The radius, r cm , of a spherical balloon at a time t seconds is given by
r = 3 + 2 / (1+t)
What is the initial radius? Find the rate of change of r (w.r.t.t) when t=3
To find the initial radius, we need to substitute t = 0 into the equation:
r = 3 + 2 / (1 + t)
When t = 0:
r = 3 + 2 / (1 + 0)
r = 3 + 2 / 1
r = 3 + 2
r = 5
Therefore, the initial radius of the spherical balloon is 5 cm.
Now, to find the rate of change of r with respect to t when t = 3, we need to compute the derivative of r with respect to t and substitute t = 3 into the resulting equation.
Given the equation:
r = 3 + 2 / (1 + t)
To find the derivative, we can use the quotient rule. The derivative of r with respect to t is:
dr/dt = [d(3) / dt] + [d(2 / (1 + t)) / dt]
The derivative of a constant, such as 3, is 0. So the first term goes to 0:
dr/dt = 0 + [d(2 / (1 + t)) / dt]
Now, we need to differentiate the second part using the quotient rule. Applying the quotient rule, we get:
dr/dt = 0 + [(2 * d(1 + t)/dt - (1 + t) * d(2)/dt) / (1 + t)^2]
Taking the derivatives of the respective terms, we get:
dr/dt = (0 + (2 * 1 - (1 + t) * 0) / (1 + t)^2)
dr/dt = 2 / (1 + t)^2
Now, substituting t = 3 into the equation:
dr/dt = 2 / (1 + 3)^2
dr/dt = 2 / 4^2
dr/dt = 2 / 16
dr/dt = 1 / 8
Therefore, the rate of change of r with respect to t, when t = 3, is 1/8 cm per second.
I'll do one, just for grins.
if you're taking calculus, you should surely be able to do
r = 3+2/(1+0)
to get the radius when t=0!
Having mastered the arcane art of plugging in a value, and knowing that
dr/dt = -2/(1+t)^2
you can probably figure what dr/dt is when t=3.