A) Let r(t) = (cos(t), sin(t))
To find the velocity vector r'(t), we take the derivative of r(t) with respect to t.
r'(t) = (-sin(t), cos(t))
For t = π/4,
r'(π/4) = (-sin(π/4), cos(π/4))
Since sin(π/4) = cos(π/4) = √2/2,
r'(π/4) = (-√2/2, √2/2)
B) Let r(t) = (t^2, t)
Again, we take the derivative of r(t) to find r'(t).
r'(t) = (2t, 1)
For t = 3,
r'(3) = (2(3), 1) = (6, 1)
C) Let r(t) = eti + e^(-3t)j + tk
Taking the derivative of r(t), we get r'(t) = e^ti - 3e^(-3t)j + k
For t = 0,
r'(0) = e^0i - 3e^(-3(0))j + k
= 1i - 3j + k
Therefore, r'(0) = (1, -3, 1)
For the given position vectors r(t)
,
compute the (tangent) velocity vector r′(t)
for the given value of t
.
A) Let r(t)=(cost,sint)
.
Then r′(π4)=
B) Let r(t)=(t2,t)
.
Then r′(3)=
C) Let r(t)=eti+e−3tj+tk
.
Then r′(0)
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