# Thread: Finding Equation of Tangent Lines

1. ## Finding Equation of Tangent Lines

Hi everyone.

I'm new to this forum so if I'm doing anything wrong, im sorry

I'm just doing some study revision, and a question popped up like iv'e never seen before. It's this:

"Find the equation of the tangent line at the given value of x on the curve y^3+xy^2-25=x+2y^2, x=2"

I'm only used to finding tangent lines given both an x AND a y.

Can someone help me solve this?

What I did was:
y^3+2y^2-25 = 2 +2y^2
y^3 + 2y^2 - 27 - 2y^2 = 0
y^3 - 27 = 0
y^3 = 27
y = 3

Is this the right way to start?

Then I differentiated y^3+xy^2-25=x+2y^2 to make:
3y^2 + 2xy = 1 + 4y

I'm lost at what to do here, if this is even right. =/
Do I do:
3y^2 + 2(2)y = 1 + 4y
= 3y^2 + 4y = 1 + 4y
= 3y^2 = 1
= y^2 = 1/3
= y = sqrt1/3

And then

y-y1 = m(x-x1)
y - 3 = sqrt1/3 (x-2)
y - 3 = sqrt1/3x - 1.154700538
y = sqrt1/3x + 1.845299462 ?

And how can I get 1.845299462 as a fraction?

Thank you!

2. ## Re: Finding Equation of Tangent Lines

Originally Posted by valdo
Hi everyone.

I'm new to this forum so if I'm doing anything wrong, im sorry

I'm just doing some study revision, and a question popped up like iv'e never seen before. It's this:

"Find the equation of the tangent line at the given value of x on the curve y^3+xy^2-25=x+2y^2, x=2"

I'm only used to finding tangent lines given both an x AND a y.

Can someone help me solve this?

What I did was:
y^3+2y^2-25 = 2 +2y^2
y^3 + 2y^2 - 27 - 2y^2 = 0
y^3 - 27 = 0
y^3 = 27
y = 3

Is this the right way to start?

Then I differentiated y^3+xy^2-25=x+2y^2 to make:
3y^2 + 2xy = 1 + 4y

I'm lost at what to do here, if this is even right. =/
Do I do:
3y^2 + 2(2)y = 1 + 4y
= 3y^2 + 4y = 1 + 4y
= 3y^2 = 1
= y^2 = 1/3
= y = sqrt1/3

And then

y-y1 = m(x-x1)
y - 3 = sqrt1/3 (x-2)
y - 3 = sqrt1/3x - 1.154700538
y = sqrt1/3x + 1.845299462 ?

And how can I get 1.845299462 as a fraction?

Thank you!
Your calculus is wrong. You're meant to use implicit differentiation to get dy/dx. Have you been taught this?

$y^3+xy^2-25=x+2y^2$

$\Rightarrow 3y^2 \frac{dy}{dx} + y^2 + 2xy \frac{dy}{dx} = 1 + 4y \frac{dy}{dx}$

where the chain rule and the product rule have been used.

Substitute x = 2 and y = 3 and solve for dy/dx. Then get the equation of the line in the usual way.

3. ## Re: Finding Equation of Tangent Lines

As mr f said $3y^2\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}=1+4y\frac{d y}{dx}$.

Here $\frac{dy}{dx}$ is the slope of the tangent. Solving the above equation will give you $\frac{dy}{dx}=\frac{-8}{27}$.

Hence the slope of the equation is $m=\frac{-8}{27}$. Now, you know the value of slope and a point through which the line passes therefore:

The equation of the tangent is $m(x-x_1)=y-y_1$ where $x_1=2,y_1=3$.