A) To find the velocity vector, we need to differentiate each component of r(t) with respect to t.
r(t) = (cos(t), sin(t))
r'(t) = (-sin(t), cos(t))
To find r'(π/4), we substitute t = π/4 into r'(t):
r'(π/4) = (-sin(π/4), cos(π/4))
= (-√2/2, √2/2)
Therefore, r'(π/4) = (-√2/2, √2/2).
B) Similarly, for r(t) = (t^2, t^3):
r'(t) = (2t, 3t^2)
To find r'(3), we substitute t = 3 into r'(t):
r'(3) = (2(3), 3(3^2))
= (6, 27)
Therefore, r'(3) = (6, 27).
C) For r(t) = eti + e^(-3t)j + tk:
r'(t) = i * d/dt(et) + j * d/dt(e^(-3t)) + k * d/dt(t)
= i * et + j * (-3e^(-3t)) + k
To find r'(0), we substitute t = 0 into r'(t):
r'(0) = (e^0)i + (-3e^(-3(0)))j + k
= i - 3j + k
Therefore, r'(0) = i - 3j + k.
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,t3)
.
Then r′(3)
= (
,
)?
C) Let r(t)=eti+e−3tj+tk
.
Then r′(0)
=
i+
j+
k
1 answer