WEBVTT

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- The differentiable
functions x and y are related

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by the following equation.

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The sine of x plus cosine of y

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is equal to square root of two.

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They also tell us that the derivative of x

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with respect to t is equal to five.

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They also ask us find the derivative of y

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with respect to t when y
is equal to pi over four

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and zero is less than x
is less than pi over two.

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So given that they are
telling us the derivative

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of x with respect to t and we wanna find

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the derivative of y with respect to t,

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it's a safe assumption that
both x and y are functions of t.

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So you could even rewrite
this equation right over here.

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You could rewrite it as sine of x,

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which is a function of t,

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plus cosine

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of y, which is a function of t,

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is equal to square root of two.

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Now, it might confuse you a little bit,

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you're not used to seeing x as a function

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of a third variable or y as a function

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of something other than x.

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But remember, x and y are just variables.

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This could be f of t,
and this could be g of t

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instead of x of t and y of t,

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and that might feel a
little more natural to you.

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But needless to say,
if we wanna find dy dt,

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what we want to do is take the derivative

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with respect to t of both
sides of this equation.

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So let's do that.

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So we're gonna do it
on the left-hand side,

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so it's gonna be we take
that with respect to t,

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derivative of that with respect to t.

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We're gonna take the derivative
of that with respect to t.

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And then we're gonna take the derivative

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of the right-hand side, this
constant with respect to t.

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So let's think about each of these things.

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So what is this.

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Let me do this in a new color.

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The stuff that I'm doing in
this aqua color right over here,

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how could I write that?

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So I'm taking the derivative
with respect to t,

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I have sine of something, which
is itself a function of t.

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So I would just apply the chain rule here.

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I'm first going to take the
derivative with respect to x of

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sine of

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x, I could write sine of x of t,

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but I'll just revert back
to the sine of x here

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for simplicity.

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And then I will then multiply
that times the derivative

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of the inside, you could
say, with respect to t

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times the derivative
of x with respect to t.

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This might be a little counterintuitive

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to how you've applied
the chain rule before

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when we only dealt with xs and ys,

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but all that's happening,
I'm taking the derivative

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of the outside of the sine of something

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with respect to the something,
in this case, it is x,

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and then I'm taking the
derivative of the something,

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in this case, x with respect to t.

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Well, we can do the same thing here,

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or this second term here.

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So I wanna take the
derivative with respect to y

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of, I guess you could say the outside,

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of cosine of y,

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and then I would multiply that

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times the derivative
of y with respect to t.

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And then all of that is
going to be equal to what?

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Well, the derivative with
respect to t of a constant,

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square root of two is a constant,

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it's not gonna change as t changes,

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so its derivative, its
rate of change is zero.

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All right, so now we
just have to figure out

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all of these things.

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So first of all, the
derivative with respect to x

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of sine of x is cosine of
x times the derivative of x

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with respect to t, I'll
just write that out here.

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The derivative of x with respect to t.

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And then we're going to have,
it's gonna be a plus here,

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the derivative of y with respect to t.

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So plus the derivative
of y with respect to t.

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I'm just flopping the order here,

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so that this goes out front.

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Now, what's the derivative of
cosine of y with respect to y?

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Well, that is negative sine of y.

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And so, actually let me
just put a sine of y here,

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then I'm gonna have a negative.

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Erase this and put a negative there.

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And that is all going to be equal to zero.

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And so what can we figure out now?

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They've told us that the
derivative of x with respect to t

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is equal to five, they tell
us that right over here.

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So this is equal to five.

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We wanna find the derivative
of y with respect to t.

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They tell us what y is, y is pi over four.

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This, y is pi over four, so
we know this is pi over four.

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And let's see, we have to figure out what,

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we still have two unknowns here.

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We don't know what x is and we don't know

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what the derivative of
y with respect to t is.

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This is what we need to figure out.

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So what would x be?

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What would x be when y is pi over four?

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Well, to figure that out,

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we can go back to this original
equation right over here.

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So when y is pi over four, you get,

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let me write down.

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Sine of x

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plus cosine of pi

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over four is equal to square root of two.

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Cosine of pi over four,

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we revert to our unit or we
think about our unit circle.

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We're in the first quadrant.

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If we think in degrees,
it's a 45 degree angle,

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that's gonna be square
root of two over two.

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And so we can subtract
square root of two over two

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from both sides, which is going to give us

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sine of x is equal to, well,
if you take square root of two

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over two from square root of two,

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you're taking half of it away,

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so you're gonna have half of it left.

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So square root of two over two.

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And so, what x value, when
I take the sine of it,

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and remember, where the angle,

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if we're thinking when the
unit circle is going to be

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in that first quadrant, x
is an angle in this case

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right over here.

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Well, that's going to be
once again pi over four.

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So this tells us that x
is equal to pi over four

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when y is equal to pi over four.

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And so we know that this
is pi over four as well.

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So let me just rewrite this,

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because it's getting a little bit messy.

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So we know that five times

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cosine of pi over four

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minus

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dy dt, the derivative
of y with respect to t,

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which is what we want to figure out,

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times sine of pi over four,

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is equal to zero,

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is equal to zero, and we
put some parentheses here,

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just to clarify things a little bit.

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All right, so let's see.

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Now, it's just a little bit of algebra.

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Cosine of pi over four,

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we already know is square
root of two over two.

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Sine of pi over four is also
square root of two over two.

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Now let's see, what if
we divide both sides

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of this equation by square
root of two over two?

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Well, what's that gonna give us?

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Well, then, this square
root of two over two

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divided by square root of two over,

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square root of two over two
divided square root of two

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over two is gonna be one.

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Square root of two over two divided

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square root of two over
two is gonna be one.

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And then zero divided by
square root of two over two

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is just still going to be zero.

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And so this whole thing simplifies to

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five times one, which is just five,

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minus the derivative
of y with respect to t

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is equal to zero,

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and so there you have it.

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You add the derivative of y
with respect to t to both sides,

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and we get the derivative of y
with respect to t is equal to

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five, when all of these
other things are true.

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When the derivative of x
with respect to t is five,

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and the derivative and y, I should say,

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is equal to pi over four.