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)
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)
A) To find the tangent velocity vector r′(t), we need to take the derivative of r(t) with respect to t.
Given r(t) = (cos(t), sin(t)), we can find r′(t) as follows:
r′(t) = (-sin(t), cos(t))
To find r′(π/4), substitute t = π/4 into r′(t):
r′(π/4) = (-sin(π/4), cos(π/4)) = (-1/√2, 1/√2)
Therefore, r′(π/4) = (-1/√2, 1/√2).
B) Given r(t) = (t^2, t), we can find r′(t) by taking the derivative:
r′(t) = (2t, 1)
To find r′(3), substitute t = 3 into r′(t):
r′(3) = (2(3), 1) = (6, 1)
Therefore, r′(3) = (6, 1).
C) Given r(t) = e^t i + e^(-3t) j + t k, we can find r′(t) by taking the derivative:
r′(t) = e^t i - 3e^(-3t) j + k
To find r′(0), substitute t = 0 into r′(t):
r′(0) = e^0 i - 3e^(0) j + k = i - 3j + k
Therefore, r′(0) = (1, -3, 1).