Question

A curve is defined by the equation x^3=3x^2y^4-4. Find the slope of the tangent line to this curve at (2,-1).
I know I have to do something with derivatives but I don't even know where to start...

Answer 1

You do know where to start. You just said the derivative. Do that. Find dy/dx.

Answer 2

But... do I have to seperate y?

Answer 3

Or plug the points in...?

Answer 4

K so I got...
3x^2 = 3x^2 (4y^3 dy/dx) + y^4 (6x). Is that right...? And now what do I do?

Answer 5

Yes that's right. Now isolate (dy/dx). Then sub in x = 2 and y = -1. That will give you the derivative at that point. The derivative is the slope of the tangent line at that point. I'm pretty sure I'm right about this.

Answer 6

Thanks.

Answer 7

No problem.

Answer 8

When I isolate, I take the whole piece, right? Like take 3x^2(4y^3dy/dx) and then divide, right?

Answer 9

yes

Answer 10

I got -3x^2(4y^3 dy/dx) = -3x^2 + y^4(6x)...

Answer 11

Subtract y^4(6x) from both sides of your equation from 6 minutes ago. Then divide both sides by (3x^2)(4y^3). That should leave dy/dx by itself on one side of your equation.

Answer 12

Alright... and then I plug in and that's it?

Answer 13

I get dy/dx=(3x^2-6xy^4)/(12x^2y^3). Is that right? o.o

Answer 14

That's what I got. Then when you put in x = 2 and y = -1, you get 0. I'll check it on a graphing program to see if that looks right.

Answer 15

Yea if I get 0...

Answer 16

Do I need to factor the equation before plugging in?

Answer 17

Or is 0 an acceptable answer? I forgot which topic it was where we had to factor when we got a 0.

Answer 18

No.

Answer 19

Factor or leave it?

Answer 20

I don't see why you'd factor it if you have a value for x and y. You plug in the x's and y's. The numerator ends up 0. The denominator isn't zero so it is an acceptable answer. It would mean that the graph flattens at that point. I just have to check a graphing program. One minute.

Answer 21

Alright. Thanks. Sorry it's late I've been doin math all day and this is an old topic we haven't touched in a while lol. I'm a little out of it. :P

Answer 22

I hope it's the right answer. I'm pretty sure it is. My grapher doesn't like the equation. Get some rest.

Answer 23

Alright. Just wanted to make sure I had this all right it's a make up quiz that I really need to pass this class. Thanks for the help. Hopefully we did this right. Thanks a lot man. Gnight.

Answer 24

Here's a link. http://www.mathsrevision.net/advanced-level-maths-revision/pure-maths/calculus/tangents-and-normals
I think we did it right.

Answer 25

Alright. Thanks. :) Gnight.

Answer 26

Closin the question. Thanks bro.