TLA Implicit Differentiation

**Biff** · December 13th 2012, 07:02 AM

Hello,

I have a problem here that I am somewhat unsure about:

"Point (-1, 1) lies on the curve $y^3 - \frac{x}{y} + \frac{y}{x} = 1$ . Use tangent line approximation to approximate the y-coordinate of the point where x = -1.01."

(A) 1.02
(B) 0.978
(C) 0.965
(D) 1.101
(E) 0.98

Here is my go at it:

First, use implicit differentiation to find the derivative:

$3y^2 \cdot \frac{dy}{dx} - \left(\frac{1}{y} - \frac{x}{y^2} \cdot \frac{dy}{dx}\right) + \left(\frac{dy}{dx} \cdot \frac{1}{x^2} - \frac{y}{x^2}\right) = 0$

$\frac{dy}{dx} = \frac{\left(\frac{y}{x^2} + \frac{1}{y}\right)}{\left(3y^2 + \frac{x}{y^2} + \frac{1}{x^2}}\right)$

Substitute (-1, 1) and solve for the tangent line:

$\frac{dy}{dx} = \frac{ \left( \frac{(1)}{(-1)^2} + \frac{1}{(1)}\right) }{ \left( 3(1)^2 + \frac{(-1)}{(1)^2} + \frac{1}{(-1)^2} \right) } = \frac{2}{3}$

Tangent line linear formula:

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

Use TLA approximation formula with a = -1.01:

$L(x) = f(x) + f'(a)(x - a) = 1 + \frac{2}{3}(-1.01)(0.01) = 1.009933 \approx 1.01$

So my guess would probably be A, or 1.02. Am I doing this right? (I could have made a mistake differentiating).

Thanks.

**emakarov** · December 13th 2012, 07:32 AM

Originally Posted by Biff

$3y^2 \cdot \frac{dy}{dx} - \left(\frac{1}{y} - \frac{x}{y^2} \cdot \frac{dy}{dx}\right) + \left(\frac{dy}{dx} \cdot \frac{1}{x^2} - \frac{y}{x^2}\right) = 0$

The last expression in parentheses should be $\left(\frac{dy}{dx} \cdot \frac{1}{x} - \frac{y}{x^2}\right)$ . I get y' = 2 and the answer is (E).

**MarkFL** · December 13th 2012, 07:37 AM

When I differentiate, I get:

$3y^2\frac{dy}{dx}-\frac{y-x\frac{dy}{dx}}{y^2}+\frac{x\frac{dy}{dx}-y}{x^2}=0$

$3y^2\frac{dy}{dx}-\frac{1}{y}+\frac{x}{y^2}\cdot\frac{dy}{dx}+\frac{ 1}{x}\cdot\frac{dy}{dx}-\frac{y}{x^2}=0$

$\left(3y^2+\frac{x}{y^2}+\frac{1}{x} \right)\frac{dy}{dx}=\frac{1}{y}+\frac{y}{x^2}$

$\frac{3xy^4+x^2+y^2}{xy^2}\cdot\frac{dy}{dx}=\frac {x^2+y^2}{x^2y}$

$\frac{dy}{dx}=\frac{y(x^2+y^2)}{x(3xy^4+x^2+y^2)}$

At the point (-1,1), we have the slope:

$\frac{dy}{dx}\left|_{(-1,1)}=\frac{1((-1)^2+1^2)}{(-1)(3(-1)1^4+(-1)^2+1^2)}=\frac{2}{1}=2$

Hence:

$y-1=2(x+1)$

$y=2x+3$

$y(-1.01)=0.98$

**Biff** · December 13th 2012, 08:03 AM

Originally Posted by emakarov

The last expression in parentheses should be $\left(\frac{dy}{dx} \cdot \frac{1}{x} - \frac{y}{x^2}\right)$ . I get y' = 2 and the answer is (E).

That makes better sense, thanks.

**Biff** · December 13th 2012, 08:06 AM

Originally Posted by MarkFL2

Hence:

$y-1=2(x+1)$

$y=2x+3$

$y(-1.01)=0.98$

Thanks Mark, I thought it might have been a differentiating mistake. Also, I see how the TLA equation isn't necessary.

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