Many students find this to be a particularly challenging section and the q_a_ was developed to provide an additional introduction.
Question:
`q001. Suppose that y(t) and x(t) are two functions of t.
What is the derivative with respect to t of the function (y(t))^2?
What is the derivative with respect to t of the function sqrt(x(t))?
Your solution:
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Given solution:
(y(t))^2 is a composite f(g(t)) of the function f(z) = z^2 and the function g(t) = y(t). For this composite
f ' (z) = 2 z and g ' (t) = y '(t). The derivative of (y(t))^2 is therefore
(f (g(t) ) ' = g ' (t) * f ' (g(t) ) = y ' (t) * 2 y(t), where ' now indicates the derivative with respect to t.
In shorthand notation, if we understand that y is regarded as a function of t, we can write this as
a notation we have used previously in implicit differentiation.
We can also write this as
Relating this to our previous f(g(t)) notation:
Similarly sqrt(x(t)) is the composite of f(z) = sqrt(z), with derivative f ' (z) = 1 / (2 sqrt(z)), and g(t) = x(t), with derivative g ' (t) = x ' (t), so that for this function
In shorthand notation we write this as
(sqrt(x)) ' = 1 / (2 sqrt(x) ) * x ', or
d(sqrt(x)) / dt = 1 / (2 sqrt(x) ) * dx/dt.
In summary:
d(y^2) = 2 y * dy/dt, expressed alternatively as 2 y y '
d(sqrt(x)) = 1 / (2 sqrt(x) ) * dx/dt, expressed alternative as 1 / (2 sqrt(x) ) * x '.
The factors y ' = dy/dt and x ' = dx/dt come from the chain rule.
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Question:
`q002. Take the derivative with respect to t of both sides of the equation
y^2 = sqrt(x),
assuming that y and x are both functions of t.
What equation do you get?
Your solution:
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Given solution:
The derivative of the left-hand side is
d (y^2) / dt = 2y * dy/dt
and the derivative of the right-hand side is
d (sqrt(x) ) / dt = 1 / (2 sqrt(x)) dx/dt.
So the resulting equation is
2y * dy/dt = 1 / (2 sqrt(x)) dx /dt.
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Question:
`q003. Solve the equation
2y * dy/dt = 1 / (2 sqrt(x)) dx /dt.
for dy/dt.
Your solution:
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Given solution:
We need only divide both sides by 2 y. We obtain
dy/dt = 1 / (2 y) * 1 / ( 2 sqrt(x) ) dx /dt, which we simplify to get
dy/dt = 1 / (4 y sqrt(x) ) dx/dt.
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Question:
Question:
`q004. We know that dy/dt is the rate at which y changes with respect to t, and that dx/dt is the rate at which x changes with respect to t. For now let's regard t as the clock time, so that speak in terms of 'how fast' each variable is changing. (There is a problem with using the concept of 'how fast', which ignores the question of whether the rate is positive or negative, but we'll keep everything positive on this example and worry about signs just a little later).
Given our previous solution
dy/dt = 1 / (4 y sqrt(x) ) * dx/dt
how many times faster is y changing than x if x = 4 and y = .1?
Your solution:
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Given solution:
Provided all quantities are positive:
According to our solution, dy /dt is 1 / (4 y sqrt(x) times as great as dx/dt.
So if x = 4 and y = .1, we evaluate 1 / (4 y sqrt(x)), obtaining
We conclude that for these values of x and y, dy/dt is 1.25 times as great as dx/dt.
So we say that y is changing '1.25 times as fast' as x.
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Question:
`q005. Continuing with our solution
dy/dt = 1 / (4 y sqrt(x) ) * dx/dt
we consider the question of positive vs. negative rates by asking how many times faster y is changing than x, if x = 25 and y = -.1.
Your solution:
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Given solution:
As in the preceding question we evaluate 1 / (4 y sqrt(x) ) , obtaining
1 / (4 y sqrt(x) ) = 1 / (4 * (-.1) sqrt(25) ) = 1 / (-.4 * 5) = -1/2.
So we are tempted to say that y is changing -1/2 times as fast as x.
We understand how y could change half as fast as x, but what does that negative mean?
Let's go back to our original solution dy/dt = 1 / (4 y sqrt(x) ) * dx/dt. For x = 25 and y = -.1, this solution gives us
dy/dt is the rate at which y changes with respect to t, and dx/dt the rate at which x changes with respect to t. So when dx/dt is positive, dy/dt is negative (and vice versa). A positive rate implies an increasing quantity, while a negative rate implies a decreasing quantity.
So our solution literally says that when x = 25 and y = -.1, then if one rate is increasing the other is decreasing, with y changing half as fast as x.
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Question:
Assuming that x and y are functions of t, take the t derivative of both sides of the equation
x^2 + y^3 = 12
and solve for dy/dt in terms of dx/dt.
Evaluate and interpret your solution for x = 4 and y = 3.
Your solution:
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Given solution:
The derivative of x^2 with respect to t is
d (x^2) / dt = 2 x dx/dt
and the derivative of y^3 with respect to t is
d(y^3) / dt = 3 y^2 dy/dt.
12 is a constant, so its derivative is zero.
Our equation is therefore
2 x dx/dt + 3 y^2 dy/dt = 0.
We easily solve for dy/dt (subtract 3x dx/dt from both sides and divide by 3 y) to get
dy/dt = - 2 x / (3 y^2) * dx/dt.
When x = 4 and y = 3 we obtain
dy/dt = - 2 * 4 / (3 * 3^2) * dx/dt, or
dy/dt = -8/27 * dx/dt.
This tells us that when x = 4 and y = 3, dy/dt is -8/27 as great as dx/dt.
