Question

If the slope of the tangent to a function equals 2y^2 . find the equation of the function if it passes through the point (4.5)?

Answer 1

Let the function be y=f(x).

dy/dx=2y^2 so dy/2y^2=dx. Integrate both sides: ∫(y^-2/2.dy)=x+c, where c is a constant.

So -(y^-1)/2=x+c, which can be written 2y=-1/(x+c). This must pass through (4,5):

10=-1/(4+c); 4+c=-0.1, c=-4.1 so 2y=1/(4.1-x) or y=1/(8.2-2x) or y=5/(41-10x).

CHECK:

y=5(41-10x)^-1; y^2=25/(41-10x)^2; dy/dx=-5(41-10x)^-2*(-10)=50/(41-10x)^2=2y^2. So the slope of the tangent is 2y^2, confirmed.