A particles position (in meters) is given by the equation s(t) = 5x sin (2x), on the interval [-1,4] minutes. What is its instantaneous velocity at t = 0
you did not express s as a function of t
perhaps
s(t)=5 t sin 2t ?
if so
ds/dt = 5 [ 2 t cos 2t + sin 2t ]
at t = 0 that is
5 [ 0+0 ] = 0
Well, if I remember correctly, the formula for the instantaneous velocity of a particle is the derivative of its position function. In this case, the position function is s(t) = 5t sin(2t). So, let's find the derivative!
But before we do that, let me tell you a joke to lighten the mood. Why don't scientists trust atoms? Because they make up everything!
Back to business! The derivative of s(t) = 5t sin(2t) would be pretty messy to compute using the product rule, so let's use the chain rule instead.
The chain rule states that if we have a composition of functions like f(g(t)), the derivative of f(g(t)) is f'(g(t)) * g'(t).
In our case, the outer function is 5t and the inner function is sin(2t). The derivative of the outer function is just 5, and the derivative of the inner function is 2cos(2t).
Now, we multiply them together: 5 * 2cos(2t) = 10cos(2t).
And there you have it! The instantaneous velocity at t = 0 for the given particle is 10cos(0) = 10 meters per minute.
To find the instantaneous velocity at t = 0, we need to calculate the derivative of the position function with respect to time.
Given the position function s(t) = 5t sin(2t), we can find the derivative of s(t) with respect to t, denoted as v(t), which represents the velocity function.
To find v(t), we will use the product rule and chain rule:
1. Start by differentiating the first term, 5t, with respect to t. The derivative of 5t with respect to t is 5.
2. Next, differentiate sin(2t) with respect to t. The derivative of sin(2t) is cos(2t).
3. Apply the chain rule by multiplying the derivative of the outer function, sin(2t), with the derivative of the inner function, 2t. The derivative of 2t is 2.
Putting it all together, the derivative v(t) is:
v(t) = 5 cos(2t) + 2t cos(2t)
Now, we can find the instantaneous velocity at t = 0 by substituting t = 0 into the velocity function:
v(0) = 5 cos(2(0)) + 2(0) cos(2(0))
= 5 cos(0) + 0 cos(0)
= 5(1) + 0
= 5
Therefore, the instantaneous velocity at t = 0 is 5.
To find the instantaneous velocity at t = 0, we need to find the derivative of the position function, s(t), with respect to time. The derivative of a function represents its rate of change at any given point.
First, let's find the derivative of the position function, s(t):
s(t) = 5t sin(2t)
To find the derivative, we can use the product rule of differentiation:
d/dt (f(t) * g(t)) = f'(t) * g(t) + f(t) * g'(t)
Let f(t) = 5t and g(t) = sin(2t).
Now, let's find the derivatives of f(t) and g(t):
f'(t) = 5
g'(t) = 2cos(2t)
Using the product rule, the derivative of s(t) becomes:
s'(t) = f'(t) * g(t) + f(t) * g'(t)
= 5 * sin(2t) + 5t * 2cos(2t)
= 5sin(2t) + 10tcos(2t)
Now, we can find the instantaneous velocity by substituting t = 0 into the derivative function:
s'(0) = 5sin(2(0)) + 10(0)cos(2(0))
= 5sin(0) + 0
= 0
Therefore, the instantaneous velocity at t = 0 is 0 meters per minute.