A person on a hang glider is spiraling upward due to rapidly rising air. The acceleration of the glider is given by a(t)=-3cost i - 3sint j +2 k. The glider departed (t=0) from the point (3,0,0) with a velocity of v(t)=3 j. Find the glider's position as a function of time, t.

Answer 1

To find the position as a function of time, we need to integrate the acceleration to get the velocity and again to get the position. We will use initial conditions of s(0) = (3,0,0) and v(0) = 3j.

Integrating a(t), we get:

v(t) = ∫ a(t) dt = ∫(-3cos(t) i - 3sin(t) j + 2k) dt
v(t) = -3sin(t) i + 3cos(t) j + 2t k

Using the initial condition v(0) = 3j, we get:

0 = -3sin(0) i + 3cos(0) j + 2(0) k
0 = 3j
j = 0

Therefore, the velocity only has a component in the x and z directions:

v(t) = -3sin(t) i + 2t k

Integrating v(t), we get:

s(t) = ∫ v(t) dt = ∫(-3sin(t) i + 2t k) dt
s(t) = 3cos(t) i + t^2 k + C

Using the initial condition s(0) = (3,0,0), we get:

(3,0,0) = 3cos(0) i + 0k + C
C = (3,0,0)

Therefore, the position function is:

s(t) = 3cos(t) i + t^2 k + (3,0,0)

Final answer: s(t) = \<3cos(t) + 3, 0, t^2>