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Question:
`q006. Suppose that for certain values of x and y we find that dy/dt = 2 dx/dt. It should be clear that this tells us that y is changing twice as fast as x.
So if x changes by .03, by about how much does y change?
Why would we say 'about'?
Your solution:
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Given solution:
If y is changing twice as fast as x, then if x changes by .03, we expect y to change by twice as much. We conclude that y changes by .06.
However, as we have seen, the values of x and y can affect the relative rates of change (for example when dy/dt = 1 / (4 y sqrt(x) ) * dx/dt, any change in x or y affects the value of 1 / (4 y sqrt(x)); so once things start changing we can't expect the relative rates to stay the same).
So we say that y change by 'about' .06. To the extent that the changes in x and y are small enough to have little effect on the relative rates, our approximate result that y changes by .06 will be accurate.
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Question:
`q007. Any equation that relates two functions x(t) and y(t) in such a way that we can take the t derivative of both sides will give us an equation which relates dy/dt and dx/dt.
We have seen two example of this.
In each case, the given equation relates the two quantities, and the derivative of the equation relates their rates of change.
So from a relation of the quantities, we get a relation of the rates.
We call the latter an expression of 'related rates'.
Let's consider an application of this concept to a simple situation.
The area and radius of a circle are related by the equation
A = pi r^2.
Let's assume that a circle is growing. As time passes, its area and radius change, so that A and r are both function of clock time t.
Assuming then thAat A and r are both functions of clock time t, what equation do we get when we take the derivative of both sides?
Why would we call this a 'related rates' equation?
Your solution:
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Given solution:
The t derivative of the left-hand side is just dA/dt.
The t derivative of r^2 is 2 r dr/dt. pi is a constant, so by the constant rule the derivative of the right-hand side us 2 pi r dr/dt.
The derivative of the equation is thus
dA/dt = 2 pi r dr/dt.
This equation relates the two rates dA/dt, the rate of change of the circle's area, and dr/dt, which is the rate of change of its radius.
It tell us that the area grows a 2 pi r times the rate at which the radius grows. Since 2 pi r increases as r increases, the relative rate at which area is added increases as the circle gets bigger. If r increases at a constant rate, for example, the circle adds area more and more quickly.
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Question:
As we saw in the previous, the time rates of change of a circle's radius and area are related by the equation
dA/dt = 2 pi r dr/dt.
Suppose that the radius is growing at the constant rate dr/dt = 3 cm / second. When its radius is 50 cm, at what rate is it adding area?
At what rate will it be adding area 10 seconds later?
Your solution:
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Given solution:
Using dr/dt = 3 cm/s and r = 50 cm we get
That is, the circle adds 100 pi square centimeters of area (approximately 314 cm^2) per second.
10 seconds later its radius, which increases at 3 cm/second, will have increased by 30 cm. The radius will therefore be 80 cm.
At this radius we have
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Question:
The radius of a sphere is increasing at 2 feet / hour. At what rate is its volume changing at the instant its radius reaches 20 feet?
Your solution:
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Given solution:
We want to relate the rate at which the volume of a sphere changes with respect to its radius.
We can start with the equation that relates volume and radius. This is just the equation for the volume of a sphere; this equation should be common knowledge:
V = 4/3 pi r^3.
As before we regard volume and radius as functions of clock time t, and we take the t derivative of both sides of the equation, obtaining
dV/dt = 4/3 pi * (3 r^2 dr/dt) = 4 pi r^2 dr/dt.
Now if r = 20 ft and dr/dt = 2 ft / hour, we have
dV/dt = 4 pi r^2 dr/dt = 4 pi * (20 ft)^2 * (2 ft / hr) = 3200 pi ft^3 / hr.
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Question:
If a sphere adds volume at the rate of 1000 ft^3 / hr, then at what rate is its area changing at the instant its volume reaches 36 000 pi ft^3?
Your solution:
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Given solution:
There are two ways to solve this problem, one more elegant than the other.
We start with the equation
V = 4/3 pi r^3
and take time derivatives, obtaining as before
dV/dt = 4 pi r^2 dr/dt.
We know the volume V, and we know dV/dt. However we don't know r or dr/dt. We might seem to be stuck, having only one equation and two unknown quantities.
However we in fact have two equations. The first is our original equation V = 4/3 pi r^3. Since we know V we can use this equation to find r:
V = 4/3 pi r^3 so
r = ( 3 / (4 pi) * V)^(1/3) = ( 3 / (4 pi) * 36 000 pi ft^3) ^ (1/3) = (27 000 ft^3)^(1/3) = 30 ft.
Armed with the value of r, we return to the related-rates equation
dV/dt = 4 pi r^2 dr/dt.
We easily solve for dr/dt, obtaining
dr/dt = (dV/dt) / (4 pi r^2) = (1000 ft^3 / hr) / (4 pi * (30 ft)^2 ) = (1000 ft^3 / hr) / (3600 pi ft^2) = .09 ft / hr, approx.
That is, the when its volume is 36 000 ft^3, the radius of the sphere is changing by about .09 ft / hr.
However the question asked for the rate at which the area of the sphere is changing. Now that we know that dr/dr = .09 ft / hr, we can use the relationship between the area and radius of a sphere, and the corresponding related rates:
The surface area of a sphere of radius r is 4 pi r^2 (which should be general knowledge), so
A = 4 pi r^2 and, taking time derivative of both sides,
dA/dt = 8 pi r dr/dt.
Recalling that at the specified instant r = 30 ft, we have
dA/dt = 8 pi r dr/dt = 8 pi * 30 ft * .09 ft/hr = 820 ft^2 / hr.
The sphere's surface area is adding 8200 square feet every hour, at the instant its volume is 36 000 ft^3 (and its radius is 30 ft).
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