A point has a position vector given by r of t equals the vector with components t squared, 4 minus , t for all time t ≥ 0 seconds. Find the speed of the object at t = 1 seconds. (10 points)
A) 1
B) 5
C) 2
D) square root of 5
why not just say r(t) = <t^2,-4,t>
then you have
v(t) = <2t,0,1>
v(1) = <2,0,1>
so the speed is just |v(1)|
sorry i wrote it wrong it is, r(t) =( t^2+1,4-t). Everything else remains the same. Please help
To find the speed of the object at t = 1 second, we need to find the derivative of the position vector r(t) with respect to time, and then evaluate it at t = 1.
The position vector given is r(t) = <t^2, 4 - t>.
To find the derivative, we take the derivative of each component of the vector separately. The derivative of t^2 with respect to t is 2t, and the derivative of 4 - t with respect to t is -1.
So, the derivative of r(t) is r'(t) = <2t, -1>.
Now, to find the speed of the object at t = 1 second, we evaluate the magnitude of the derivative vector r'(t) at t = 1.
r'(1) = <2(1), -1> = <2, -1>.
The magnitude of this vector is given by:
|<2, -1>| = sqrt((2^2) + (-1^2)) = sqrt(4 + 1) = sqrt(5).
Therefore, the speed of the object at t = 1 second is the square root of 5.
Hence, the correct answer is option D) square root of 5.