Brief notes for each week

Math 006 - Hints

If students email me questions about assignment, or other matters, I can give hints and explainations, and in case other people have similar questions, I'll paste my answer here too.

Question about assignement 2

Someone asked about what I wanted for assignment two, for tables. Here's the answer:

In class we went over how the tangent is the limit of secants, and the slope of the tangent is the limit of the slope of secants. For question 8.c, page 36, it asks for the equation of the tangent at (2,5). To show this as a limit of secants, I mean a table which looks like

h	start of secant	end of secant	change in x,	change in y,	slope of secant
.5	(2,5)	(2.5, ? )	0.5	?	?
.1	(2,5)	(2.1, ? )	0.1	?	?

And maybe two or three more rows, to show secants closer to the tangent. (I've put questions marks where you have to fill in numbers by using the formular for the function in this case, which is (2x+1)/(x-1), and then you calculate change in x and y, and slope, as in class)

For 2.a, just do the question as it asks; then to put it in the same format as in class, you have a table that looks like:

	start of secant,		end of secant,		change in t,	change in s,	average speed
	t1,	f(t1)	t2,	f(t2)	t2-t1	f(t2)-f(t1)	(f(t2)-f(t1))/(t2-t1)
(i)	3,	f(3)	5,	f(5)	2
(ii)	3		4
(iii)	3		3.5
(iv)	3		3.1

Note here slope of secant just measures average time over an interval, and you are told where to start and end (in class we choose whatever change in x we liked, as long as it got smaller (the change in x was called "h" in class). Here the variable is t instead of x, and the change is just the length of time you measure for, eg, from 3 to 5, for 2.a.i, the change in t is 2 (5-3=2). I have just put a bit of data in the table, hopefully it should be clear how to fill in the rest from the column headings.

For part d, you draw the tangent at t=2, this is just a line that touches the curve at t=2, but does not go through the curve.

For question 8, page 44: Rate of change means "change in how much water flows out"/"time taken to flow out". To find the instanteneous rate of change means to keep making the time interval smaller and smaller. You start the measurements at time t=10. you get a table like this:

Start time,	start vol,	end time,	end vol,	change in time,	change in vol,	rate
10		12		2
10		11		1
10		10.5		.5
10		10.1		.1
10		10.01		.01

the second column is for "start volume", which is calculated from the formular given in the exercise. In the above table, you'll work out how much the volume changes over a time interval of 2 minutes for the first row. you can use this to find the rate of change over that interval, which you write in the last column. Keep choosing a smaller and smaller change of time, to get a rate over a smaller and smaller interval (eg, down to change of time .001). Then you look at what the numbers in the last column look like they are getting closer to - and that will be the instantaneous rate of change.

Limits of functions:

The limit of a function f(x) as x tends to a means the value f(x) gets closer and closer to as x gets closer and closer to a. Eg, the limit of x² as x gets closer to 2 is 4. Another example is, if f(x) = 9, then what is the limit of f(x) as x tends to 4? Well, it must be 9, since f(x) is always 9 in this case.

This is pretty obvious, so you might wonder what's the point of drawing a table to show this. The point is that by looking at simple examples of limits, hopefully it's easier to understand more complicated examples, eg the derivative.

Here are tables for the above two examples:

The numbers in the limit column are getting closer and closer to 4, so the limit of f(x)=x² as x approaches 2 is 4.

The numbers in the limit column are getting closer and closer to 9, so the limit of f(x)=9 as x tends to 2 is 9.

Question: Why do the above tables only have three columns, and the tables in class for calculating the derivative had a lot more?

Well, really all you need is two columns, the column for x, and the column for f(x), and you need that the column for x gives values of x getting closer and closer to some thing, eg, 2 in the above tables; you can work out the value of h from the x column, so if I gave you a table without the h column, you could fill it in; the h column is just to be sure the numbers are getting closer and closer to some limit, since they tell you exactly how far away you are from that limit, so as long as h gets closer and closer to zero, then the x value must be getting closer and closer to what you want - 2 in the above cases.

Now, in the class, if we just had three columns, the table would look like this:

I've not filled in the last column, since I've not decided what function to use. I decided to take the derivative at x=7, but I could be using any other value for the limit x tends to.

We used more than just three columns in class so we could calculate the last column in several steps, instead of doing it all at once in your head.

We had a column for f(x), a column for f(x+h), a column for f(x+h) - f(x) - this was all to calculate the numerator of the slope of the secant (that's the top part of the secant); then we had a column to tell us the change in x, which is the denominator (bottom part) of the slope of the secant. Then putting this together gives the last column.

See the calculation in the week three notes too.

Question about assignment 5

A hint for Question 10 page 145

This is a question about related rates, very similar to example 3 in the text book, page 143.

Question A man 2m tall walks away from a lamppost whose light is 5m above the ground. If he walks at a speed of 1.5 m/s, at what rate is his shadow growing when he is 10m from the lamp post.

Hints for answer

You need to work out what all the variables are, and what are the relationships between them. A good diagram helps. The rates of change measure something that's changing; lable the two things that are changing on a diagram, and work out what is the relation between them. Differentiate to get the relation between the rates of change. Use this to find the answer.

More hints

Here is a picture of the situation:

This picture gives the essentials of the question. But we are interested in rates. What rates of change? One is the change of the person's position, and the other is of the size of his shadow. So, I'll draw a more abstract diagram showing these things:

Here we've chosen variables S and L:

Now we know that the speed of the man is 1.5m/s. This means that dS/dt = 1.5m/s.

We want to find dL/dt when S=10. So we need to find a relationship between S and L, and differentiate that to find a relationship between dS/dt and dL/dt.

