1 §3.2 related rates. the student will learn about related rates
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Introduction to Related RatesAll of our relationships involve two variables. We have found the derivative with respect to one of those variables in order to find a rate of change in that variable.
Related rate problems. Find the rate of change of
both variables (implicitly) with respect to time and solve for one rate with respect to the other.
There will be two variables and thus two related rates. You will know three of the four parts and find the fourth!
x y dx/dt dy/dt
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Example 1A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 2 feet per second, how fast is the area changing when the radius is 10 feet? [Use A = π R 2 ]
The formula for the area of a circle is A = πr2.Both the area A and the radius r (two variables) of the circle increase with time (two rates), so both are functions of t.
We are told that the radius is increasing by 2 feet per second (dr/dt = 2), and we want to know how fast the area is changing (dA/dt).
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Example 1A rock is thrown into a still pond and causes a circular ripple. If the radius of the ripple is increasing at 2 feet per second, how fast is the area changing when the radius is 10 feet? [Use A = π R 2 ]
Making a drawing.
Note dR/dt = 2 and R = 10 are given. Find dA/dt.
dA1
dt
dA2 10 2 40
dt
A = π R 2
Therefore, at the moment when the radius is 10 feet, the area of the circle is growing at the rate of about 126 square feet per second.
Important! Do you understand it?
dR2R
dt
≐ 126
Substitute what we know.
2d dA R
dt dt
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Solving Related Rate ProblemsStep 1. Identify the quantities that are changing with time. Express these rates as derivatives. There will be 2 of them.
Step 2. Identify all variables, including those that are given and those to be found. There will be 2 of them.
Step 3. Find an equation connecting variables. Sketch a figure if helpful.
Step 4. Implicitly differentiate this equation wrt time.
Step 5. Substitute into the differentiated equation any given values for the variables and the derivatives.Step 6. Solve for the remaining derivative and interpret the answer as the unknown rate.
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Example 2A weather balloon is rising vertically at the rate of 5 meters per second. An observer is standing on the ground 300 meters from the point where the balloon was released. At what rate is the distance between the observer and the balloon changing when the balloon is 400 meters high?
Consider making a drawing.
Note from the drawing that y and z are the variables.
What is an equation that relates y and z?
z 2 = y 2 + 300 2
y
300
z
dz/dt is the unknown.
dy/dt is given as 5 meters per second.
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Example 2 - continuedA weather balloon is rising vertically at the rate of 5 meters per second. An observer is standing on the ground 300 meters from the point where the balloon was released. At what rate is the distance between the observer and the balloon changing when the balloon is 400 meters high? dy/dt = 5 and z 2 = y 2 + 300 2 Find dz/dt.
Implicitly differentiate the equation.
and solving for dz/dt yields,
2 2 2d d dz y 300 and
dt dt dt
2z
dz 2y dy y dy
dt 2z dt z dt
y
300
z
Divided by 2z and cancelled the 2’s.
dz
dt 2y
dy
dt0
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Example 2 - continuedA weather balloon is rising vertically at the rate of 5 meters per second. An observer is standing on the ground 300 meters from the point where the balloon was released. At what rate is the distance between the observer and the balloon changing when the balloon is 400 meters high?
And dy/dt = 5 meters/sec.But when y = 400, z = 500. Why?
dz y dy
dt z dt
5dz
4 meter / sec400
.dt 500
z 2 = 400 2 + 300 2
y
300
z
dy/dt = 5 and z 2 = y 2 + 300 2 Find dz/dt.
We know y and dy/dt but not z!
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Example 3The number x of handbags that a manufacturer will supply per week and the price p (in dollars) are related by the following equation. If the price is rising at $2 per week, find how the supply will change if the current price is $100.
5x 3 = 20,000 + 2p 2
Note. We know dp/dt = 2 and we know p = 100 and we can find dp/dt and dx/dt implicitly, but first we need to find x.
Put 100 into the equation for p and find x.3 25x 20000 2 100 40000 3x 8000x 20
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Example 3 - continuedThe number x of handbags that a manufacturer will supply per week and the price p (in dollars) are related by the following equation. If the price is rising at $2 per week, find how the supply will change if the current price is $100.
5x 3 = 20,000 + 2p 2 dp/dt = 2 x = 20 p = 100
Implicitly differentiate the equation.
and solving for dx/dt yields,
3 2d d d5x 20000 2p and
dt dt dt
215x
2
dx 4p dp
dt 15x dt
dx
dt 0 4p
dp
dt
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Example 3 - continuedThe number x of handbags that a manufacturer will supply per week and the price p (in dollars) are related by the following equation. If the price is rising at $2 per week, find how the supply will change if the current price is $100.
5x 3 = 20,000 + 2p 2 dp/dt = 2 x = 20 p = 100
Now substitute and solve for dx/dt.
2
dx 4p dp
dt 15x dt
2 2
dx 4p dp 4
dt 15x d
100
20t 152
The supply is increasing by 0.13 per week.
0.13
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Summary.
• We learned that related rate problems occur in life and in business and they need calculus for their solutions.
Test Review
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§ 2.1
Know the basic derivative formula.
If f (x) = C then f ’ (x) = 0.
If f (x) = xn then f ’ (x) = n xn – 1.
If f (x) = k • u (x) then f ’ (x) = k • u’ (x) = k • u’.If f (x) = u (x) ± v (x), then
f ’ (x) = u’ (x) ± v’ (x).
Test Review
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§ 2.2
Know the Product Rule. If f (x) and s (x), then
f • s ' + s • f ' df s
dx
Know the Quotient Rule. If t (x) and b (x), then
2
d t b t ' t b'
dx b b
continued
Test Review
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§ 2.2
Know
Marginal average costd
C'(x) C(x)dx
Marginal average revenue dR'(x) R (x)
dx
Marginal average profit dP'(x) P(x)
dx
Test Review
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§ 2.3Know the chain rule.
n n 1d duu nu
dx dx
Know that some functions are not differentiable.
Test Review
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§ 2.5
Know how to do implicit derivatives with respect to x, including the product, quotient and chain rules.
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