Let r(t) = e ^t i + e²^t j + (e³^t + 3)k. Find T(t) when t = 0. B) 6i - 6j + 2k E) i

Question

Quizplus · Accepted Answer

T(t) is the unit tangent vector, which can be found by taking the derivative of r(t) and dividing by its magnitude. When t = 0, we have:

r(0) = S1U1P0 i + S1S1P0 j + (3S1U1P13S1S1P0 + 3)k

r'(t) = S1U1P1 e S1U1P1tS1S1P0 i + (S1U1P12 e S1U1P12tS1S1P0 + S1U1P1 e S1U1P1tS1S1P0)j + S1U1P13 e S1U1P1tS1S1P0 k

| r'(t) | = √( (S1U1P1)^2 e^(2S1U1P1tS1S1P0) + (S1U1P12)^2 e^(2S1U1P12tS1S1P0) + (S1U1P1)^2 e^(2S1U1P1tS1S1P0) + (S1U1P13)^2 e^(2S1U1P1tS1S1P0) )

| r'(0) | = √( (S1U1P1)^2 + (S1U1P12)^2 + (S1U1P1)^2 + (S1U1P13)^2 ) = √(3)

T(0) = r'(0) / | r'(0) | = (eS1U1P1S1S1P0 i + (eS1U1P12S1S1P0 + eS1U1P1S1S1P0)j + eS1U1P13S1S1P0 k) / √(3) = 1/√(3) [ (eS1U1P1S1S1P0)i + (eS1U1P12S1S1P0 + eS1U1P1S1S1P0)j + eS1U1P13S1S1P0 k ]

Simplifying, we get:

T(0) = 1/√(3) [ (0)i + (2j) + (1)k ] = 2/√(3) j + 1/√(3) k

Therefore, the correct answer is (C) 11ead0bc_8e04_0adb_99a0_1d32485d2ae2_TB6988_11.

Let R(t) = E t I + E2t J

Let R(t) = E ^t I + E²^t J