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Finding the derivative of a function allows you to calculate the rate of change at a point by finding the slope at such point. . . . Okay, I was going to attempt to answer my own question, but I'm totally blank.

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Finding the derivative of a function allows you to calculate the rate of change at a point by finding the slope at such point. . . . Okay, I was going to attempt to answer my own question, but I'm totally blank.

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uart

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There are so many applications that any list I could think of would be certain to be incomplete. So let me just pick one really common example from physics. We all know the simple distance-speed-time relationships, *distance = speed x time* for example.

Great formula*but*, it only works when speed is a constant! This is extremely limiting and means we can't handle any type problem involving accelerated motion. Fortunately calculus let us salvage that equation and make it much more general.

Say that we are trying to apply that equation, [itex]x = vt[/itex], (where "x" is distance), except the velocity "v" is not constant. Since the equation "*dist = speed x time*" only applies for constant velocity, it looks like we're stuck? Imagine however that we divide up time into lots of really small slices, say for example one microsecond increments*. Now over such a small time increment any changes in velocity will be very small so we can take "v" as approximately constant within any given time slice. That is, we can now use our formula despite the fact we have accelerated motion.

The new problem however, is that we are only finding the distance traveled one microsecond at a time, and we have to add up all of those one microsecond slices to find the total distance traveled. This is what integration does. We write the distance traveled in one time slice "dx" as,

[tex]dx = v \, dt[/tex]

Where "dt" is the small time slice and "dx" is the small distance traveled. This of course integrates to,

[tex]x = \int v \, dt[/tex]

So we've taken a simple speed time formula that only works for constant speed, and made it into a much more general integral equation that works for velocity as an arbitrary function of time. Physics and Engineering are absolutely full of applications very like this.

* The example of one microsecond is just to help you visualize it. Of course differentiation and integration are defined in terms of the limit as this time slice goes to zero.

Great formula

Say that we are trying to apply that equation, [itex]x = vt[/itex], (where "x" is distance), except the velocity "v" is not constant. Since the equation "

The new problem however, is that we are only finding the distance traveled one microsecond at a time, and we have to add up all of those one microsecond slices to find the total distance traveled. This is what integration does. We write the distance traveled in one time slice "dx" as,

[tex]dx = v \, dt[/tex]

Where "dt" is the small time slice and "dx" is the small distance traveled. This of course integrates to,

[tex]x = \int v \, dt[/tex]

So we've taken a simple speed time formula that only works for constant speed, and made it into a much more general integral equation that works for velocity as an arbitrary function of time. Physics and Engineering are absolutely full of applications very like this.

* The example of one microsecond is just to help you visualize it. Of course differentiation and integration are defined in terms of the limit as this time slice goes to zero.

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uart

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The equation [itex] W = F x[/itex] ("x" is distance) which is only true if "F" is a constant, becomes,

[tex] W = \int F \, dx[/tex]

Test yourself on an actual problem.

Q. A heavily braking car moves in a straight line with a velocity versus time given by [itex]v(t) = 36 - t^2[/itex] m/s, for t=0 to t=6 seconds.

How far does the car travel in the six seconds that it takes to stop?

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First, yes the units for work is Joules. However, after integrating you can have different types of units. uart's example after integrating velocity (units: m/s) you get just (m). So integration is not strictly used for work problems. In general, integration is a simple way to find out the area under a non-linear function.I'm assuming you could tell the amount of work done of an accelerating car? What units is that measured in? Joules?

Example:

If f(x) = x^2. Integrating that without a boundaries gives you 1/3 x^3 + c. But with limits you are able to determine the area under the curve. That area under the curve corresponds to the amount of (something). That is all it represents.

Things to take away: the integral of a function does not mean anything unless the units are known.

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I'm not sure if it's right. It looks weird. Here's what I did: first, I integrated the formula for the car that is breaking.Test yourself on an actual problem.

Q. A heavily braking car moves in a straight line with a velocity versus time given by [itex]v(t) = 36 - t^2[/itex] m/s, for t=0 to t=6 seconds.

How far does the car travel in the six seconds that it takes to stop?

[itex]\int[/itex][itex]^{6}_{0}[/itex] 36-t

Then I used the fundamental theorem of calculus.

[36t-[itex]\frac{1}{3}[/itex]t

Which would be

-(36(6)-[itex]\frac{1}{3}[/itex](6)

I'm barely into Calculus 2, so i'm not sure how to handle units. Would this mean that the car traveled 144 meters? Does the negative sign mean anything?

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The car does travel 144 meters. However, there shouldn't be a negative sign in the answer or in front of the expression.-(36(6)-[itex]\frac{1}{3}[/itex](6)^{3})=-144

Does the negative sign mean anything?

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