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Calculus Exam III Review
Topics:
1
...
Motion
3
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Related Rates
5
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Mean Value Theorem
7
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Linear Approximations
1
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NOT vice versa
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absolute value)
(Not differentiable: cusps or sharp corners)
• Critical Number: an x-value where the graph could change direction
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Where f ’(x)= 0 or = undefined
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Come from critical #’s
• Extrema (absolute): THE highest and lowest points
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Existence Theorem: for absolute extrema, the function must be continuous (ie: no holes)
NOTE: Not all critical numbers are relative extrema
NOTE: Relative extrema cannot come from endpoints
NOTE: when simplifying functions, anything that can be cancelled becomes a hole
...
Evaluate the domain of the function
a
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To find vertical asymptotes: goes with the domain, where the function is undefined (set
the denominator = 0 and solve for x)
ii
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If the degree is the same, use ratio
of coefficients
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NOTE: Functions CAN touch horizontal asymptotes, to check if
it does set the original function = horizontal asymptote y-value
iii
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NOTE: if given a domain, make sure critical numbers are in the given domain! If the
critical number is on the domain, it is an endpoint (remember, endpoints are not relative max’s or
min’s)
Take the derivative of the function
Set the derivative equal to 0, solve for x
0
...

This also gives you critical numbers
1
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Negative sign = function is decreasing
1
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These are x-values
...
Max’s and Min’s, plug the x-values
into the original function
Using the Second Derivative:
The second derivative is the derivative of the first derivative
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1
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Set the second derivative equal to 0, solve for x
a
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O
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Negative sign = function is concave down
a
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Where the signs change = where the concavity changes = a P
...
I
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Motion
Position Function: s(t) = -16t + vt + h
-16 = acceleration due to gravity (32 ft/sec ) x ½
v = initial velocity
h = initial height
t = time
2

2

First derivative = Velocity function

v(t)

Second derivative = Acceleration function
a(t)
VERTICAL
To find…
• How long it takes the object to reach the ground:
o Set the position function [s(t)] = 0, solve for t
• Average velocity of the object:
o Plug in each point (given) to the velocity function and average them [v(1) + v(2)
/ 2]
• The speed of the object when it hits the ground:
o NOTE: speed has no direction associated with it
o Speed = absolute value of v(t) when s(t) = 0
1
...
Plug into v(t) {the derivative of s(t)}
3
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Subtract the number of feet fallen from the initial height
2
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Plug this answer into the v(t) function
• The max height of the object:
o NOTE: vertical motion functions look like parabolas
o For down functions: max height = initial height
o For up functions: set v(t) = 0

HORIZONTAL
To Find…
• How many times it changed direction
o Think of critical numbers
1
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Set v(t) = 0
3
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Remember: negative numbers are not included in the function
• When it is increasing
o When the derivative is positive {v(t) > 0 }
o Set v(t) = 0
o Use number line (positive = increasing)
• Speed
o Same rule applies: absolute value of v(t)
• When v is increasing
o Where acceleration is positive {a(t) > 0 }
o Set a(t) = 0
o Use number line (positive = increasing)
• When speed is increasing
o When v(t) and a(t) have the same sign
! Going up, pulled down = speed is decreasing
! Going down, pulled down = speed is increasing
o Use the number line of v(t) and the number line of a(t) stacked on top of each
other to see where the signs line up
• The total distance on an interval
o Plug interval numbers into the original [s(t)] function
o Plug the locations where it changed direction into the original [s(t)] function
o Add up total units travelled
! Keep signs in mind! Visualize a number line
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5 → +20 = 107
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When the derivative is negative, the function is decreasing
...

When the second derivative is negative, the function is decreasing and concave down
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Optimization
Steps:
1
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Write the equation of this variable
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Plug in given information
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Solve for 1 variable if 2 exist in the equation
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Find the derivative and set it equal to 0 (use distance formula if needed)
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Check using the second derivative
a
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If you’re looking for a min: 2 derivative = positive (concave up)
nd

nd

4
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Write down the variables that change with respect to time
2
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Take the derivative with respect to time
4
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Solve
6
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Determine the rate at which
the radius of the balloon is increasing when the diameter of the balloon is 20 cm
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Before we do that let’s notice that both the volume of the balloon and the
radius of the balloon will vary with time and so are really functions of time, V(t) and r(t)
...
This is the rate at
which the volume is increasing
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We want to determine the rate at which the radius is changing
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3

t

Now that we’ve identified what we have been given and what we want to find we need to relate
these two quantities to each other
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V(t) = 4/3 𝝅[r(t)]
2

Now we don’t really want a relationship between the volume and the radius
...
We can do this by differentiating both sides with
respect to t
...

Doing this gives: V’ = 4 r r’
2

Note that at this point we went ahead and dropped the (t) from each of the terms
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The bottom is initially 10 feet away from the wall
and is being pushed towards the wall at a rate of ¼ ft/sec
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We’ve defined the distance of the bottom of the ladder from the wall to be x and the distance of
the top of the ladder from the floor to be y
...

However, as is often the case with related rates/implicit differentiation problems, we don’t write
the (t) part just try to remember this in our heads as we proceed with the problem
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We know that the rate at which the bottom of the ladder is moving towards the wall
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We always
need to be careful with signs with these problems
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This is y’
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As with the first example we first need a relationship between x and y
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x + y = (15) = 225
All that we need to do at this point is to differentiate both sides with respect to t, remembering
that x and y are really functions of t and so we’ll need to do implicit differentiation
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2xx’ + 2yy’ = 0
2

2

2

Next, let’s see which of the various parts of this equation that we know and what we need to find
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Determining x and y is actually fairly simple
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We know that:
distance
= (rate)(time) = (¼)(12) = 3
So, the end of the ladder has been pushed in 3 feet and so after 12 seconds we must have x = 7
Note that we could have computed this in one step as follows,
x = 10-¼(12) = 7
To find y (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the
values of x that we just found above
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Notice that we got the correct sign for y’
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5
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Indeterminant forms include:

Steps:
1
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To make a fraction, use negative exponents: x = 1/x
Take the derivative of the numerator and the denominator individually

...

For example, if x → 4 is beneath the limit function, plug in 4
If the limit is still in indeterminant form, apply the rule again (take the derivative of the
numerator and the denominators respectively again)

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Mean Value Theorem
• Function must be continuous and differentiable
• For any continuous, differentiable function that has two equal values at two distinct
points, the function must have a point on the function where the first derivative is zero
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Determine if the function is continuous and differentiable (if not, you can’t use Rolle’s
Theorem!)
2
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Make sure it is also continuous!
Plug in interval values

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Rolle’s Theorem
f '(c) =0
The special case in mean value theorem when f(a) = f(b)
= the points have the same height:

Steps:
1
...

3
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Plug in each value given in the interval to make sure they exist at the same height
Take the derivative of the function
Set the derivative = 0
Make sure the new x-value exists in the interval

8
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Pick an x-value closest to the specific x-value in question but easier to work with
a
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1, use x = 1
Use Tangent Line:

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Find the derivative of the function (= slope)
b
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Use point-slope to find the equation for the slope
i
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This gives you the approximate value
Concavity determines whether the approximate value is more or less than the actual
value
0
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If the second derivative is negative:
concave down = over approximates
2



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