A bicycle wheel has a diameter D = 60 cm and the tyre has a cross sectional radius of r = 2.5 cm. The volume V of the inner tube is given by
V = pi^2 D r^2 + (8/3*pi*r^3)
If air is being pumped into the tyre so that the volume of the inner tube is increasing at the rate of 12 cm3/sec, what is the rate at which the cross sectional radius r of the inner tube is increasing in cm/sec? (Assume the diameter D is not changing.)
Correct a decimal number to 3 significant figures.
given V as above,
dV/dt = (2pi^2 D r + 8pi r^2) dr/dt
now plug in the numbers, assuming that at the time of interest, r = 2.5cm :
12 = (2pi^2*60*2.5 + 8pi*2.5^2) dr/dt
dr/dt = 6/(25pi(6pi+1))= 0.0038 cm/s
Hi thanks for helping. But when differentiate pi^2 isn't that is 2pi? How come your answer is 2pi^2? and when different r^2 isn't that is 2r? How come your equation is dv/dt = 2pi^2 D r + 8/3*pi*r^2? Because my answer is 2pi D 2r + 8/3*pi*r*^2. anyone can explain? Thanks!
i get the answer already! thank you!
To find the rate at which the cross-sectional radius r of the inner tube is increasing, we need to differentiate the volume V with respect to time and solve for the rate of change of r.
Given:
Diameter D = 60 cm
Cross-sectional radius r = 2.5 cm
Rate of change of volume dV/dt = 12 cm³/sec
Using the given formula for the volume V, we can differentiate it with respect to time:
dV/dt = d/dt (π²Dr² + (8/3)πr³)
To find dV/dt, we need to consider the chain rule when differentiating both terms in the volume equation:
dV/dt = d/dt (π²Dr²) + d/dt ((8/3)πr³)
Let's differentiate each term separately:
For the first term, d/dt (π²Dr²):
Since D is constant and r is a function of time, we can treat D as a constant and apply the power rule for differentiation:
d/dt (π²Dr²) = π²D * d/dt (r²)
Applying the power rule, we get:
d/dt (π²Dr²) = π²D * (2r * dr/dt)
For the second term, d/dt ((8/3)πr³):
Here, r is a function of time, so we need to apply the chain rule:
d/dt ((8/3)πr³) = (8/3)π * d/dt (r³)
Again, applying the power rule, we get:
d/dt ((8/3)πr³) = (8/3)π * (3r² * dr/dt)
Now, let's substitute these values back into the expression for dV/dt:
dV/dt = π²D * (2r * dr/dt) + (8/3)π * (3r² * dr/dt)
Simplifying further:
dV/dt = 2π²Dr * r' + 8πr² * r'
We know that dV/dt = 12 cm³/sec, so let's substitute that value in the equation:
12 = 2π²(60)(2.5) * r' + 8π(2.5)² * r'
Simplifying:
12 = 15π² * r' + 50π * r' = (15π² + 50π) * r'
Now, we can solve for r':
r' = 12 / (15π² + 50π)
Calculating the value:
r' ≈ 0.019 cm/sec (rounded to 3 significant figures)
Therefore, the rate at which the cross-sectional radius r of the inner tube is increasing is approximately 0.019 cm/sec.