URL http://www.mast.queensu.ca/~helena/math006/week5.html

Math 006 - Introduction to Calculus


Lecture 5

New Stuff

We will cover the following topics in this class:

Main Part of class

We will cover some more ideas from chapter 2 - higher derivatives, and more examples of the chain rule. Then we'll go on to applications of derivatives, from chapter 3. We will be covering 3.1 and 3.5 of the text book.

In chapter 3, we see applications of derivatives, so this means we will be spending time on working out how to convert a problem given in words into a mathematics problem.


Example

If you pull down at speed 30 m/s, for the above situation with the pulley and the weight, at what speed does the weight rise?

Answer

The proceedure for this kind of question is very similar to that for max/min problems, which is outlined on page 185 of the text book.

1. Understand the question
2. Draw a diagram
3. Work out what are all the variables and constants, give them names.
4. Lable the diagram with variables, constants.
5. Find relationships between variables, constants.
6. Differentiate this relationship to get an equation relating the rates you are interested in.
7. Substitute values given in question.
8. Answer question (eg, give the speed of something at a certain time.)

In this example, we're interested in two speeds. Speed of the wieght, speed of the end being pulled. How can you express speeds as derivatives? A speed is a rate of change of a distance, so you need to find some appropriate distances, and lable the diagram accordingly:

In the diagram, dX/dt is the speed of the weight, and dY/dt is the speed you pull at. In the end, we want to find how these two speeds are related; that means first finding out how X and Y are related.

How are X, Y, and L related? This is shown in the following diagram:

we have

5X + Y = L

Differentiate this, and we get:

5dX/dt + dY/dt = dL/dt

But the length of the rope is fixed, so L is not changing, so the rate of change of L is zero, so dL/dt = 0, so:

5dX/dt + dY/dt = 0
5dX/dt = - dY/dt
dX/dt = - (1/5)dY/dt

We are told that dY/dt = 30, so this means

dX/dt = - (1/5)30 = -6

So this means:

the weight moves up at 6m/s.

(Note, the minus sign just means that the weight is moving in the opposite direction from thhe direction you pull in.)


Tutorial and Examples

Homework

The following questions from the text book should be done for practice, but are not to be handed in for marking.

Assignment

The following questions should be handed in Wednesday 4 March:

1. Find the second derivatives of the following functions:
(x2+3)1/2
5x4 + 3x2 + 2
(5x+2)3
(x+1)/(x-1)
2. Page 126, 3.1 Ex 9
3. Page 145, 3.5 Ex 10

See the hints page for some hints. Let me know if you need any more hints!


Handout


Please send corrections to spelling mistakes, and any other comments, to me.



• Welcome Home • Origami •

Queen's Mathematics and Statistics
Queen's University