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) ?
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.
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.
To find the rates at which the volume and surface area of the balloon are increasing, you need to differentiate the formulas for volume and surface area with respect to time.
For the volume of a sphere, V, the formula is V = (4/3) π r^3.
To differentiate this equation, you need to use the chain rule, which states that if y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). In this case, the function f(g) is (4/3) π g^3, and g(x) is the radius, r.
So, to find dV/dt (the rate at which the volume is changing with respect to time), you can apply the chain rule:
dV/dt = (4/3) π * [(d(r^3)/dt)]
Now, to find the surface area of a sphere, A, the formula is A = 4π r^2.
To differentiate this equation, you again need to apply the chain rule. The function f(g) is 4π g^2, and g(x) is the radius, r.
So, to find dA/dt (the rate at which the surface area is changing with respect to time), you can apply the chain rule:
dA/dt = 4π * [(d(r^2)/dt)]
Now, to find the rates at which the volume and surface area are increasing when the radius is 5 cm, you can substitute the given value into the rate of change equations.
However, note that before doing that, you need to find the derivative of (r^3) and (r^2) with respect to time (dt), since the radius is changing with time.
Using the power rule, the derivative of (r^3) with respect to t is 3r^2 * (dr/dt), and the derivative of (r^2) with respect to t is 2r * (dr/dt).
Substituting these derivatives into the rate of change equations, you get:
dV/dt = (4/3) π * [3(5^2) * (dr/dt)]
dA/dt = 4π * [2(5) * (dr/dt)]
Thus, you need to plug in the radius of 5 cm into the derived equations, along with the given constant rate of change of 2 cm/min, to find the rates at which the volume and surface area of the balloon are increasing.