The discussion centers on the relationship between the rates of change of a circle's area and circumference as its radius increases. Specifically, it establishes that when the rate of increase in area (da/dt) equals the rate of increase in circumference (dc/dt), the equation 2πr (dr/dt) = 2π (dr/dt) simplifies to indicate that the radius must be 1. This conclusion is reached through the application of calculus and the properties of derivatives related to the circle's geometry.
PREREQUISITESStudents studying calculus, particularly those focusing on related rates, as well as educators seeking to clarify concepts of geometry and differentiation.
This is incorrect. Since you have converted from a and c to functions of r, the derivatives are both dr/dt: 2\pi r dr/dt= 2\pi dr/dtDollarBill said:Homework Statement
The radius of a circle is increasing at a nonzero rate, and at a certain instant, the rate of increase in the area of the circle is numerically equal to the rate of increase in its circumference
The Attempt at a Solution
c=circumference
a=area
If the rate of change in the circumference and area are equal,
da/dt = dc/dt
πr2=2πr
2πr da/dt = 2π dc/dt
What was the question the problem asked?So would the radius just be 1?
"The radius of a circle is increasing at a nonzero rate, and at a certain instant, the rate of increase in the area of the circle is numerically equal to the rate of increase in its circumference. At this instant, the radius of the circle is:"HallsofIvy said:This is incorrect. Since you have converted from a and c to functions of r, the derivatives are both dr/dt: 2\pi r dr/dt= 2\pi dr/dt
What was the question the problem asked?