 # Using Limits to Calculate the Derivative

If you know the position of someone as a function of time, you can calculate the derivative — the velocity of that person — as a function of time as well. Use the definition of the derivative and your knowledge of limits to do just that in this lesson.

## Review of Derivatives

Remember how the derivative is a mathematical way to describe rate of change? When looking at some function on a graph, it’s the slope. For some function y=f(x), we say that the derivative is dy/dx. That’s how much y is changing as x changes. We also call this y` or f`(x), because x here is our independent variable. Formally, we write this as f`(x) equals the limit, as some delta x goes to zero, of f(x + delta x) – f(x), all divided by delta x.

Now, really, all this is saying is that we’re going to calculate the slope on some graph as delta x goes to zero. This is just calculating the tangent. So let’s calculate the derivative using this formal definition.

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Let’s say you’re in your car, and you’re going exactly 60 miles per hour. You know already that this is your rate of change, or how your position is changing as a function of time. You expect that 60 is going to be your derivative when you’re looking at your position as a function of time. If you’re driving for one hour at exactly 60 mph, you go 60 miles. So here is your position as a function of time. We can write this mathematically as position equals speed times time. For this example, it’s x = 60t, where x is your position in miles and t is the number of hours that you’ve been driving.

Let’s find the derivative of this. If I write the equation for your position, x=60t, I could say that this is x=f(t). Now t here is your independent variable. The derivative, then, is going to be dx/dt or x` or f`(t). Formally, we write this as the limit, as delta t goes to zero (because t is your independent variable) of (f(t + delta t) – f(t)) / delta t. So let’s plug in our function: f(t + delta t) is 60(t + delta t) and f(t) is 60t. We can expand these terms and simplify, then calculate dx/dt as 60, which is exactly what we thought it’d be because we’re going 60 mph. Notice that 60 does not depend on time. Your x`, or velocity, is always 60, no matter if it’s time = 0, or time = 1 or 15 hours from now. It is always going to be constant.

## Derivative of a Not-So-Constant Velocity Let’s look at a case where that’s not true. Let’s look at Super C, the human cannonball. We can graph his height as a function of time, where height is in feet and time is in seconds. Let’s say that the equation for his height as a function of time is -16t^2 + 36t. We know that the derivative is the slope on this graph. And the slope isn’t the same at every point in time. Here, the slope is different from here. In fact, at the beginning of this graph, the slope is actually increasing. It’s a positive slope. He’s going up. As time moves on, the slope is negative. He’s falling back down to the ground.

So how can we write this? How can we find his velocity, his upward velocity, as a function of time? We need to calculate the derivative of this. So if the function is height as a function of time (h= f(t)), then the derivative is dh/dt. Formally, I write this as the limit, as delta t goes to zero, of (f(t + delta t) – f(t)) / delta t. Let’s plug in f(t + delta t) and f(t) and simplify. So we plug in f(t + delta t) and we plug in f(t): (-16(t + delta t)^2 + 36(t + delta t)) – (-16t^2 + 36t). Now we have to expand this (t + delta t)^2: (-16(t^2 + 2(delta t) * t + delta t^2) + 36(t + delta t)) – (-16t^2 + 36t) Then we can simplify and remove some terms, like the -16t^2 + 16t^2 and the + 36t – 36t.

When we simplify, we get this: (36(delta t) – 32(delta t) * t – 16(delta t^2)) / delta t. If we remember how to find the limit of a polynomial divided by another polynomial, we know that we have to divide both the top and the bottom by delta t, because that’s the highest order here in the bottom. If we do that, we find that the derivative, dh/dt, is equal to the limit, as delta t goes to zero, of 36 – 32t – 16(delta t), all over 1. Well, this is just 36 – 32t. So we have the height as a function of time, h(t), equals -16t^2 + 36t. The derivative, how fast Super C is going up and then down, is h`(t) = -32t + 36.

## Graphing the Derivative Let’s graph these. We’ve got h(t), h as a function of t, and we’ve got the derivative, h`(t). You can calculate that Super C is going to hit the ground after 2.25 seconds. So we have to graph the derivative all the way to t=2.25. This graph goes through zero; at some point, his upward velocity is zero. This makes sense, because when Super C reaches the apex of his curve, he has no upward velocity. He just kind of sits there for an instant. We can find out when this happens by finding out when h`(t) equals zero. And h`(t) equals zero at t=1.125 seconds. So right about 1 second into his flight, he’s not moving, just before he starts falling back down.

Now how high does he get? Well, his peak height is when he stops moving. So his peak height is at t=1.125. Let’s plug 1.125 into the function for his height as a function of time: h= -16(1.125)^2 + 36(1.125). If I look at how high Super C is at t=1.125 seconds, I find out that he’s actually about 20 feet up, or really 20 feet and 3 inches up.

## Lesson Summary

Now we can find the slope or rate of change or derivatives using the formal definition of the derivative. So for y=f(x) (x is your independent variable), y` is the limit, as delta x goes to zero, of (f(x + delta x) – f(x)) / delta x.

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