I would like to begin with a review of some of the midterm problems.
People had trouble with problem 3.
Here is the text of problem 3:
Problem 0.1. Little Lizzie is sliding down the Great Green Run, her favourite
straight slide at the local playground. The highest point of the slide is 4 metres
above the ground and the slide is 8 metres long. When she is half way down
the slide, she is losing elevation at 3 metres per second. How fast is she sliding
down at that instant?
Now, in order to do this problem, it really helps to draw a picture. Now,
many of you drew a picture of the slide as a right triangle, which is the right
idea. However, you forgot an important part of the picture: Lizzie.
Most of you, upon seeing a picture of a right triangle, immediately attempted
to use the Pythagorean theorem. This makes sense; the Pythagorean theorem
is probably the most important fact about right triangles, and you often need
to use the Pythagorean theorem when solving related rates problems involving
right triangles. If you leave the drawing as it is, without including Lizzie, then
it’s easy to get sidetracked into thinking that the Pythagorean theorem is the
correct strategy for this problem.
However, placing Lizzie on the slide (and drawing a perpendicular to the
ground) suggests a different method: similar triangles.
Let ℓ represent the distance from Lizzie to the end of the slide, and h represent the vertical distance from Lizzie to the ground. By similar triangles, we
get
8m
ℓ
=
=2
h
4m
So
ℓ = 2h
and
dℓ
dh
=2 .
dt
dt
dℓ
We know that dh
dt is −3m/s, so dt is −6m/s, and Lizzie is sliding at a speed
of 6 meters per second.
There are two things that I want to make sure you all take away from this
problem:
1. Make sure you draw a complete picture. If you hear about Lizzie sliding
down a slide, make sure you draw Lizzie in addition to the slide. Here,
this was critical to coming up with the right plan to solve the problem.
2. You need to know what your variable names represent. Way too many
students got confused about the length of the slide and Lizzie’s distance
to the end of the slide. To prevent this error, try explicitly writing out the
meaning of your variable names. You should do so using more than just
a single word. Write out the meaning explicitly. For many of you, simply
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labelling the picture isn’t enough to resolve the ambiguity. So instead of
saying “h is the height” say “h is the height of the oil drum (in meters)”
or “h is the height of the oil in the drum (in meters).” Say “t is the time
elapsed since noon(in seconds)” instead of “t is the time.” Say “d is the
distance (in km) from car A to car B,” not “d is the distance.” This may
seem laborious, but it’s better than getting confused about your variable
names. If you got problem 3 wrong, chances are you got confused about
the meaning of your variables at some point, so something needs to change.
I’m completely serious about these two points. If you got problem 3 wrong (or
the boat problem from quiz 1) you should get in the habit of always providing a
detailed explanation of what your variables represents. This may sound harsh,
but if you missed these problems, that’s a lot of evidence that your method of
keeping the meaning of your variables straight is not working. The easiest way
to remember something is to write it down.
Now, we’ll go over part (ii) of problem 5. This problem was intended to be
kind of difficult.
Here’s the text of the problem:
Problem 0.2. Consider a function f differentiable everywhere. Suppose that
the derivative f ′ satisfies the following conditions:
• f ′ (x) = 0 at x = 2 and x = 6 and f ′ (x) 6= 0 anywhere else,
• f ′ is increasing for x < 5 and decreasing for x > 5.
Based on the information above,
(a) Determine all intervals where the function f is increasing and where it is
decreasing.
(b) Determine the x-coordinates of all local maximum and minimum values of f
(specify which values correspond to a maximum and which ones correspond
to a minimum).
Now, this problem is sort of easier using the material from this week. Let’s
go over the problem.
The point here is that f ′ can only change sign at x = 2 and at x = 6.
Therefore, we can make a table:
Region x < 2 2 < x < 6 x > 6
f ′ (x)
f (x)
We need to determine where f ′ is positive and where f ′ is negative. Note
that f ′ (x) is increasing at x = 2 and 0 at x = 2; therefore, f ′ (x) is negative
if x < 2 and positive if 2 < x < 6. Therefore, f (x) is decreasing for x < 2
and increasing for 2 < x < 6. Similarly, f ′ is decreasing at x = 6, so f ′ (x) is
negative if 2 < x < 6 and positive for x > 6. So we fill in our chart:
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Region
x<2
2<x<6
x>6
f ′ (x)
negative
positive
negative
f (x)
decreasing increasing decreasing
It is clear from the table that f (x) has a local minimum at x = 2 and a local
maximum at x = 6.
So we were able to do this problem without using the second derivative
stuff we covered last Friday. However, we can determine that f (x) has a local
minimum at x = 2 and a local maximum at x = 6 using the second derivative
test: f ′ (x) is increasing at x = 2, so f ′′ (x) is positive at x = 2 and f has a local
minimum at x = 2. f ′ (x) is decreasing at x = 6 so f ′′ (x) is negative at x = 6
and f has a local maximum at x = 6.
Technically, the above argument isn’t quite right: It’s possible for f ′′ (x) to
be equal to zero at x = 2 or x = 6, and it’s possible that f (x) isn’t twice
differentiable, but we’ll ignore these concerns.
Let’s do one more practice problem on concavity and inflection points:
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Problem 0.3. Sketch the graph of f (x) = xex . Pay particular attention to
where f (x) is increasing and decreasing, and where f (x) is concave up and
concave down.
Let’s take the derivative:
2
f ′ (x) = ex + 2x2 ex
2
2
So we get that this is zero when 2x2 + 1 = 0 or when ex = 0. Notice that
this never happens so there are no local maxima or minima. In particular, this
function is always increasing because f ′ is a positive function.
Let’s look at the concavity now:
2
2
f ′′ (x) = 2xex + 4xex + 4x3 ex
2
2
2
which simplifies to (4x3 + 6x)ex . Again, ex is nonzero (and always positive).
So we just need to determine where 4x3 + 6x is positive and where it is negative.
But 4x3 + 6x is x(4x2 + 6). This is zero when x = 0 and nonzero elsewhere.
Plugging in x = −1 gives f ′′ (−1) = −10 and plugging in x = 1 gives f ′′ (1) = 10.
So there is an inflection point at x = 0. f (x) is concave down for x < 0 and
concave up for x > 0.
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