There are various ways to find a relationship, probably the easiest is to use similar triangles. There are two similar triangles in the picture - see if you can find them. (Similar means that all the angles of the two triangles are the same - so this means that the ratios of side lenghts are the same... if you need more hints about which triangles to use, and about using similar triangles in this kind of problem, let me know, and I'll try and add some more details.)

This means you can find an equation to relate S+L and L, then rearange that equation to find an equation relating L and S, then differentiate to find the relation between dL/dt and dS/dt. Since you're told dS/dt, that means you can find dL/dt, which is what you want.

Note about similar triangles

Related rates problems are often based on finding similar triangles in a diagram, so it's a good idea to (1) practice turning a description in words into a diagram, and (2) finding similar triangles in a diagram. For an example to test whether you can find sets of similar triangles, try finding all sets of similar triangles in the following diagram:

There are at least 3 sets of similar triangles in this picture. Click here to see them - triangles the same colour are similar.

Hints/examples for second derivatives

Second derivative just means taking the derivative two times, so if you know how to take a derivative one time, you shouldn't have problems doing it twice.

Here are some examples:

Example 1: Find the second derivative of f(x) = (x⁵+7)^1/4

Answer: Well the first derivative can be found using a combination of the chain rule and power rule. You get:

Simplifying and rearranging this a bit I get:

The second derivative means just take the derivative of this thing, so we'll have done this differentiation opperation twice by the time we finish.

To take the derivative of (5/4)(x⁴)(x⁵ + 7)^-3/4, we have to use several rules: power rule, chain rule, product rule.

If you want to see this in a few steps, note that

Now use the product rule to put these two things together, and we get that the second derivative of f(x) (which is the first derivative of f'(x)) is given by:

I could simplify if I wanted to... but I'm feeling lazy and the question didn't ask for simplification, so I'll leave it like that. The answer is 5x³(x⁵ + 7)^-3/4 + (5/4)x⁴ (-15/4)x⁴(x⁵ + 7)^(-7/4)

Example 2: Find the second derivative of x¹¹ + 6x⁵ + 43.

Answer:

The first derivative is

(note, I did simplify a bit, since I want to do some work with this, ie, differentiate it - and so it's simpler to start with something simpler.)

Then I take the derivative of what I've got above, and get:

Simplifying a little, we have 110x⁹ + 120x³. This is the final answer.

Example 3: Find the second derivative of (3x + 2)² /(x² -1)

The first derivative is found by using the chain rule, power rule, quotient rule(!) Quite a lot, so I'll do it in several steps.

Then the first derivative of f(x)/g(x) is given by (gf' - fg')/g²

So we find the derivative of f and g and then put it all together:

f'(x) = 2(3x + 2)*(3)

g'(x) = 2x

So the first derivative of the function (3x + 2)² /(x² -1) is given by:

The next step is to find the derivative of this, to get the second derivative. Before starting to take the derivative, I'm going to simplify a little bit, so it's easier to work with. The above is the same as

We use the same kind of method of find the derivative of this function, and what we get is:

The above is the second derivative of the original function, and this completes the question. If you want, you can simplify, and end up with:

Example 4: Find the second derivative of (7x + 9)⁵⁶

The first derivative is 56(7x + 9)⁵⁵*(7) = 392(7x + 9)⁵⁵

So the second derivative is 392*55*(7x + 9)⁵⁴*(7) = 150920(7x + 9)⁵⁴

Let me know if you need more examples.

Hints for finding intervals where the function is increasing or decreasing

To find the critical values, you find where f'(x) = 0; these are local max and min. Between these points the function is either increasing or decreasing.

If the function has no infinities, it's fairly easy to see what has to happen.

Example:

Where are the intervals where f(x) = x⁴ - 2x² + 5 is increasing or decreasing?

First, we can find where it's neither increasing nor decreaing; this is where

This means x = 0, -1, or 1 at local max or min:

We don't yet know whether the function goes up or down between these points; calculate some values of the function, which we can plot:

Plot these points:

Since there are no infinities of the function x⁴ - 2x² + 5, we can just join up the dots:

So we can see from the diagram that the function is decreasing on [-infinity,-1), increasing on (-1,0), decreasing on (0,1), and increasing on (1,infinity).

Thinks can be a little more complicated if the function has some points where it is infinite.

Example Where is the function f(x) = 1/x = x^-1 increasing or decreasing?

This function does not have any local max or min; the second derivative is

The only way to get f'(x) = 0 would be x = infinity; but this is not allowed.

To find where the function is increasing, we find where f'(x) is positive; to find where the function is decreasing, we find where f'(x) is negative.

Where is -1/x² positive? Well, it's never positive, since to get it positive, you need x² negative; but this is always positive; so the function is always decreasing; expect at o, where is is infinite (so we can't really say what it's doing, and it's not really properly defined). This function is increasing on the intervals [-infinity,0) and (0,infinity]:

Sometimes there are several factors to a derivative, in which case it can be useful to make a table to work out where it the function is increasing or decreasing.

ExampleWhere is the function f(x) = (x² - 6x)⁶ increasing or decreasing?

We have, by using the chain rule:

Whether the whole of f'(x) is positive or negative depends on the sign of the factors. The factors are 5, which is always poisitive, so we don't need to worry about it, (x² - 6x) which occurs 5 times, which is the same sign as if it just occurred one time; and then there is (2x - 6). The function (x² - 6x) has two factors, x and (x-6). The function (2x-6) = 2(x-3), so the sign depends on the sign of (x-3)

So we are interested in where is x(x-6)(x-3) positive or negative. Well, it's zero at x = 0, x= 6, and x = 3; so we look at the signs of each factor on each of the intervals [-infinity,0), (0,3), (3,6), (6,infinity); we have a table like this:

So from the table, the function is increasing on (0,3) and (6, infinity), and decreasing on (-infinty, 0) and (3,6).

For more hints, see this page.