Posted by **Allison** on Saturday, March 12, 2011 at 8:14pm.

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

- Calculus -
**bobpursley**, Saturday, March 12, 2011 at 8:28pm
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.

- Calculus -
**Allison**, Saturday, March 12, 2011 at 8:46pm
I just figured it out! thank you!

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