Finding tangent line to 8x³y² + x²y⁵ + 6 = 4y⁴ - 3x⁴ at (1,3)

[Cod]

Here is the problem:

Find the equation of the line which is tangent to the curve at the point (1,3): 8x^3y^2 + x^2y^5 + 6 = 4y^4 - 3x^4


Here is what I've done so far (I'm stuck now):

(24x^2)(y^2) + (8x^3)(2y dy/dx) + (2x)(y^5) + (x^2)(5y^4 dy/dx) = (16y^3 dy/dx) + 12x^3


Where do I go from here? Collect like terms? Any help is greatly appreciated.

 

[Doc Al]

Originally posted by Cod
Where do I go from here? Collect like terms? Any help is greatly appreciated.

Start plugging in numbers and solve for dy/dx.

 

[Cod]

Originally posted by Doc Al
Start plugging in numbers and solve for dy/dx.

So I just plug in a 1 where x's are and a 3 where the y's are?

 

[Doc Al]

Originally posted by Cod
So I just plug in a 1 where x's are and a 3 where the y's are?

You got it.

 

[Cod]

But the differentiation process isn't done. I haven't even factored out the dy/dx yet to get the derivative. I thought I plugged in the points AFTER I found the derivative...?

 

[Doc Al]

Originally posted by Cod
But the differentiation process isn't done.

Sure it is. All you need is dy/dx at one point, not at all points.

I haven't even factored out the dy/dx yet to get the derivative.

No need to do that!

I thought I plugged in the points AFTER I found the derivative...?

That would be the hard way. Don't do it.

I think you are confusing what you need to do here, which is find the slope at a particular point, with a much harder problem: finding dy/dx as a function of x for all x. Lot's of luck with that one.

 

[paul11273]

Here is what I've done so far (I'm stuck now):
(24x^2)(y^2) + (8x^3)(2y dy/dx) + (2x)(y^5) + (x^2)(5y^4 dy/dx) = (16y^3 dy/dx) + 12x^3

Unless I am mistaken, shouldn't the last term be -12x^3 ?
You have a + there.

I figured dy/dx=-34.
That would make the equation of the tangent line y=-34x+37.

Can someone confirm this? I am rusty...

Whoops, caught an error myself...edited to add the +37 in my answer!

 

[Doc Al]

Originally posted by paul11273
Unless I am mistaken, shouldn't the last term be -12x^3 ?
You have a + there.

Good catch.

That would make the equation of the tangent line y=-34x.

No. The line had better contain the point (1,3).

 

[paul11273]

No. The line had better contain the point (1,3).

Yeah, I have to think a little longer next time.

y=-34x+37.