Find the equation of the tangent line at the given value of x on the curve y^3+xy^2-5=x+3y^2;x=3

Question

The equation is written in slope intercept form then you distribute the slope anfd get y=

Answer 1

Differentiate wrt x: 3y^2dy/dx+y^2+2xydy/dx=1+6ydy/dx. Putting x=3 in the cubic equation: y^3+3y^2-5=3+3y^2, so y^3=8 and y=2. The tangent is therefore at (3,2). At this point we can work out dy/dx: 12dy/dx+4+12dy/dx=1+12dy/dx and 12dy/dx=1-4=-3 so dy/dx=-1/4. The equation of the tangent is y=mx+c where m=-1/4. We can find c by putting in the point (3,2): 2=-3/4+c, so c=11/4. The equation is y=11/4-x/4 or 4y=11-x.