# calculus

when finding the tangent line to the curve using T(t)= r(t)prime/ /r(t)prime/ and the eq. of the curve being r(t)= (e^t, te^t, te^(t^2)), at point (1,0,0) why does t=o. How is the parameter for t found or calculated?

1. 0
2. 1
1. I think maybe you mean a surface defined by
x(t) = e^t
y(t) = t e^t
z(t) = t e^(t)^2

Now at (1,0,0)
x(t) = 1
y(t) = 0
z(t) = 0
Where is that true?
well e^t = 1 only where t = 0
and at t = 0, sure enough y = 0 and z = 0
So go ahead, find tangent to the surface at t = 0

posted by Damon
2. let's put this in the form
z = f(x,y)
I will assume that t e^t^2 means
t e^t * e^t
so z = x * y
dz/dx = x dy/dx + y
dz/dy = x + y dx/dy
where
dy/dx = dy/dt*dt/dx =[t+1]e^t*1/(e^t)
dx/dy = dx/dt*dt/dy = e^t * 1/[e^t(t+1)]
in terms of t
dz/dx =[t+1]e^t +t e^t = [2t+1]e^t
dz/dy = e^t + te^t/(t+1) = e^t
The equation of the plane is then
(x-xo)/A =(y-yo)/B =(z-zo)/C
where
A = dz/dx at t=0 = 1
B = dz/dy at t = 0 = 1
C = -1
so
(x-1)/1 = y/1 = z/-1
or in terms of t
e^t - 1 = t e^t = -t e^t * e^t

posted by Damon

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