if x=e^t and y=(t-3)^2 , find an equation y=mx+b of the tangent to the curve at (1,9).

Answer 1

dy/dx=2(t-3)dt/dx

but
dx/dt=e^t, so

dy/dx=2(t-3)e^-t
so m= that. Now when x=1, (x=e^t), t must be zero, so

m=2(-3)=-6

y=-6x+b

now, at point 1,9
9=-6+b,or b=15, so equation for line must be

y=-6x+15

check that.

Answer 2

I just figured it out! thank you!