Hi guys. A question for you.

Question

The weight in metres of a ball dropped from the top of the CN tower is given by h(t) = -4.9t^2 + 450 where t is the time in elapses in seconds.

-4.9(0)^2 + 450 = 450

-4.9(1)^2 + 450= 445.1

d) use the secant method to approximate the instantaneous velocity at t = 1 ssecond

Can somebody talk a little bit about the secant method and help guide me with this question?

Thank you so much i appreciate your help

Answer 1

The graph of this function is a parabola, and the instantaneous velocity is the slope of the line tangent to the parabola at t=1.

We only know one point on the tangent line, (1, 445.1), so we can't compute the slope in the usual y₂-y₁/x₂-x₁ way. So to approximate the slope, we pick points on the parabola very close to (1, 445.1), and use these two points to compute the slopes of the lines. These lines that pass through two points on the curve are called secant lines.

So for a mediocre approximation, we might take t=2 for our second point. h(2)=430.4, so the slope is

(445.1-430.4)/(2-1)=14.7.

For a better approximation, we can take t=1.1. h(1.1)=444.071, and the slope is

(445.1-444.071)/(1.1-1)=10.29

So you can pick t very close to 1 approximate the instantaneous velocity as closely as you like. And if you take the limit as the second t-value approaches 1, you will get the instantaneous velocity exactly; this is how you compute derivatives.