The area A = πr2 of a circular puddle changes with the radius. At what rate does the area change with respect to the radius when r = 5ft?
A = pi r^2
A/dr = 2 pi r
(which is the circumference of course, draw a circle and a ring of thickness dr on the outside and look at the added area)
if r = 5
dA/dr = 10 pi
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this is most interesting if the radius is changing with time
r = k t
dr/dt = k
dA/dt = dA/dr * dr/dt
dA/dt = k (2 pi r)
To find the rate at which the area of a circular puddle changes with respect to the radius, we need to take the derivative of the formula for the area of a circle with respect to the radius.
The formula for the area of a circle is A = πr^2, where A represents the area and r represents the radius.
To find dA/dr, the rate at which the area changes with respect to the radius, we differentiate the formula A = πr^2 with respect to r:
dA/dr = d/dx (πr^2)
Using the power rule of differentiation, we get:
dA/dr = 2πr
Now, we can substitute the given radius r = 5ft into the above expression to find the rate at which the area changes with respect to the radius when r = 5ft:
dA/dr = 2π(5) = 10π
Therefore, when the radius of the circular puddle is 5ft, the rate at which the area changes with respect to the radius is 10π square feet per foot.
To find the rate at which the area changes with respect to the radius, we need to differentiate the area formula with respect to the radius.
Given that the area of a circle is A = πr^2, where A is the area and r is the radius, we can differentiate this formula with respect to r.
dA/dr = d/dr (πr^2)
Applying the power rule of differentiation, we get:
dA/dr = 2πr
Now, we can substitute the given radius value, r = 5 ft, into this expression to find the rate at which the area changes with respect to the radius when r = 5 ft.
dA/dr = 2π(5)
Simplifying further:
dA/dr = 10π
Therefore, when the radius r = 5 ft, the rate at which the area changes with respect to the radius is 10π square feet per foot.
a = pi r^2
da/dr = 2pi r