Question 1

For Problems 2 and 3:  Correct me if I'm wrong, but what you're really asking is, "Which *type of* equation (below) should be used *as part of the process of solving* the problem?"

The problem had me confused for several minutes. The first time I saw it, I was perplexed and thought, "None of these equations is the right one because there are no derivatives in any of them."

After all, you can plug into the equation the numbers you've been given but you're still several steps short of having the solution.

However, what you're asking is which of the equations provided forms the proper structure by which you can calculate the correct answer. Am I understanding that right?

Accepted Answer

Yes, that was the question. The task was to figure out what the relationship between rates was given a certain word problem.

Question 2

A 20-meter ladder is leaning against a wall. The distance x(t), between the bottom of the ladder and the wall is increasing at a rate of 3 meters per minute. At a certain instant t0 the top of the ladder is y0, 15m from the ground. What is rate of change of the angle between ground and ladder.

Whouldnt you use Sin instead of Cos? You have the opposite - 15m, you have the hyp -20m, you need to find the angle, which should be opp/ hyp?

Whouldnt you use Sin instead of Cos? You have the opposite - 15m, you have the hyp -20m, you need to find the angle, which should be opp/ hyp?

Accepted Answer

You can't, because the question didn't tell you the change of y(t0) and we are looking for the dirivative. The question told us that x(t)=3t so we can use this and the constant that the ladder is 20m to solve for it's derivative.

Question 3

If rate of change of the radius over time is true for every value of time. If radius changes to 17, then does the new radius affect the rate?

Accepted Answer

There can be instances of that, but in pretty much all questions  the rates are going to stay constant.  Especially early on.

Question 4

Hello, can you help me with this question, when we relate the rate of change of radius of sphere to its rate of change of volume, why is the rate of volume change not constant but the rate of change of radius is...? Is it because they aren’t proportional to each other ? Kinda urgent ..thanks

Accepted Answer

It's because rate of volume change doesn't depend only on rate of change of radius,  it also depends on the instantaneous radius of the sphere. We know that volume of a sphere is (4/3)(pi)(r)^3.
Now if we differentiate volume with respect to time,  we will get the rate of change of volume. So, 
dV/dt = d/dt((4/3)(pi)(r)^3)
applying chain rule(i.e. differentiating first wrt r and then wrt t. Remember constant can be taken out. 
dV/dt = 4/3*pi(d/dt(r^3))
      = 4/3*pi*(3r^2)(dr/dt)
Therefore rate of change of volume of a sphere with constant rate of change of radius(dr/dt) (let's call it C for constant) is 
dV/dt = (4piC)(r)^2 where r is constantly changing and hence volume rate isn't constant.

Question 5

Heello, for the implicit differentation of A(t)'=d/dt[pi(r(t)^2)]. I undertsand why the result was 2piR but where did you get the dr/dt come from, thank you.

Accepted Answer

The dr/dt part comes from the chain rule.   2pi*r was the result of differentiating the right side with respect to r.  But we need to differentiate both sides with respect to t (not r).     Think of it as essentially we are multiplying both sides of the equation by d/dt

So to do that for the right side, you'd need to first take the derivative with respect to r    d/dr (which you did to get 2pi*r, but then that result has to be multiplied by dr/dt so that the END result would be you finding out what d/dt is (differentiating that side with respect to time).

If you looked at just the derivative operators with no numbers or symbols there for the right side, it may be more clear.     d/dr * dr/dt...   The dr terms cancel out, and you're left with d/dt.        And both sides of the equation have to be multiplied by the same thing (which was d/dt).     So it ends up being equal to what you did on the left side to get dA/dt.

Hope that helps.

Question 6

for the 2nd problem, you could also use the following equation, d(t)=sqrt ((x^2)+(y^2)), and take the derivate of both sides to solve the problem.

it then seems like the unit of dd(t)/dt (the rate of change of the distance) is km^3/h? how do we make sense of this unit  of the rate of change of the distance? Thanks!

Accepted Answer

True, but here, we aren't concerned about how to solve it. We're only seeing the setup. But yeah, that's how you'd solve it.

(Also, a small tip from my side. When you have a square on one side, don't take it to the other side and make it a square root. Sometimes, keeping it as a square makes stuff much better, both in terms of solving and the amount of writing. I've shown it later on where I take the derivative of s^2 instead of just s)

For the other question, I'm not quite sure how you ended up with those units. I've done a dimensional analysis below (I'll just take d(t) = s to avoid confusion between distance and the differential operator)

So, we have s^2 = x^2 + y^2

Differentiating both sides w.r.t. time, we have

d(s^2)/dt = d(x^2)/dt + d(y^2)/dt

2s(ds/dt) = 2x(dx/dt) + 2y(dy/dt)

Now, dx/dt and dy/dt are instantaneous rates of change of position in the x and y directions respectively. So, I'll call them v_x and v_y. So,

2s(ds/dt) = 2x(v_x) + 2y(v_y)

Simplifying a bit, we have

s(ds/dt) = x(v_x) + y(v_y)

Solving for ds/dt, we have

(ds/dt) = (x(v_x) + y(v_y))/s

Now, s = sqrt(x^2 + y^2). Using this substitution, we have,

(ds/dt) = (x(v_x) + y(v_y))/(sqrt(x^2 + y^2))

Now, let's see the dimensions for this. Both x(v_x) and y(v_y) have the dimensions [L^2T^(-1)]. And the denominator has the dimensions of [L]. So, the dimensions of ds/dt would be [L^2T^(-1)]/[L] = [LT^(-1)]. As the length is in km and time in hours, the units would be km/hr.

	$m$ ‍	$m/h$ ‍	$m^{2}$ ‍	$m^{2} /h$ ‍
$b^{'} (t)$ ‍
$A (t_{0})$ ‍
$h (t_{0})$ ‍
$\frac{d A}{d t}$ ‍

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 4

Analyzing problems involving related rates

Worked example of solving a related rates problem

Making sense of the quantities and their rates

Making sense of the given information

Relating the area and the radius

Differentiating

Solving

Common mistake: Confusing which expressions are variables and which are constants

Common mistake: Selecting an equation that misrepresents the given problem

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