This is actually a related rates question, so you want rates of change: so the rate of change of volume, the rate of change of surface area.
This is represented as the derivative, but before you go finding the first derivative, think on how you should use the change of radius provided in the question.
A spherical balloon is being inflated and the radius is increasing at a constant rate of 2 cm per minute. At what rates are the volume and surface area of the balloon increasing when the radius is 5 cm?
For this problem do I plug in the 5 cm into the Volume formula ( 4/3 pi r^3) or the Area formula (4 pi r^2) ?
3 answers
So would dr/dt = 5 cm? And the after I find the first derivative of the volume do I put the 5 cm into the first derivative and multiply it by 2cm/minute?
No, they said that
the radius is increasing at a constant rate of 2 cm per minute
That means dr/dt = 2.
dv/dt = 4πr^2 dr/dt
da/dt = 8πr dr/dt
You have r and dr/dt, so crank it out.
the radius is increasing at a constant rate of 2 cm per minute
That means dr/dt = 2.
dv/dt = 4πr^2 dr/dt
da/dt = 8πr dr/dt
You have r and dr/dt, so crank it out.
1 answer