10(3)^3)/3

substitution

Same steps for 0.005s

B) I'm not sure how to approach this problem.

Please help me understand the problem. Thank you very much. :)

 

[eumyang]

I'm not exactly sure if I'm approaching this problem at the right angle...

No, you don't want to take an integral. You're given a position function, and integrating it doesn't give you the velocity function. What you want to do is to use the average rate of change formula (from x = a to x = b):
\frac{f(b) - f(a)}{b - a}
So, given the position function s(t) = -10t² + 49t, evaluate
\frac{s(3.01) - s(3)}{3.01 - 3}

B) I'm not sure how to approach this problem.

Do you know the formula for finding a derivative at a point (I'm talking about the one with the limit)?

 

[silverbell]

Do you know the formula for finding a derivative at a point (I'm talking about the one with the limit)?

Actually, I don't know the formula for finding a derivative at a point.

 

[eumyang]

Actually, I don't know the formula for finding a derivative at a point.

Sure you do:
f&#039;(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}

 

[silverbell]

Sure you do:
f&#039;(a) = \lim_{h \rightarrow 0} \frac{f(a + h) - f(a)}{h}

Thanks. :D