A spherical balloon is to be deflated so that its radius decreases at a constant rate of 12 cm/min. At what rate must air be removed when the radius is 5 cm? Must be accurate to the 5th decimal place.
It seems like an easy question but no matter how many times i attempt it I get it wrong.
Most questions of this type appear to disregard pressure, so in the following solution I will do the same (physics expert will surely groan)
V = (4/3)π r^3
dV/dt = 4πr^2 dr/dt
so when r = 5 and dr/dt = -12
dV/dt = 4π(25)(-12)
= appr -3769.91118
Well, it looks like you're really "deflated" by this question! Don't worry, I'm here to help put a smile on your face. Let's pump up our knowledge and solve this together.
To find the rate at which air must be removed, we need to use the formula for the volume of a sphere, V = (4/3)πr^3, where V is the volume and r is the radius.
Now, to find the rate at which air is being removed, we'll take the derivative of the volume with respect to time, dV/dt. This will give us the rate at which the volume is changing with respect to time.
Remember, the rate at which the radius is decreasing is given as dr/dt = -12 cm/min.
Differentiating the volume equation gives us dV/dt = 4πr^2(dr/dt).
Plugging in the numbers, when the radius is 5 cm, we have:
dV/dt = 4π(5^2)(-12) cm^3/min
Calculating this, we get:
dV/dt = -120π cm^3/min
Now, to be accurate to the 5th decimal place, we need to approximate π to 5 decimal places, which is 3.14159.
Therefore, the rate at which air must be removed when the radius is 5 cm is approximately:
dV/dt = -120(3.14159) cm^3/min
Simplifying this, we get:
dV/dt ≈ -376.99112 cm^3/min
So, the rate at which air must be removed when the radius is 5 cm is approximately -376.99112 cm^3/min.
Keep up the good work, and remember that even if answering this question was a bit like being caught in a deflating balloon, you learned something new!
To find the rate at which air must be removed from the balloon when the radius is 5 cm, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
where V represents the volume and r represents the radius.
Differentiating both sides with respect to time (t), we can find the rate of change of the volume:
dV/dt = (4/3) * π * 3r^2 * dr/dt
Given that dr/dt (the rate at which the radius is changing) is -12 cm/min (as the radius is decreasing), and r is 5 cm (when we want to find the rate at which air must be removed), we can substitute these values into the equation:
dV/dt = (4/3) * π * 3(5 cm)^2 * (-12 cm/min)
Simplifying:
dV/dt = (4/3) * π * 3 * 25 cm^2 * (-12 cm/min)
dV/dt = - 4π * 25 cm^2/min
So, the rate at which air must be removed from the balloon when the radius is 5 cm is approximately - 100π cm^2/min.
Now, let's calculate the numerical value, accurate to the 5th decimal place:
dV/dt = - 4 * 3.14159 * 25 cm^2/min
dV/dt ≈ -314.159 cm^2/min
Therefore, the rate at which air must be removed from the balloon when the radius is 5 cm is approximately -314.159 cm^2/min, accurate to the 5th decimal place.
To solve this problem, we can use the formula for the volume of a sphere:
V = (4/3)πr^3
Where V is the volume and r is the radius of the balloon.
To find the rate at which air must be removed, we need to find the rate of change of the volume with respect to time (dV/dt) when the radius is 5 cm.
We are given that the radius is decreasing at a constant rate of 12 cm/min, so the rate of change of the radius with respect to time (dr/dt) is -12 cm/min (negative because it is decreasing).
Now, we need to find the relationship between the rate of change of the radius and the rate of change of the volume. We can use the chain rule of differentiation to relate these rates:
dV/dt = dV/dr * dr/dt
To find dV/dr, we can differentiate the volume formula with respect to r:
dV/dr = 4πr^2
Substituting the given value of r = 5 cm:
dV/dr = 4π(5^2) = 100π cm^2
Now, substituting the given value of dr/dt = -12 cm/min:
dV/dt = (100π cm^2) * (-12 cm/min) = -1200π cm^3/min
To get the answer accurate to the 5th decimal place, we can approximate π as 3.14159:
dV/dt = -1200(3.14159) cm^3/min = -3770.946 cm^3/min
Therefore, air must be removed from the balloon at a rate of approximately 3770.946 cm^3/min when the radius is 5 cm.