She went off on a [b]tangent[/b] again (haha, very cheesy)

L

Lizzie

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Joined
Sep 8, 2005
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317
The tangent line equation to the curve 3x<sup>2</sup>+y<sup>2</sup>=1 at the point (0,1) has a slope whose value is: ?

Maybe I'm just having a brain fart again, but another nudge would be appreciated, lol.[/b]
 
The slope of the tangent line is the value of the derivative at that point. To find this, you need dy/dx. Oh no! The variable "y" isn't by itself? However shall we find dy/dx?!? :wink:

And this is where implicit differentiatation comes into play.

You have the x-value and the y-value for the point in question. So far, so good. Now differentiate implicitly with respect to "x". The 3x<sup>2</sup> term is simple enough. Do the y<sup>2</sup> term the way I showed before. And the derivative of "1" is just "0".

Subtract the x stuff to the right-hand side. Plug in the known values for x and y; dx/dx is just 1, of course, so you're ignoring that.

Now solve for dy/dx. This is your slope.

Eliz.
 
ok...basically, implicit differentiation is used when y isn't by itself to take a derivative? or am I wrong about that?
 
ok, not understanding this...sorry, Maybe I should read a bit more of those websites before buggin ya again, lol.
 
\(\displaystyle {\rm{3x}}^{\rm{2}} + y^2 = 1\quad \Rightarrow \quad 6x + 2yy' = 0\)
\(\displaystyle 6(0) + 2(1)y' = 0\)

Solving for y’, what sort of tangent line do we have?
 
Lizzie said:
ok...basically, implicit differentiation is used when y isn't by itself to take a derivative?
Click to expand...
Exactly. Or when (as in the "two ships" example in the other thread) you're differentiatiating with respect to some entirely other variable.

Eliz.
 
Ah ha! Got it stapel! now I feel like I am actually a little bit in the loop, hehe.
 
Yes, y’=0. The tangent is y−1=0(x−0) or y=1.
That is a horizontal line.

BTW. What does “lmao @pka” mean? I have no clue!
 
I thought what you said about mad_mathematician was funny, so I said lmao@pka. lmao= Laughing my *butt* off, @=at, and then pka. I was just saying that what you said made me laugh!
 
Just keep in mind, you can differentiate implicitly with respect to any variable.

When differentiating with respect to x, you get

\(\displaystyle \
\begin{array}{l}
y^2 + x^2 = 4 \\
2y\frac{{dy}}{{dx}} + 2x = 0 \\
\end{array}
\\)

However, if you were doing a related rates question, you would most likely differentiate with respect to t (for time) which would yield

\(\displaystyle \
\begin{array}{l}
y^2 + x^2 = 4 \\
2y\frac{{dy}}{{dt}} + 2x\frac{{dx}}{{dt}} = 0 \\
\end{array}
\\)

See the difference? You must always look at which variable you are differentiating with respect to.
 
Thanks for everyone's help and haha happy. You know what it means. :roll: lol
 
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