# Derivatives as Slope of a Curve

What do snickerdoodles and velocity have in common? Derivatives! No, that wasn't a bad attempt at a joke. Get on our level by watching this speedy video.

 AP Calculus Derivatives Language English Language MPAC 1 Reasoning with definitions and theorems MPAC 3 Implementing algebraic/computational processes Math Calculus

### Transcript

00:22

go in his new golf cart. He starts at his house and decides to have

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the finish line at his friendÕs house. His friend, Mr. Macadamia Nut, Macnut for short,

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decides to help him, by drawing a graph of his distance traveled in feet versus time

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elapsed in seconds. Because Mr. SnickerdoodleÕs favorite number

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is 10, he wants to figure out how fast the car can go at time equals 10 seconds.

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How fast is Mr. Snickerdoodle going at 10 seconds?

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One way to express how fast he is going is by using velocity, which equals the change

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in position over change in time.

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We know the function for position of the golf cart can be written as f of x equals x-squared.

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If we can find the rate of change, or slope, of the position graph at 10 seconds, we will

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find how fast he is going.

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Typically, we can find the slope of a line by taking two points on the line, and taking

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the quantity y2 minus y1 over x2 minus x1. However, our position graph is a curve, not

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a line. This means that the slope of the graph varies at different points in time.

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Even though the slope is changing, we can find the slope at a single x-value by finding

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the slope of a line tangent to the curve at that point, when t equals 10.

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To find the tangent line, we start by finding the line that goes through t equals 10 and

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another point on the graph. This is called the secant line. To find the tangent line,

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we can slide the other point closer and closer to t equals 10Éand eventually weÕll get

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a line that is tangent to the graph at t equals 10.

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A mathematical term we can use to describe getting closer and closer to something is

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a limit. In other words, we want to find the limit as the second point approaches 10.

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As you may recall, slope is y2 minus y1 over x2 minus x1. To find the slope at x equals

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ten, we take the limit of the slope of the secant line as the second point approaches

02:25

02:29

distance between the two points, which we can call the variable h, approaching zero.

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Since we want to express everything in terms of variables, we can label the initial point

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with the variable x, which we will later substitute 10 for. Our formula for slope now becomes

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the limit as h approaches zero of f of x plus h minus f of x over x plus h minus x.

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The xÕs on the bottom cancel out, which leaves us with the limit as h approaches zero of

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f of x plus h minus f of x all over h.

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This formula will show up again and again when you do derivatives, so MEMORIZE IT.

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Plugging in Mr. SnickerdoodleÕs equation, we have f of x equals x squared. For the first

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part, we plug in x plus h instead of x..to get the limit as h approaches zero of x plus

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h squared minus x squared over h. If we FOIL x plus h squared, we get x squared plus 2xh

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plus h-squared. We can cancel the x squaredÕs, leaving us with the limit as h approaches

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zero of two x h plus h squared all over h. We can factor an h out of the numerator..to

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get h times 2x plus h. The hÕs cancel on the top and bottom. WeÕre left with the limit

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as h approaches 0 of 2x plus h.

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Now we can plug h equals zero in, giving us that the slope equals two x.

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What we just found is the rate of change of the position over time, which, if you think

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about it, is just velocity. This is also called TAKING THE DERIVATIVE of a function. Which

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is presumably why you came to watch this video.

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So back to Mr. Snickerdoodle. To find the velocity at ten seconds, we just plug in ten

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for x into 2x. to find that Mr. SnickerdoodleÕs speed is 2 times 10, or 20 feet per second.

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For those of you not familiar with feet and seconds, heÕs going a blazing 13.6 miles

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per hour. His race car driving dreams are finally fulfilled.