PLEASE HELP!!
The x and y cooridinates of an object are given by
x(t) = 10t + 5t^2 + 3t^3
y(t) = 20t - 10t^2 + 2t^3
where x and y are in meters and t is in seconds. How fast is the object traveling at t = 1 s?
Please explain in detail how to do this problem.
Thanks
Double post. Already answered
To find how fast the object is traveling at t = 1 s, we need to find the magnitude of its velocity vector at that time.
The velocity vector is the derivative of the position vector, so let's find the derivatives of x(t) and y(t).
x(t) = 10t + 5t^2 + 3t^3
To find dx(t)/dt, we differentiate each term with respect to t:
dx(t)/dt = d(10t)/dt + d(5t^2)/dt + d(3t^3)/dt
= 10 + 10t + 9t^2
Similarly, for y(t) = 20t - 10t^2 + 2t^3,
dy(t)/dt = d(20t)/dt - d(10t^2)/dt + d(2t^3)/dt
= 20 - 20t + 6t^2
Now we have the components of the velocity vector at any given time t.
To find the velocity at t = 1 s, substitute t = 1 into the expressions for dx(t)/dt and dy(t)/dt:
vx = 10 + 10(1) + 9(1)^2
= 10 + 10 + 9
= 29 m/s
vy = 20 - 20(1) + 6(1)^2
= 20 - 20 + 6
= 6 m/s
The magnitude of the velocity vector is given by the Pythagorean theorem:
v = sqrt(vx^2 + vy^2)
= sqrt(29^2 + 6^2)
≈ 29.97 m/s
Therefore, the object is traveling at approximately 29.97 m/s at t = 1 s.