View Full Version : Implicit Differentiation Problem (stuck at a point)
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.
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.
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?
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.
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...?
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. [:D]
paul11273
Feb17-04, 12:52 PM
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!!!
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
Feb17-04, 01:05 PM
No. The line had better contain the point (1,3).
Yeah, I have to think a little longer next time.
y=-34x+37.
vBulletin® v3.8.7, Copyright ©2000-2012, vBulletin Solutions, Inc.