Question

find the derivative of r(t)=(t^(2)+4,sin(2t),(t)/(2t^(t)+4)) and find the vector equation of the tangent line to the curve r(t) at the point where t=0

Answer 1

I believe the derivative is \[r'(t)=(2t,2\cos(2t),\frac{ 2t^2+4-4t^2 }{ (2t^2+4)^2 }\]

Vector derivatives are slightly confusing to me.

Answer 2

if the third coordinate really has a \(t^t\) then I would expect to see a natural log in the derivative:
you write \( t^t= e^{ln(t)t} \)

Answer 3

ahhhh is see, no sorry let me correct that:

\[r(t)=(t^2+4,\sin(2t),\frac{ t }{ 2t^2+4 })\]

2t^2.

took me a little to see where we got such different answers.

Answer 4

and then the third coordinate simplifies to

\[\frac{ -2t^2+4 }{ (2t^2+4)^2 }\]

Answer 5

so the derivative is: r'(t)=(2t,2cos(2t),\frac{ -2t^2+4 }{ (2t^2+4)^2 })

Now i just need to find the vector equation of the tangent line to the curve where t=0

Answer 6

That makes sense, as t^t at t=0 is indeterminate.

now that you have the derivative, plug in t=0 to get the "slope vector"

also, plug in t=0 into r(t) to get the point on the curve at r(0)

the equation of the line is going to be
p(t)= r(0) + t * v
where v is r'(t) evaluated at t=0

Answer 7

I won't type out the whole answer but i got p(t)=<4i,(2j)t,(0.25k)t>

Thanks for your help.

Answer 8

Looks good, but I would write it either as
(4,2t,0.25t) or
4i + 2t j + 0.25t k

but not as (4i,2t j, 0.25t k) which is a mixture of the two forms