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Newbie
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Joined: 23 Jan 2004 Posts: 2247

Posted: 14 Dec 2009 12:58:22 pm Post subject: 


If I have an equation like x³3x²y+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? 

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GloryMXE7 Puzzleman 3000
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Joined: 02 Nov 2008 Posts: 604

Posted: 14 Dec 2009 03:00:17 pm Post subject: 


well first you would set the function to zero so that in your example F(x,y)=x³3x²y+3xy²26=0
the equation of the tangent line is
f_{x}(x_{0},y_{0})*(xx_{0})+f_{y}(x_{0},y_{0})*(yy_{0})=0
where f_{x}(x_{0},y_{0}) is the derivitive of F(x,y) in terms of x
f_{y}(x_{0},y_{0}) is the derivative of F(x,y) in terms of y
x_{0} is the x coordinate of the given point, and y_{0} is the y coordinate of the given point 

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Newbie
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Posted: 14 Dec 2009 03:24:41 pm Post subject: 


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.
Last edited by Guest on 14 Dec 2009 03:26:50 pm; edited 1 time in total 

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GloryMXE7 Puzzleman 3000
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Posted: 14 Dec 2009 05:33:18 pm Post subject: 


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. 

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Newbie
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Posted: 14 Dec 2009 06:52:47 pm Post subject: 


GloryMXE7 wrote: 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.
Calc 3? What? I'm only in Calc 1 in college. Again in my book the only thing was just finding the tangent line and y was already separate with only x variables in the equation. It's just now that I'm seeing this new form on the final study guide. I'm pretty sure I haven't encountered it on anything else until now with an equation written like above. 

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thornahawk μολών λαβέ
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Joined: 27 Mar 2005 Posts: 569

Posted: 14 Dec 2009 10:36:36 pm Post subject: 


x³3x²y+3xy²26=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 

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Newbie
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Posted: 16 Dec 2009 08:48:14 am Post subject: 


Thanks for the help thornahawk. 

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IAmACalculator In a state of quasihiatus
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Joined: 21 Oct 2005 Posts: 1571

Posted: 16 Dec 2009 07:16:38 pm Post subject: 


Thornahawk is a math ninja. Seriously. 

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Graphmastur
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Joined: 25 Mar 2009 Posts: 360

Posted: 16 Dec 2009 08:12:42 pm Post subject: 


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? 

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GloryMXE7 Puzzleman 3000
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Posted: 16 Dec 2009 08:16:33 pm Post subject: 


well m equals N/c^2 

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simplethinker snjwffl
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Joined: 25 Jul 2006 Posts: 700

Posted: 16 Dec 2009 08:21:07 pm Post subject: 


Graphmastur wrote: 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?
The equation of a line passing through the point (x_{0}, y_{0}) is yy_{0} = m * (xx_{0}) where m is the slope. If it's tangent to f(x) at x_{0}, then m=f'(x_{0}), so... (fill in the rest )
Last edited by Guest on 01 Jul 2010 10:00:42 am; edited 1 time in total 

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thornahawk μολών λαβέ
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Joined: 27 Mar 2005 Posts: 569

Posted: 16 Dec 2009 09:36:08 pm Post subject: 


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=x³3xy²
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 xx_{0}.
f(x)=f(x_{0})+f'(x_{0})(xx_{0})+…
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 

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IAmACalculator In a state of quasihiatus
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Joined: 21 Oct 2005 Posts: 1571

Posted: 16 Dec 2009 10:35:37 pm Post subject: 


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

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