Question

find the equation of the line tangent to cos(x+y) = 2x+2y at the point (pi,pi)

Answer 1

Are you comfortable with implicit differentiation at all, or do we need help on that part?

Answer 2

I need help with implicit differentiation

Answer 3

Alright. Well, as always, we want to find dy/dx, or y' if you will. Except this time we don't have y by itself. Usually, if we have something like y = x^2, we get dy/dx = 2x and we're done. This time, though, we have y's scattered about and everytime we differentiate one of those y's we're going to get a dy/dx. Well, the idea is to differentiate everything, get all of those dy/dx's collected together and then move everything else to the other side, essentially treating dy/dx as a variable in which we're trying to solve for. So that's the plan.

So to start out, the derivative of cos(x+y). I'll need to differentiate this with chain rule, taking the derivative of cosine first and then multiplying it by the derivative of x+y. Think you could differentiate cos(x+y)?

Answer 4

so first we need to take the derivative of cos (x+y) ?

Answer 5

I know the derivative of cos is -sinx

Answer 6

We'll need to take the derivative of all of the terms in the problem, using any chain rule, product rule, or whatever necessary. While doing this, if we ever apply a derivative rule to a y variable, we'll need to tack on dy/dx.

Answer 7

Right, -sinx. But we want to maintain the inner function. Whenever we take derivative, any inner function or inner portion doesn't get changed. So we will get -sin(x+y), the x+ y part unchanged. But now we must apply chain rule. x+y is an inner function, so we must find its derivative and multiply it by -sin(x+y). So what would the derivative of x+y be?

Answer 8

(y' (x)+1)(-sin(x+y)

Answer 9

Yep, that works. I'll be writing that as -sin(x+y)(1+y') if you don't mind, just to try and not have a ton of clutter. But exactly, you got it. So now the next two derivatives are simple. We know the derivative of 2x = 2 and the derivative of 2y =2, but since it was a y derivative, we attach y'. So all together we get:
\[-\sin(x+y)(1+y') = 2x+2y'\]

So now we have two y's in our equation. We have to solve for y' and get it all by itself. Think you can do that part?

Answer 10

so am I supposed to factor the equation out?

Answer 11

Yes. Just like if we had something like 2x - 5 = 3x + 2, we would want to gather all the x's and then get it by itself. We do the same thing with y', get them all factored out and on one side of the equation by itself.

Answer 12

ok I understand your example but I get confused with the sin part

Answer 13

Well, whatever is inside of the angle of sin cannot be touched. -sin(x+y) is all one unit, it cannot be broken apart without identities, and we don't want to do that. So when we distribute, we will have\[-\sin(x+y)(1+y') = -\sin(x+y) - y'\sin(x+y)\]

Answer 14

ok how did you get -sin (x+y) on the right side?

Answer 15

Well, I was only focused on the sin(x+y) portion of the problem since that was the part that seemed to confuse you. The whole equation I would have is:
\[-\sin(x+y) -y'\sin(x+y) = 2x + 2y'\]

Answer 16

ohh ok I see! that was only half of it thanks!

Answer 17

Yeah, exactly :)

Answer 18

Now we just need to isolate y' on one side of the equation

Answer 19

ok i will do that

Answer 20

Show me what ya got when you finish and we can continue :)

Answer 21

ok so the first step to isolating y' would be to subtract 2x on both sides right? and then we can add y' on both sides to leave y' only on the right?

Answer 22

subtract 2x from both sides and add all of -y'sin(x+y) to both sides. That will get everything y' on the right and everything else on the left.

Answer 23

so we have -sin(x+y) -2x = -y'sin(x+y) +2y' ?

Answer 24

Careful, its now a positive y'sin(x+y)

Answer 25

true thanks!, and then can we divide both sides by sin (x+y)?

Answer 26

did you guys ever differentiate the 2x term?

Answer 27

Ah, didnt even notice, lol. Good catch.

Answer 28

Yeah, should be -2 on the left.

Answer 29

so just to be clear what do you have so far as the equation?

Answer 30

\[-\sin(x+y) -2 = y'\sin(x+y) + 2y'\]

Answer 31

ok and can we divide both sides by sin(x+y)?

Answer 32

Well, we want to factor out y' on the right hand side.

Answer 33

so, y'(sin x+y+y') ?

Answer 34

y'[sin(x+y) + 2]
Do you see how I get that?

Answer 35

yes I got it :)

Answer 36

So then we would divide both sides by [sin(x+y) + 2]

Answer 37

so, y' = -sin (x+y) -2/ [sin (x+y) +2 and can I cancel most of that out?

Answer 38

Which would leave you with? :)

Answer 39

y'= 0 ?

Answer 40

Division can only equal 0 if the numerator is 0.

Answer 41

-sin (x+y)/ sin (x+y)

Answer 42

\[y' = \frac{ -\sin(x+y)-2 }{ \sin(x+y)+2}\]

Your 2's are still there.

Answer 43

\[y' =\frac{ -[\sin(x+y)+2] }{ \sin(x+y) + 2 } =?\]

Answer 44

ok thanks

Answer 45

And what would that cancel out to?

Answer 46

1

Answer 47

-1

Answer 48

oh ok -1

Answer 49

So if y' = -1, do you know what that tells us?

Answer 50

I'm not sure can we plug that number in to the original equation? not sure

Answer 51

Well, we don't have to plug anything in. The idea is that a derivative gives us an equation for slope at any x-value. But we differentiated and got simply got -1. This means the slope of our original function is ALWAYS -1, no matter which point we choose. That's why we take the derivative, so we can figure out a slope.

But now it wants us to construct a tangent line. Do you recall what we need to do to construct one?

Answer 52

we needed our slope which we got, and we need to get the y intercept at (pi,pi)

Answer 53

this way we can get our tangent line equation

Answer 54

Well, a y-intercept would help us construct a line, but that doesn't always help us. We need a line through a "specific point", namely (pi,pi). Now since the slope is always -1, the point doesn't end up mattering as much, but we still always want to use the point we're given.

The idea is we want to use the point-slope form formula, this will help us get a line through a specific point. So the point-slope equation is:
\[y-y_{1} = m(x-x_{1})\] So just plug in the given point and you're slope into the equation and simplify :)

Answer 55

I ended up with y = -x+2pi

Answer 56

Exactly, that's your tangent line.

Answer 57

ok thank you so much!

Answer 58

Yep, np :)