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Handout Tuesday 11/11
Today we will start exploring the connection between velocity and distance
Continuing our exploration of moving from velocity to distance:
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Example: suppose you drop a ball down a well. Since acceleration on earth is a constant \(32\) feet per second squared, the velocity after \(t\) seconds \(v(t) = 32t\) (assuming that the initial velocity was 0). We are thinking of positive velocity as moving downward here, otherwise we would make the 32 negative. Can we find the distance traveled by the ball? Graph this, look at the antiderivative, and look at area,.
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Another question we asked that first day: if you recorded your speeds every 5 minutes, can you estimate how far you have traveled? We said after 5 minutes, you noticed speeds of 75mph, 60mph, 65mph and 50mph. Or in terms of miles/minute, these numbers are 1.25, 1, 1.08, and 0.84.
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In fact, for any rate of change of some quantity, the total change in that quantity will be the rate of change multiplied by the change in independent variable. Thinking about units here is helpful.
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A very useful observation about derivatives or instantaneous rates of change, is that these can be pictured graphically as slopes. What is the graphical interpretation of going the other way?
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Consider a graph of a constant function that represents a velocity. What does the product of velocity and time look like? It is exactly the area under the graph. That’s because the velocity is the height of the function, and the time is the width of the interval on the horizontal axis.
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But what if the function is not constant? Well, we can still think of distance as being an area under the curve: just think of the curve as infinitely many small horizontal line segments, sort of like a staircase that when you zoom out looks like a smooth curve.
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A big idea: we also have developed a symbolic way to represent rates of change: the derivative function. For example, if the distance traveled is given by the function \(f(t) = t^2 - 3t + 8\) then the velocity is \(f'(t) = 2t - 3\text{.}\) Okay, well now if I told you the velocity was \(v(t) = 2t-3\text{,}\) what is the function that describes the distance? We can take an antiderivative.
