A stone is thrown into a calm pond and circular ripples are formed at impact. If the radius expands at the rate of 0.5 feet/second, how fast is the circumference and the area of the ripples growing when the radius is 3 feet?
C = 2πr
dC/dt = 2π dr/dt
A = πr^2 dr/dt
dA = 2πr dr/dt
you have the values to plug in for both rates.
Oh, so you want to know how fast things are growing when a stone is thrown into a pond? Well, brace yourself because we're about to dive into some math!
Let's start by finding the formulas for the circumference and area of a circle. The circumference of a circle is given by the formula C = 2πr, where r is the radius. The area of a circle is given by the formula A = πr^2.
Now, we know that the radius is expanding at a rate of 0.5 feet/second. We want to find out how fast the circumference and area are growing when the radius is 3 feet.
To find the rate of change of the circumference, we can differentiate the circumference formula with respect to time (t):
dC/dt = 2π(dr/dt).
Since dr/dt is given as 0.5 feet/second, we can substitute it into the equation when the radius is 3 feet:
dC/dt = 2π(0.5) = π feet/second.
So, the circumference of the ripples is growing at a rate of π feet/second when the radius is 3 feet.
To find the rate of change of the area, we can differentiate the area formula with respect to time (t):
dA/dt = 2πr(dr/dt).
Again, we can substitute the values when the radius is 3 feet:
dA/dt = 2π(3)(0.5) = 3π feet^2/second.
Therefore, the area of the ripples is growing at a rate of 3π feet^2/second when the radius is 3 feet.
I hope that was enlightening, or at least didn't make you feel too waterlogged with information!
To find the rate at which the circumference and area of the ripples are growing, we need to differentiate the formulas for circumference and area with respect to time.
1. Circumference (C) of a circle is given by the formula: C = 2πr
Taking the derivative with respect to time (t), we get: dC/dt = 2π(dr/dt)
2. Area (A) of a circle is given by the formula: A = πr^2
Taking the derivative with respect to time (t), we get: dA/dt = 2πr(dr/dt)
Given that dr/dt = 0.5 ft/second and r = 3 feet, we can substitute these values into the above equations.
1. To find the rate at which the circumference is growing, substitute r = 3 and dr/dt = 0.5 into the equation: dC/dt = 2π(0.5) = π ft/second.
2. To find the rate at which the area is growing, substitute r = 3 and dr/dt = 0.5 into the equation: dA/dt = 2π(3)(0.5) = 3π ft^2/second.
Therefore, when the radius is 3 feet, the circumference is growing at a rate of π ft/second and the area is growing at a rate of 3π ft^2/second.
To find the rates at which the circumference and the area of the ripples are growing, we can apply the formulas for the circumference and the area of a circle, and then differentiate them with respect to time.
The formula for the circumference C of a circle is C = 2πr, where r is the radius. By differentiating this equation with respect to time t, we can find the rate of change of the circumference, which is denoted as dC/dt.
Differentiating C = 2πr with respect to t, we get:
dC/dt = 2π(dr/dt) (1)
The formula for the area A of a circle is A = πr^2. By differentiating this equation with respect to time t, we can find the rate of change of the area, which is denoted as dA/dt.
Differentiating A = πr^2 with respect to t, we get:
dA/dt = 2πr(dr/dt) (2)
Given that the radius is expanding at a rate of 0.5 feet/second, we have dr/dt = 0.5 ft/s.
Substituting this into equations (1) and (2), we can find the rates of change of the circumference and the area:
For the circumference:
dC/dt = 2π(dr/dt)
dC/dt = 2π(0.5 ft/s)
dC/dt = π ft/s
So the circumference of the ripples is growing at a rate of π feet/second.
For the area:
dA/dt = 2πr(dr/dt)
dA/dt = 2π(3 ft)(0.5 ft/s)
dA/dt = 3π ft^2/s
So the area of the ripples is growing at a rate of 3π square feet/second.
Therefore, when the radius is 3 feet, the circumference of the ripples is growing at a rate of π feet/second, and the area of the ripples is growing at a rate of 3π square feet/second.