Question

Plz tell me how to find the slope of the tangent and the normal to the curve x 3 +3xy+y 3 =5 at (1,1).

Answer 1

The equation of curve is x³ + 3xy + y³ = 5  ...............(1)
Now, slope of tangent = dy/dx
      Differentiating (1) with respect to x ,we get
        d(x³)/dx + 3[yd(x)/dx + xdy/dx] + d(y³)/dx  = d(5)/dx
      ⇒3x² + 3[y + xdy/dx] + 3y²dy/dx = 0 ..............(∵d(constant)/dx = 0)
      ⇒3x² + 3y + 3xdy/dx + 3y²dy/dx = 0
      ⇒dy/dx(3x + 3y²) = -(3y + 3x²)
      ⇒ dy/dx = -3(y + x²)/3(x + y²)
      ⇒dy/dx = -y - x² / x + y²
at (1,1) i.e. x = 1 & y = 1
  dy/dx = (-1 - 1)/1 + 1
⇒dy/dx = -2/2 = -1
∴slope of tangent = -1
Now,
slope of normal = -dx/dy
                            = 1............................(∵dy/dx = -1)