Tangent Line

[Newbie]

If I have an equation like x³-3x²y+3xy²=26 how do I find an equation of the tangent line at the point (2,3)?

Am I suppose to make the equation say "y=" and then find the derivative? And then solve for it at the given point?

[GloryMXE7]

well first you would set the function to zero so that in your example F(x,y)=x³-3x²y+3xy²-26=0
the equation of the tangent line is
f_x(x₀,y₀)*(x-x₀)+f_y(x₀,y₀)*(y-y₀)=0
where f_x(x₀,y₀) is the derivitive of F(x,y) in terms of x
f_y(x₀,y₀) is the derivative of F(x,y) in terms of y
x₀ is the x coordinate of the given point, and y₀ is the y coordinate of the given point

[Newbie]

Any example i've seen in my math book hasn't shown two variables or how that would be worked unless I'm completely missing something.

"fy(x0,y0) is the derivative of F(x,y) in terms of y"

What are the steps to find the derivative in terms of y?

At least it's asking to find y'. I guess what I don't understand is separating the variables out?

Thanks for your help.

[GloryMXE7]

to take the derivative of F(x,y) in terms of y you treat x like a constant instead of a variable
i learned how to find the equation of the tangent line in calculus 3 if there is another way to find the equation of the tangent line either i cant remember it or i haven't learned it.

[Newbie]

[thornahawk]

x³-3x²y+3xy²-26=0

If you start from that implicit equation, you take the chain rule into account. To use the third term of the left hand side as an example, taking the derivative of y with respect to x of 3xy² gives 3x*2y*y'+3y² or 6xyy'+3y² ; do it to all the terms of the left hand side and solve for y', an expression in terms of y and x. Replace y and x with the coordinates of your point, and you have the slope of your tangent line. You should know what to do next to get the equation of the tangent line. ;)

HTH,

thornahawk

[Newbie]

Thanks for the help thornahawk.

[IAmACalculator]

Thornahawk is a math ninja. Seriously.

[Graphmastur]

So if f(x)=N/X, and the tangent line goes through (c,0), what is m and b in the tangent line in the form y=mx+b?

[GloryMXE7]

well m equals -N/c^2

[simplethinker]

[thornahawk]

I should add, for completeness, that the approach I gave in my previous post for implicit equations doesn't always work, especially if you're trying to find tangents through singular points (signaled by a zero denominator or similar behavior). For instance, if you wish to find tangents through the origin of the implicit equation

(x²+y²)²=x³-3xy²

the method I gave won't work, and special tricks are needed.

For explicit functions (y=f(x)), it is also a useful viewpoint to think of the tangent line as the Taylor expansion, cut off at the first power of x-x₀.

f(x)=f(x₀)+f'(x₀)(x-x₀)+…

thornahawk

P.S. @IAmACalculator

Well, I used to be a member of that order, but was suitably defrocked after my Spandex suit shrunk by two sizes in the wash. Now they won't take me back. ;P

[IAmACalculator]

Dang. I've always wanted to learn how to kill someone using only a slide rule and a logarithm table.