(Compute all instantaneous rate of change using the limit of the difference quotient)

2) A football's path is represented by the function h(t)=-4.9t^2 +10t+2, where h is its height, in meters, after t seconds. Find the rate of change of the football's height at 1s.

lim (-4.9(t+dt)^2+10(t+dt)+2)/dt

= -4.9(t^2+2tdt+dt^2)+10t+10dt+2 +4.9t^2-10t-2) /dt

= -4.9 (0t^2 + 0t+2tdt+dt^2+10dt) /dt
= -9.8t +10 as the limit of dt>>>zero

at t=1, then rate of change=.2 m/s

dh/dt = -9.8 t + 10

when t = 1
dh/dt = 10-9.8 = 0.2

thats 0.2m/s right?

ok got it thanks

To find the rate of change of the football's height at 1 second, we need to compute the instantaneous rate of change of the function h(t) at t = 1 using the limit of the difference quotient.

The instantaneous rate of change is the slope of the tangent line to the graph of the function at a specific point. We can approximate this slope by considering the average rate of change over a small interval around the point in question and then taking the limit as the interval approaches zero.

In this case, we want to find the rate of change of h(t) at t = 1, so let's use an interval around t = 1. We'll choose a small interval of width Δt centered at t = 1:

Δt represents this small change in time. We'll let Δt approach zero to get a more accurate approximation of the instantaneous rate of change.

The average rate of change over this interval is given by:

average rate of change = (h(1 + Δt) - h(1)) / Δt

We can plug in the function h(t) = -4.9t^2 + 10t + 2 and simplify the expression:

average rate of change = (-4.9(1 + Δt)^2 + 10(1 + Δt) + 2 - (-4.9(1)^2 + 10(1) + 2)) / Δt

Simplifying further, we expand and simplify the expression:

average rate of change = (-4.9(1 + 2Δt + (Δt)^2) + 10 + 10Δt + 2 + 4.9 - 10 - 2) / Δt
average rate of change = (-4.9 - 9.8Δt - 4.9(Δt)^2 + 10Δt) / Δt

Now we can simplify the expression by canceling out Δt terms:

average rate of change = -4.9 - 4.9(Δt) + 10

Taking the limit as Δt approaches zero, we find the instantaneous rate of change:

instantaneous rate of change = lim(Δt→0) (-4.9 - 4.9(Δt) + 10)

Evaluating the limit, we get:

instantaneous rate of change = -4.9 + 10
instantaneous rate of change = 5.1

Therefore, the rate of change of the football's height at 1 second is 5.1 meters per second.