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
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.
A) Given r(t) = (cos(t), sin(t)), to find r'(π/4), we need to compute the derivative of r(t) with respect to t and evaluate it at t = π/4.
Taking the derivative of each component separately, we get:
r'(t) = (-sin(t), cos(t))
Substituting t = π/4 into r'(t), we have:
r'(π/4) = (-sin(π/4), cos(π/4))
Using the trigonometric identities sin(π/4) = cos(π/4) = 1/√2, we get:
r'(π/4) = (-1/√2, 1/√2)
Therefore, r'(π/4) = (-1/√2, 1/√2).
B) Given r(t) = (t^2, t^3), to find r'(3), we need to compute the derivative of r(t) with respect to t and evaluate it at t = 3.
Taking the derivative of each component separately, we get:
r'(t) = (2t, 3t^2)
Substituting t = 3 into r'(t), we have:
r'(3) = (2(3), 3(3)^2)
Simplifying, we get:
r'(3) = (6, 27)
Therefore, r'(3) = (6, 27).
C) Given r(t) = e^t i + e^(-3t) j + t k, to find r'(0), we need to compute the derivative of r(t) with respect to t and evaluate it at t = 0.
Taking the derivative of each component separately, we get:
r'(t) = e^t i - 3e^(-3t) j + k
Substituting t = 0 into r'(t), we have:
r'(0) = e^0 i - 3e^(-3(0)) j + 0k
Simplifying, we get:
r'(0) = i - 3j + 0k
Therefore, r'(0) = i - 3j + 0k.