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 increase in its circumstance. At this instant, the radius of the cirle is...
Here is what I did: (unsure)
pi(r^2)=2pi(r)
r^2-r=0
r(r-2)=0
r=2
no, it is the rates of increase that are equal, so you have to set the derivatives equal.
dA/dr = 2pi(r)
dC/dr = 2pi
so 2pi(r) = 2 pi
r = 1
To find the radius of the circle at the instant when the rate of increase in the area is equal to the increase in its circumference, you can set up an equation based on the given information.
Let's assume the radius of the circle at the instant in question is represented by "r". The rate at which the radius is increasing is nonzero, so let's denote it as "dr/dt" (rate of change of radius with respect to time).
The area of a circle is given by A = πr^2, and the circumference is given by C = 2πr.
According to the problem, the rate of increase in the area is numerically equal to the increase in the circumference. Mathematically, this can be expressed as:
dA/dt = dC/dt
Differentiating both equations with respect to time (t), we get:
d/dt(πr^2) = d/dt(2πr)
2πr(dr/dt) = 2π(dr/dt)
Canceling out 2π from both sides, we are left with:
r(dr/dt) = dr/dt
Now, we have two possibilities:
1) dr/dt is not equal to zero:
In this case, we can divide both sides of the equation by (dr/dt), yielding:
r = 1
So, the radius of the circle at the instant in question is 1.
2) dr/dt is equal to zero:
If the rate of change of the radius is zero, then the radius is not increasing, and the condition mentioned in the problem cannot be satisfied.
Therefore, the only valid solution is when the radius is equal to 1.
To solve the problem, we first need to understand the formulas for the area and circumference of a circle.
The area of a circle is given by the formula A = πr^2, where A represents the area and r represents the radius.
The circumference of a circle is given by the formula C = 2πr, where C represents the circumference and r represents the radius.
Now, let's use the given information to find the radius of the circle at the instant when the rate of increase in the area is numerically equal to the increase in its circumference.
We are told that the rate of increase in the area is numerically equal to the increase in the circumference. Mathematically, this can be written as:
dA/dr = dC/dr
Differentiating the formulas for the area and circumference with respect to the radius, we get:
dA/dr = 2πr
dC/dr = 2π
Setting these two expressions equal to each other, we have:
2πr = 2π
Dividing both sides by 2π, we get:
r = 1
Therefore, at the instant when the rate of increase in the area is numerically equal to the increase in its circumference, the radius of the circle is 1.