please help with this math prob???

Question

The position vector for a particle moving in the xy plane for t being greater or equal to 0 is (10ln (1+t), 16sqrt(t)). The slope of the tangent line to the path of the particle at t=4 is????

Answer 1

dy/dx=(dy/dt)/(dx/dt)
=(8/sqrt(t))/(10/(1+t))

for t=4

the answer is
4/2 over 5/5
=2/1=2

Answer 2

x = 10Ln(1 + t)
y = 16t^(1/2)
1 + t = e^(x/10)
t = e^(x/10)
t = (y/16)^2 = e^(x/10)
y/16 = e^x/20
y = 16e^x/20
y' = (4/5)e^x/20
y'(4) = (4/5)e^(4/20)
y'(4) = 0.97712

Answer 3

umm well idk i olny kno 7th grade stuf sorry