Calculus
posted by Allison .
if x=e^t and y=(t3)^2 , find an equation y=mx+b of the tangent to the curve at (1,9).

dy/dx=2(t3)dt/dx
but
dx/dt=e^t, so
dy/dx=2(t3)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. 
I just figured it out! thank you!
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