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# Instantaneous Velocity - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

By finding the slope between two points that are very close together on a position vs. time graph, you can find an approximate value for the instantaneous velocity of an object. One method of finding a point close to one single point is by using the "trace" button on a graphing calculator to find a value of t and f(t) that are sufficiently close enough to the point to get an accurate estimate of the instantaneous velocity of that object. Instantaneous velocity can then be estimated using the same methods as finding the average velocity. However, this is closer to instantaneous velocity because the points are chosen to be very close together.

I’m talking about instantaneous velocity and I’ve got my pumpkin problem from the previous episode. Pumpkins catapulted into the air time t is in seconds, height f(t) is in feet. Here’s a graph of the function and I actually have a formula this time. f(t) equals 4 plus 130t minus 16t².

I want to discuss the instantaneous velocity at t equals 2. And I want to represent that velocity on the graph. We’ve computed this instantaneous velocity in the previous episode, but I want to show you how you represent these instantaneous velocity graphically. Now remember, instantaneous velocity, we obtain that by taking average velocities of smaller and smaller intervals of time. I want to show you what this looks like when we do it on my calculator.

We’re looking at the TI 84. This is a graph of the position of the pumpkin over time. Remember that equation, I have it up here; 4 plus 130, I have to use x on the TI 84 minus 16x ² and what I want to do for starters is play around with this a little bit. Let’s hit the trace button. When you hit the trace button, you can move around. And different points in the curve, represent different points in time and the position of the pumpkin. The height of the pumpkin at that point in time.

For example, I can cursor over to x equals 2. It’s a lot slower to do with arrows. Let me just type in 2, enter. And there we are. x equals 2 when time is 2, the height is 200. I’m interested in finding the instantaneous velocity here. One thing I can do is I can find the average velocity over a really small interval. I could do say x equals 2 and x equals 2.1. That’s pretty small. I have 206.44.

Let’s say I want to get much more precise than that. What I’m going to do is zoom in on this point. So I hit zoom and zoom in. One of the tricks I use when I zoom in, is I trace over that point first, because it takes a long time to cursor this little point over. So I’m going to hit enter and it will zoom in. Now you’ll notice that I’m much closer to the curve.

As I get closer, the curve has seemed to straighten out a bit. I’m focusing more on the part of the curve that’s right near t equals 2. Let me zoom in again. I hit enter again and I zoom in further. I’m going to get a little closer to the curve. Zoom in a third time, and then a little closer.

Let me actually hit trace here. I’ll make sure that I’m right on 2. I’m going to zoom in on that. So I’m centered on t equals 2, I’ve zoomed in 4 times. Look how straight the curve is? This is really important because what it means is that, what I’m finding here, when I’m taking an average velocity over a really short interval of time is that essentially the slope of the curve. As the interval of time gets smaller and smaller, the slope of the secant line, is much, much closer to the actual slope of the curve.

Let me pick, x equals 2 is my one point. Y equals 200. Let me pick another point, 2.001. I’m going to remember this value; 200.066, it's rounded. Let’s go back to the board and I want to calculate the average velocity over this tiny little interval. That will be our approximation for instantaneous velocity.

Let’s put those two points on the board. We had 2,200. This means when t equaled 2, the height was 200 feet. We have 2.001, 200.0066. So when time was 2.001, just 1/10,000 to the second later, the height was 200.0066. So I’m going to find the instantaneous velocity by approximating it using the average velocity between these points.

So instantaneous velocity is approximately the difference in height, the change in position, 200.0066 minus 200 and this will be in feet. Over 2.001 minus 2, and that will be in seconds. So my numerator is going to give me 0.0066 feet over 0.0001 seconds.

To simplify this, I can multiply the top and bottom by 10,000. Basically what I’m doing is I’m multiplying by a power of 10 big enough to get the decimal point all the way over to the right. So move the decimal point over four places and get 66 on top and 1 on the bottom and that’s feet per second. 66 feet per second. That’s my instantaneous velocity.

This is an approximation but I used a really tiny interval so this is really, really close to the instantaneous velocity. Now, notice what we did to get this instantaneous velocity. We took two points on the curve that were really, really close together. So this value represents the slope of the secant line drawn between those two points. It’s effectively very, very close to the slope of the tangent line, or the slope of the curve at that point. Let’s represent that on our graph.

Remember, when we were looking for the instantaneous velocity here, we found that it was 66 feet per second. The way we represent that graphically is with a tangent line. This curve tells me height of the pumpkin over time and the slope of this line tells me the velocity at this point, 66 feet per second.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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