# Finding the equation of a tangent line for an implicit function

• Nov 18th 2009, 01:35 PM
mbacarella
Finding the equation of a tangent line for an implicit function
Given:

$
xy^{2} + 5x^{2} = 4y
$

Find the equation of the tangent line at x = 1

Say what?

Deriving the implicit function of y with respect to x yields:
$
\frac{dy}{dx} = \frac{5}{y-2}
$

but I'm completely stuck on how to find the equation of the tangent line from this (assuming I did this correctly).

This equation doesn't seem to be a function. I get multiple values of y for x equal 1.

Was I tricked?
• Nov 18th 2009, 01:36 PM
e^(i*pi)
Quote:

Originally Posted by mbacarella
Given:

$
xy^{2} + 5x^{2} = 4y
$

Find the equation of the tangent line at x = 1

Say what?

Deriving the implicit function in terms of x yields
$
\frac{dy}{dx} = \frac{5}{y-2}
$

but I'm completely stuck on how to find the equation of the tangent line from this (assuming I did this correctly).

Find $f(1)$ (note this is the original function!) to give you the corresponding $y$ coordinate.

You can then sub the ordered pair $(1, f(1))$ into $\frac{dy}{dx}$
• Nov 18th 2009, 02:44 PM
mbacarella
My algebra is broken
Thanks.

btw, that implicit function should have read:

$
xy^{2} - 5x^{2} = 4y
$

Anyway, here's what I got:
$
(1, f(1)) = (1, (1)y^{2} - 5(1)^{2} = 4y) = (1, -1)
$

Re-evaluating the derivative:
$
\frac{d}{dx}( xy^{2} -5x^{2} = 4y ) = \frac{dx}{dx}2y^{1}\frac{dy}{dx} - 10x\frac{dx}{dx} =4y^{0}\frac{dy}{dx} = (1)2y\frac{dy}{dx} - 10x(1) = 4(1)\frac{dy}{dx}
$

Now substituting y and x
$
= (1)2(-1)\frac{dy}{dx} - 10(1) = 4(1)\frac{dy}{dx} = \frac{dy}{dx} = \frac{10}{2(-1)-4} = -\frac{5}{3}
$

AKA
$
f^{\prime}(1) = m_{tan} = -\frac{5}{3}
$

Using the point-slope form
$
y - y_{1} = m_{tan}(x - x_1) = y - (-1) = -\frac{5}{3}(x - 1) = y = -\frac{5}{3}x + \frac{5}{3} - 1
$

Or finally?

$
y = -\frac{5}{3}x + \frac{2}{3}
$

I'd appreciate any comments or criticism on my thought processes or approach, even if the answer isn't wrong.
• Nov 18th 2009, 03:59 PM
Calculus26
For

http://www.mathhelpforum.com/math-he...ed113311-1.gif

Use the product rule on the first term

y^2 - 2xydy/dx -10 x = 4dy/dx

dy/dx(-2xy-4) = 10x - y^2

dy/dx = [ y^2 -10 x]/[2xy+4]

When x =1 y^2 -4y -5 = 0 (y-5)(y +1) = 0

There are 2 possibilities (1,5) and (1,-1)

Now finish
• Nov 19th 2009, 06:10 AM
mbacarella
Oops
Thanks for the correction. How did I miss that?

From the implicit derivation:
$
\frac{d}{dx}\left[ xy^{2} -5x^{2} = 4y \right]
$

$
\frac{dx}{dx}y^{2} + x2y^{1}\frac{dy}{dx} - 10x\frac{dx}{dx} =4y^{0}\frac{dy}{dx}
$

$
(1)y^2 + 2xy\frac{dy}{dx} - 10x(1) = 4(1)\frac{dy}{dx}
$

$
y^2 + 2xy\frac{dy}{dx} - 10x = 4\frac{dy}{dx}
$

$
2xy\frac{dy}{dx} - 4\frac{dy}{dx} = 10x - y^{2}
$

$
\frac{dy}{dx}(2xy - 4) = 10x-y^2
$

We seem to disagree on this point:
$
\frac{dy}{dx} = \frac{10x-y^{2}}{2xy-4}
$
• Nov 19th 2009, 07:24 AM
Calculus26
yes--- I had a sign mistake and have changed my last post to reflect this.
• Nov 19th 2009, 11:49 AM
mbacarella
Ah, thank you.

Substituting for x and y:
$
\frac{dy}{dx} = m_{tan} = \frac{10(1) - (-1)^{2}}{2(1)(-1)-4} = -\frac{9}{6} = -\frac{3}{2}
$

Equation of a tangent line in point-slope form
$
y - y_{1} = m_{tan}(x - x_1)
$

Performing the substitutions:
$
y - (-1) = -\frac{3}{2}(x - 1)
$

$
y = -\frac{3}{2}x + \frac{3}{2} - 1
$

Or finally?
$
y = -\frac{3}{2}x + \frac{1}{2}
$

I'm curious why you said there were two possibilities (1, -1) and (1, 5). Wouldn't that make the dependent value y not a function?
• Dec 5th 2009, 09:53 AM
mbacarella
n/m
For completeness sake:

I don't know why I expected to be unable to find tangent lines of relations. Please disregard my blathering on "not a function".