Answer to Question #109251 in Calculus for mya 6

Question

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

Answer 1

Let's find $\frac{dy}{dx}$

$3y^2y'+y^2+2xyy'=1+4yy'$

$y'(3y^2+2xy-4y)=1-y^2$

$y'=\frac{1-y^2}{y(3y+2x-4)}$

Let's find

$y(2)=y_0:y_0^3+2y_0^2-25=2+2y_0^2$

$y_0^3=27;y_0=3$

$y'(2)=\frac{1-3^2}{3(3*3+2*2-4)}=-\frac{8}{27}$

The equation of the tangent line

$y=y'(2)(x-2)+y_0$

Therefore

$y_t=-\frac{8}{27}(x-2)+3$