T^(1/3)=e^(r/10)m^(3/2)
If the mass and radius are given as m= 2 mg and r= 5mm, and the rates of change of the temperature and radius are given as 3 deg/sec and 1 mm/sec, what is the rate of change of mass (m)?
To find the rate of change of mass (m), we can use the chain rule from calculus. The chain rule states that if we have a function of two independent variables, let's call them x and y, and we want to find the derivative of the function with respect to x, we can use the following formula:
dy/dx = (dy/dt)/(dx/dt)
In this case, we have the equation T^(1/3) = e^(r/10) * m^(3/2), where T is the temperature, r is the radius, and m is the mass. We are given the rates of change of temperature and radius, so we can differentiate both sides of the equation with respect to time (t):
(dT/dt) * (1/3) * T^(-2/3) = (1/10) * e^(r/10) * (dr/dt) * m^(3/2) + (3/2) * e^(r/10) * (dm/dt) * m^(1/2)
Now, we can rearrange the equation and solve for (dm/dt) to find the rate of change of mass:
(dm/dt) = [(dT/dt) * (1/3) * T^(-2/3) - (1/10) * e^(r/10) * (dr/dt) * m^(3/2)] / [(3/2) * e^(r/10) * m^(1/2)]
Let's substitute the values given in the problem statement: dT/dt = 3 deg/sec, dr/dt = 1 mm/sec, T = T0 (initial temperature), r = r0 (initial radius), and m = 2 mg.
(dm/dt) = [(3 deg/sec) * (1/3) * T0^(-2/3) - (1/10) * e^(r0/10) * (1 mm/sec) * (2 mg)^(3/2)] / [(3/2) * e^(r0/10) * (2 mg)^(1/2)]
Simplify and substitute the known values:
(dm/dt) = [1 deg/sec * T0^(-2/3) - (1/5) * (1 mm/sec) * (2 mg)^(3/2)] / [(3/2) * (2 mg)^(1/2)]
Now, you can plug in the values for T0 and mg to calculate the rate of change of mass (dm/dt).