Suppose that a ball is rolling down a ramp. The distance...

Suppose that a ball is rolling down a ramp. The distance traveled by the ball is given by the function in each exercise, where $t$ is the time, in seconds, after the ball is released, and $s(t)$ is measured in feet. For each given function, find the ball's average velocity from a. $t_{1}=3$ to $t_{2}=4$. b. $t_{1}=3$ to $t_{2}=3.5$. c. $t_{1}=3$ to $t_{2}=3.01$. d. $t_{1}=3$ to $t_{2}=3.001$. $$s(t)=10 t^{2}$$

Suppose that a ball is rolling down a ramp. The distance traveled by the ball is given by the function in each exercise, where $t$ is the time, in seconds, after the ball is released, and $s(t)$ is measured in feet. For each given function, find the ball's average velocity from
a. $t_{1}=3$ to $t_{2}=4$.
b. $t_{1}=3$ to $t_{2}=3.5$.
c. $t_{1}=3$ to $t_{2}=3.01$.
d. $t_{1}=3$ to $t_{2}=3.001$.
$$s(t)=10 t^{2}$$

Key Concepts

Limit Process for Instantaneous Rate of Change

The limit process involves investigating what happens to the average rate of change as the time interval shrinks towards zero. This concept underlies the derivative in calculus, which, when applied to a distance function, yields the instantaneous velocity at a specific time. It’s important in understanding how instantaneous quantities are derived from average measurements.

Average Velocity

Average velocity is a fundamental concept in kinematics that measures the total displacement over a specified time interval divided by the length of that interval. It is computed as (s(t2) – s(t1))/(t2 – t1) and provides a simple way to understand motion over a finite period, indicating the ball’s overall rate of motion between two time points.

Difference Quotient

The difference quotient is a mathematical formulation used to calculate the average rate of change of a function over an interval. In the context of motion, it takes the form (s(t2) – s(t1))/(t2 – t1), representing the slope of the secant line connecting two points on the distance-time graph. This concept is essential in transitioning from average to instantaneous rates of change through the process of taking limits.

Transcript

00:01 We're going to find the average rate of change of this function, s of t, on four different intervals.

00:06 So just as a refresher, the average rate of change is going to be defined as, for any function f, f of b minus f of a divided by b minus a, that is between points b and a.

00:24 So with that covered we can move on to our first point.

00:27 That is, we're going to be looking at the average rate of change between 3 and 4.

00:35 So our average rate of change is going to be, well, f of b, that is s of 4, so that is 10 times 4 squared, and then we subtract f of a, so that is s of 3, so that's 10 times 3 squared.

00:54 So we've just plugged in our values of a and b into the function here as t, then we divide that by b minus a, so that's going to be 4 minus 3.

01:06 Now i'm just going to do a little trick here to save us some time.

01:10 You can see that there's a 10 multiplied by both terms on the top here.

01:14 So what i'm going to do is just pull that 10 out front.

01:17 This is a perfectly legitimate algebraic move, and it'll make things a little bit easier to write.

01:22 So that's going to be 10 times 4 squared minus 3 squared divided by 4 minus 3, which is of course 1.

01:33 So that's going to be 10 times, well, 4 squared is 16, 3 squared is 9, so that's going to be 10 times 16 minus 9, or 10 times 7, which is 70.

01:52 Alright, that wasn't so bad.

01:53 Things are going to get a little bit funnier now that we're going to be dealing with these decimals, but it's alright as long as you have a calculator on hand.

02:00 I've already done these calculations, so i'll just be writing down the numbers that i got.

02:04 But, so we're going to start out with, once again, we're just going to be pulling out this 10, doing that same trick as before.

02:10 We have 10 times, then b squared minus a squared.

02:12 This time b is going to be 3 .5.

02:17 A is once again 3, and a is actually going to be 3 in all of these.

02:21 And then this is divided by b minus a, so b is 3 .5 minus a, which is 3.

02:28 Then, running this through your calculator, we'll get you, well, still 10 times, of course.

02:34 3 .5 squared is 12 .25, and then minus 9, because once again, 3 squared.

02:42 This is divided by 3 .5 minus 3, which is just 0 .5.

02:47 Alright, that's not so bad.

02:50 Then, crunching a few numbers, well, you'll see that this 12 .25 minus 9, this is going to be equal to 3 .25...

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