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Study Guide

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Derivatives are, believe it or not, used a lot. The field of differential equations is an area of math that studies equations with derivatives in them.

Differential equations are used to investigate problems in physics, model the spread of disease ("mathematical epidemiology"), clarify blurry computer images, and protect us from zombie apocallypse.

### I Like Abstract Stuff; Why Should I Care?

We've started getting into "real" math in the last couple of units, with theorems, hypotheses, and counterexamples. It's about time we actually did a proof.

**Rolle's Theorem**says:Let

*f*be a function that- is continuous on the closed interval [
*a*,*b*]

- is differentiable on the open interval (
*a*,*b*), and

- has
*f*(*a*) =*f*(*b*).

Then there is some

*c*in the open interval (*a*,*b*) with*f*' (*c*) = 0.### Proof of Rolle's Theorem

We'll prove this in a couple of different cases.

**Case 1 (the boring case):**If*f*is constant on [*a*,*b*], then*f*(*a*) =*f*(c) for every*c*in (*a*,*b*):A horizontal line has a derivative (slope) of zero everywhere, so

*f*' (*c*) = 0 for all*c*in (*a*,*b*).If

*f*is constant on any sub-interval of [*a*,*b*], the same argument applies:Now suppose

*f*is not constant on any sub-interval of [*a*,*b*], since we've already taken care of that case. The extreme value theorem says*f*must have a maximum and a minimum on [*a*,*b*]. The max and the min can't both equal*f*(*a*) since then we would be back in Case 1.**Case 2:**The maximum value of*f*on [*a*,*b*] is greater than*f*(*a*). Let*c*be a value of*x*where this maximum occurs:Calculate

*f*' (*c*) by looking at the limit definition:.

We know

*f*(*c*+*h*) <*f*(*c*) since*f*(*c*) is the maximum value of*f*on our interval. The numerator of the quotient will be negative for any value of*h*close to 0. If*h*is less than zero we'll have a negative over a negative, so the quotient will be positive:.

If

*h*is greater than zero we'll have a negative over a positive, so the quotient will be negative:.

We're assuming

*f*is differentiable on (*a*,*b*), which means*f*' (*c*) needs to exist. We're getting positive numbers as*h*approaches 0 from the left, and negative numbers as*h*approaches 0 from the right. The only way the limit can exist is if it's to equal 0. Therefore**Case 3:**The minimum value of*f*on [*a*,*b*] is less than*f*(*a*).### Sample Problem

Finish the proof in Case 3.

This is the same as Case 2. Let

*c*be a value of*x*where a minimum occurs:Since

*f*(*c*) is a minimum,*f*(*c*+*h*) will be larger and therefore*f*(*c*+*h*) –*f*(*c*) is positive. If*h*is negative then the quotientis negative; if

*h*is positive then the quotient is positive. As in Case 2, the only way for the limitto exist is to have

*f*' (*c*) = 0, since we find negative numbers as*h*approaches from one side and positive numbers as*h*approaches from the other side.- is continuous on the closed interval [
### How to Solve a Math Problem

There are three steps to solving a math problem.

- Figure out what the problem is asking.

- Solve the problem.

- If you get stuck, figure out why you're stuck.

- Check the answer.

### Sample Problem

What does the Mean Value Theorem say about the function

*f*(*x*) = 4*x*+*x*^{2}on the interval (4, 6)?**Figure out what the problem is asking.**We need to understand what all the words in the problem mean, and what any theorems mentioned say. The MVT says if

*f*is continuous on [*a*,*b*], and*f*is differentiable on (*a*,*b*),then there is some

*c*in (*a*,*b*) with.

The problem is actually two problems. Are we allowed to apply the MVT to the given function on the given interval? If so, what does the conclusion of the MVT say for this particular function on this particular interval?

**Solve the problem.**To solve the problem we need to do two smaller problems.

Are we allowed to apply the MVT to the given function on the given interval?

First, we need to understand what the problem is asking. This problem is asking "do the hypotheses/assumptions of the MVT hold in this case?"

Next, we solve the problem. The answer is yes.

*f*is a polynomial, since it's continuous and differentiable everywhere. In particular,*f*is continuous on [4, 6] and differentiable on (4, 6). There's not anything to check for this answer.Now what does the conclusion of the MVT say for this particular function on this particular interval? First we need to understand what the problem is asking. This problem is asking "if we plug in the right values of

*a*and*b*and use the conclusion of the MVT, what does it say?" To solve the problem, we plug in*a*= 4 and*b*= 6. The MVT says there is some*c*in (4, 6) withAgain, there's not anything to check for this answer.

**Check the answer.**Problems like this that don't involve much calculation, don't have answers to check. A good thing to do, though, would be to read through the answer again and make sure it's believable.

- Figure out what the problem is asking.
### Appendix: Speed v. Velocity

It's possible to mix the words "speed" and "velocity," especially since we almost never use the word "velocity" in everyday life.

**Speed is a number.**Some examples of speeds are 55 mph, 100 km/hr, and 33 ft/sec.**Velocity is a vector.**For our current purposes, velocity is a speed with some kind of direction attached to it. Some examples of velocities are 55 mph due East, 100 km/hr away from home, and 33 ft/sec downwards.The direction associated with velocity gives us additional information and lets us answer questions like this:

### Sample Problem

Jenna travels at 50 mph away from home for one hour, and then at 50 mph toward home for 30 minutes. How far away from home is she at the end of this time?

First, Jenna travels at 50 mph away from home for one hour, therefore she's 50 miles from home:

Then Jenna travels back towards home for half an hour, covering 50/2 = 25 miles:

She ends 25 miles away from home.

If we didn't have the information about direction, we couldn't answer the question.

### Sample Problem

Jenna travels at 50 mph for one hour, then at 30 mph for one hour. How far from home is she at the end of this time?

We can't answer this question. If Jenna went in the same direction for two hours, she would be 80 miles from home.

If she went out and then back, she'd be only 20 miles from home.

And if she went in funny directions, who knows where she would be?

Calculus classes often use problems in which velocity has only two possible directions: positive or negative. For example, a ball moving upwards would have positive velocity, and a ball falling downwards would have negative velocity. A driver driving away from home would have positive velocity, and driving toward home would have negative velocity. In such cases, speed is the absolute value of velocity. More generally,

*speed is the magnitude of velocity*.### Sample Problems

Find the speed associated to each velocity.

- 70 mph due North

- -16 ft/sec

- 15 meters per second at an angle of 45
^{°}to the ground and straight ahead.

For each of these, leave out the words that tell about direction and take absolute values of the numbers.

- Omit the words "due North." The number left over is 70 mph.

- Take the absolute value of -16 ft/sec to find 16 ft/sec.

- Ignore all the words except the unit.

The answer is 15 meters per second.

- 70 mph due North

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