I have a practice question that I am stuck on and need help.
A certain insect pest has infested a cornfield. The pests are spreading through the field in a circular pattern at the rate of 3m/day. When the radius of the infested area is 70m. How fast is the area of the infested region increasing at that moment?
The answer is 420(pi)m^2 /day
I know it has to do with derivatives and rates of change, but I'm having trouble finding the connection for this one.
area-PI r^2
d area/dtime= 2PI r dr/dt
plug the numbers, and solve.
is that area - pir^2?
as in negative?
we use the ^ to say exponent.
No, the "^" symbol in math is used to represent exponentiation, not negativity. In this context, "r^2" means "r squared," which is the radius of the infested region squared. The formula for the area of a circle is A = πr^2, where A represents the area and r is the radius.
To find how fast the area is changing with respect to time, we need to take the derivative of the area equation with respect to time. This is written as dA/dt, where dA represents the change in area and dt represents the change in time.
In this case, we are given that the radius is changing at a rate of 3 m/day, so dr/dt = 3 m/day. We need to plug this information into the formula for the rate of change of the area, which is given by dA/dt = 2πr(dr/dt).
Substituting the given values, we have r = 70 m and dr/dt = 3 m/day. Plugging these values into the equation, we get:
dA/dt = 2π(70)(3) = 420π m^2/day.
Therefore, the rate at which the area of the infested region is increasing at that moment is 420π m^2/day.