Calculus help (tangent lines)

a curve is defined parametrically by x=t^3 - (1/2)t^2 and y=t^4 +1
write equations for all tangent lines with slope 2.

I got to the point where
dy/dx = [4(y-1)^1/4]/3 - 4(y-1)^1/2 = 2

I don't know what to do next..

This is now a quadratic equation.

-4(y-1)^(1/2) + (4/3)*(y-1)^(1/4) - 2 = 0

Multiply both sides by (-3/2)

6*(y-1)^((1/2) - (2)*(y-1)^(1/4) + 3 = 0

Let u = (y-1)^(1/4)

6u^2 - 2u + 3 = 0

Since the determinant of this equation is 4 - 72 = -68 the answers are complex.

Perhaps we should check the derivatives?

dy/dt = 4t^3
dx/dt = 3t^2 - t

dy/dx = (4/3)t - 4t^2 ............ x=t^3 - (1/2)t^2

I don't like this so let's just find t where dy/dx = 2
BTW this is the easiest way to find the answer to this type of problem 9 times out of 10.

(4t^3) / (3t^2 - t) = 2

(4t^2 / 3t - 1) = 2

4t^2 = 6t - 2

2t^2 - 3t + 1 = 0

(2t - 1) * (t - 1) = 0

So t = (1/2) and t = 1 cause dy/dx to equal 2.

x=t^3 - (1/2)t^2 and y=t^4 +1

When t = 1 (x,y) = ((1/2),2) When t = (1/2) (x,y) = (0,(17/16)) <------------ Points

Now we have the slopes (2) and the points.

I won't insult your intelligence by finding the equations by
finding a line given the slope and a point. You can do it. :)

At least consider trying this method in the future. :)

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