A little parametric review help?

Question

I just started taking this calculus course and I'm a little rusty so if you could help me understand this material a little more then I'd be grateful.

2. Determine a cartesian expression for the given parametric curve

x = 2e ^(2t) y= 3e^(-t)

I know that I should be eliminating the parameter by solving for t, but I'm rusty with natural logs.

7. When a mortar shell is fired with an initial velocity Vo ft/sec at an angle a above the horizontal, then its position after t seconds is given by:

x = Vo (cos a) t ; y = Vo(sin a) t - 16t²

If the mortar shell hits the ground 3600 feet from the mortar when a=75º, determine it's Vo

Answer 1

2.

x = 2e^(2t)

x/2 = e^(2t)

ln(x/2) = 2t

½ ln(x/2) = t = ln( √(x/2))

y = 3e^-t

ln(y/3) = -t

-ln(y/3) = t = ln(3/y)

3/y = √(x/2)

9/y² = x/2

y² = 18/x

7. also solve for t:

x = (V cos a)t

t = x / (V cos a) ......... then plug in to other equation

y = (V sin a)[ x / (V cos a)] - 16[x / (V cos a)]²

y = x tan a - 16x² / (V²cos² a)

since y = 0

x tan a = 16x² / (V²cos² a) .......... divide by x, mult by cos² a

sin a cos a = 16x / V²

V² = 16x / (sin a cos a)

now plug in 3600 for x and 75° for a, compute V², and take √.

V² = 230,400

V = 480 ft/s