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
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?
As mr f said .
Here is the slope of the tangent. Solving the above equation will give you .
Hence the slope of the equation is . Now, you know the value of slope and a point through which the line passes therefore:
The equation of the tangent is where .