# Implicit differentiation find points where slope is ___

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Oct 8th 2013, 01:28 PM
Jonroberts74
Implicit differentiation find points where slope is ___
I have two problems dealing with this. I am getting myself all screwed up and could use help.

the first is at what points is the slope of the tangent line equal to 1 on [x^2/9] + [y^2/16] = 1

So I got the derivative dy/dx = -[16x/9y]

so now I solve for -[16x/9y] = 1 ; correct?

The other question is find the points at which the tangent is horizontal for

x^3 + 4x^2 +xy^2 - 4y^2 = 0

I got dy/dx = [3x^2 + 8x +y^2]/6

so same thing, solve for dy/dx to be zero?
• Oct 8th 2013, 01:43 PM
SlipEternal
Re: Implicit differentiation find points where slope is ___
Quote:

Originally Posted by Jonroberts74
I have two problems dealing with this. I am getting myself all screwed up and could use help.

the first is at what points is the slope of the tangent line equal to 1 on [x^2/9] + [y^2/16] = 1

So I got the derivative dy/dx = -[16x/9y]

so now I solve for -[16x/9y] = 1 ; correct?

Yes, when you have $\displaystyle -\dfrac{16x}{9y} = 1$, you solve for one of the variables (either x or y) and plug it back into the original equation for the eclipse: e.g. $\displaystyle y = -\dfrac{16x}{9}$, so $\displaystyle \dfrac{x^2}{9} + \dfrac{\left(-\dfrac{16x}{9} \right)^2}{16} = 1$.

Quote:

Originally Posted by Jonroberts74
The other question is find the points at which the tangent is horizontal for

x^3 + 4x^2 +xy^2 - 4y^2 = 0

I got dy/dx = [3x^2 + 8x +y^2]/6

so same thing, solve for dy/dx to be zero?

Yes, you do the same thing to get a relationship between x and y. But you may want to check what you got for the derivative once more. You should have gotten $\displaystyle \dfrac{dy}{dx} = \dfrac{3x^2 + 8x + y^2}{8y- 2xy}$.
• Oct 8th 2013, 03:20 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
ah yeah, oops. got the correct derivative the second time around.
• Oct 8th 2013, 03:33 PM
SlipEternal
Re: Implicit differentiation find points where slope is ___
Quote:

Originally Posted by Jonroberts74
ah yeah, oops. got the correct derivative the second time around.

Also for that second problem, when you plug in zero for the derivative, you can solve for $\displaystyle y^2$ since the original formula does not have any terms with only $\displaystyle y$ in them. This way, you avoid square roots.
• Oct 8th 2013, 03:49 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
Can you show me how you would solve to get that, I'm not getting it. And I tried using wolfram but it's losing me. It's saying multiple both sides by the denominator to clear the fraction. Which I understand in other problems fine but it's confusing me with this.

thanks.
• Oct 8th 2013, 03:57 PM
SlipEternal
Re: Implicit differentiation find points where slope is ___
$\displaystyle \dfrac{dy}{dx} = \dfrac{3x^2 + 8x + y^2}{8y-2xy}$

Set the derivative equal to zero and solve for $\displaystyle y^2$:

$\displaystyle 0 = \dfrac{3x^2 + 8x + y^2}{8y-2xy}$

$\displaystyle 0 = 0\cdot(8y-2xy) = \dfrac{3x^2 + 8x + y^2}{8y-2xy}(8y-2xy) = 3x^2+8x+y^2$

$\displaystyle y^2 = -3x^2-8x$

Plug this into the original equation $\displaystyle x^3 + 4x^2 + xy^2 -4y^2=0$:

\displaystyle \begin{align*}x^3 + 4x^2 + x(-3x^2-8x) -4(-3x^2-8x) & = 0 \\ x^3 + 4x^2 -3x^3 - 8x^2 + 12x^2 + 32x & = 0 \\ -2x^3+8x^2+32x & = 0 \\ -2x(x^2-4x-16) & = 0 \end{align*}
• Oct 8th 2013, 04:51 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
okay perfect, then input the value of x into the original to find the y point. correct?
• Oct 8th 2013, 05:04 PM
SlipEternal
Re: Implicit differentiation find points where slope is ___
Or plug them into $\displaystyle y^2 = -3x^2-8x$
• Oct 8th 2013, 05:06 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
I was just going to ask that haha
• Oct 8th 2013, 05:14 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
should have mentioned there was a diagram showing that the x value wouldn't be exceeding 4 for the the right value for x is x = 2 - 2sqrt5

so plugging that in I get

y^2 = -3[2-2sqrt5] - 8[2 - 2sqrt5]

distribute those out for y = sqrt[-22 + 22sqrt5] but thats not the right answer. what am I missing?
• Oct 8th 2013, 05:16 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
the answer comes out to (2 - 2sqrt5, 2 +/- sqrt[10sqrt(5) - 22])
• Oct 8th 2013, 05:18 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
oh I missed a square. try this now
• Oct 8th 2013, 05:31 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
nope, still not getting it.
• Oct 8th 2013, 05:44 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
I get to -56 +40sqrt5 = y^2 from in the value of x into y^2 = -3x^2 -8x
• Oct 8th 2013, 05:51 PM
Jonroberts74
Re: Implicit differentiation find points where slope is ___
solved, got it, thank you
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last