# Understanding Higher Order Derivatives Using Graphs

Coming up next: Calculating Higher Order Derivatives

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• 0:06 Rate of Change
• 1:45 Acceleration
• 2:43 Higher Order Derivatives
• 4:08 Second Derivatives
• 6:54 Lesson Summary

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Lesson Transcript
Instructor: Robert Egan
The derivative is a rate of change, like velocity. What happens, though, when your velocity - that is, your rate of change - is changing? Explore the changing changes in this lesson.

## The Rate of Change of Super C

Consider for a minute the trajectory of Super C, the human cannonball. Super C is shot out of a cannon, has some trajectory flying through the air and eventually, hopefully, lands in a net. When Super C is shot out of the cannon, he has some vertical velocity taking him upward. We know from physics that the only thing pulling him down is acceleration due to gravity. So what does this all mean?

Consider Super C's height as a function of time: h(t) = -16t^2 + 36t. This is his rate of change in the upward direction. We know that the derivative of his height is his upward velocity, and we can calculate his velocity using our derivative rules: d/dt(h(t)) = d/dt(-16t^2 + 36t) is the derivative of his height. We can divide this up into -16d/dt(t^2) + 36d/dt(t). Our derivative, which we can call h`, equals -32t + 36. Let's be specific here and say that h(t) is his change in height (feet), and h` is his change in height (feet) / his change in time (seconds). So h` is in ft/s. Let's look at graphs for his change in height and for his upward velocity.

## Acceleration

We know that the graph for his change in height is a parabola. His upward velocity is a straight line. Super C starts out going up at 36 ft/s. His velocity keeps dropping and eventually becomes negative; that's when he starts falling back down toward Earth. But hang on a second; his upward velocity is changing as function of time. The function h` depends on t, so what about looking at the rate of change of his velocity? This is the derivative of his velocity. That's like saying d/dt(ft/s), so what's in ft/s^2? Well, that's his acceleration; the rate of change in his velocity is his acceleration. How do we calculate this? We know that h` = -32t + 36. The derivative of h` is d/dt(h`) = d/dt(-32t + 36). This is easy to calculate: d/dt(-32t) + d/dt(36). That's just -32 + 0 or -32.

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