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## Introduction to related rates

0 energy points

# Worked example: Differentiating related functions

AP.CALC:

CHA‑3 (EU)

, CHA‑3.D (LO)

, CHA‑3.D.1 (EK)

, CHA‑3.D.2 (EK)

## Introduction to related rates

## Video transcript

- The differentiable
functions x and y are related by the following equation. The sine of x plus cosine of y is equal to square root of two. They also tell us that the derivative of x with respect to t is equal to five. They also ask us find the derivative of y with respect to t when y
is equal to pi over four and zero is less than x
is less than pi over two. So given that they are
telling us the derivative of x with respect to t and we wanna find the derivative of y with respect to t, it's a safe assumption that
both x and y are functions of t. So you could even rewrite
this equation right over here. You could rewrite it as sine of x, which is a function of t, plus cosine of y, which is a function of t, is equal to square root of two. Now, it might confuse you a little bit, you're not used to seeing x as a function of a third variable or y as a function of something other than x. But remember, x and y are just variables. This could be f of t,
and this could be g of t instead of x of t and y of t, and that might feel a
little more natural to you. But needless to say,
if we wanna find dy dt, what we want to do is take the derivative with respect to t of both
sides of this equation. So let's do that. So we're gonna do it
on the left-hand side, so it's gonna be we take
that with respect to t, derivative of that with respect to t. We're gonna take the derivative
of that with respect to t. And then we're gonna take the derivative of the right-hand side, this
constant with respect to t. So let's think about each of these things. So what is this. Let me do this in a new color. The stuff that I'm doing in
this aqua color right over here, how could I write that? So I'm taking the derivative
with respect to t, I have sine of something, which
is itself a function of t. So I would just apply the chain rule here. I'm first going to take the
derivative with respect to x of sine of x, I could write sine of x of t, but I'll just revert back
to the sine of x here for simplicity. And then I will then multiply
that times the derivative of the inside, you could
say, with respect to t times the derivative
of x with respect to t. This might be a little counterintuitive to how you've applied
the chain rule before when we only dealt with xs and ys, but all that's happening,
I'm taking the derivative of the outside of the sine of something with respect to the something,
in this case, it is x, and then I'm taking the
derivative of the something, in this case, x with respect to t. Well, we can do the same thing here, or this second term here. So I wanna take the
derivative with respect to y of, I guess you could say the outside, of cosine of y, and then I would multiply that times the derivative
of y with respect to t. And then all of that is
going to be equal to what? Well, the derivative with
respect to t of a constant, square root of two is a constant, it's not gonna change as t changes, so its derivative, its
rate of change is zero. All right, so now we
just have to figure out all of these things. So first of all, the
derivative with respect to x of sine of x is cosine of
x times the derivative of x with respect to t, I'll
just write that out here. The derivative of x with respect to t. And then we're going to have,
it's gonna be a plus here, the derivative of y with respect to t. So plus the derivative
of y with respect to t. I'm just flopping the order here, so that this goes out front. Now, what's the derivative of
cosine of y with respect to y? Well, that is negative sine of y. And so, actually let me
just put a sine of y here, then I'm gonna have a negative. Erase this and put a negative there. And that is all going to be equal to zero. And so what can we figure out now? They've told us that the
derivative of x with respect to t is equal to five, they tell
us that right over here. So this is equal to five. We wanna find the derivative
of y with respect to t. They tell us what y is, y is pi over four. This, y is pi over four, so
we know this is pi over four. And let's see, we have to figure out what, we still have two unknowns here. We don't know what x is and we don't know what the derivative of
y with respect to t is. This is what we need to figure out. So what would x be? What would x be when y is pi over four? Well, to figure that out, we can go back to this original
equation right over here. So when y is pi over four, you get, let me write down. Sine of x plus cosine of pi over four is equal to square root of two. Cosine of pi over four, we revert to our unit or we
think about our unit circle. We're in the first quadrant. If we think in degrees,
it's a 45 degree angle, that's gonna be square
root of two over two. And so we can subtract
square root of two over two from both sides, which is going to give us sine of x is equal to, well,
if you take square root of two over two from square root of two, you're taking half of it away, so you're gonna have half of it left. So square root of two over two. And so, what x value, when
I take the sine of it, and remember, where the angle, if we're thinking when the
unit circle is going to be in that first quadrant, x
is an angle in this case right over here. Well, that's going to be
once again pi over four. So this tells us that x
is equal to pi over four when y is equal to pi over four. And so we know that this
is pi over four as well. So let me just rewrite this, because it's getting a little bit messy. So we know that five times cosine of pi over four minus dy dt, the derivative
of y with respect to t, which is what we want to figure out, times sine of pi over four, is equal to zero, is equal to zero, and we
put some parentheses here, just to clarify things a little bit. All right, so let's see. Now, it's just a little bit of algebra. Cosine of pi over four, we already know is square
root of two over two. Sine of pi over four is also
square root of two over two. Now let's see, what if
we divide both sides of this equation by square
root of two over two? Well, what's that gonna give us? Well, then, this square
root of two over two divided by square root of two over, square root of two over two
divided square root of two over two is gonna be one. Square root of two over two divided square root of two over
two is gonna be one. And then zero divided by
square root of two over two is just still going to be zero. And so this whole thing simplifies to five times one, which is just five, minus the derivative
of y with respect to t is equal to zero, and so there you have it. You add the derivative of y
with respect to t to both sides, and we get the derivative of y
with respect to t is equal to five, when all of these
other things are true. When the derivative of x
with respect to t is five, and the derivative and y, I should say, is equal to pi over four.

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