Area of a circle incresses by a constant rate of 6?

Question

the area of a circle increases by a constant rate of 6 mi^2/h. How fast is the radius increasing when the area is 9 mi^2? I have been working on this problem for the last four hours (though, I know that it will be some small thing that I have missed).

Answer 1

Let, area = A = pi * r^2
where, r = radius.

given, dA/dt = 6. (rate of change ofd area)
we need to find dr/dt when A = 9, that is, r = root over (9 / pi)=1.69

now, dA/dt = dA/dr * dr/dt
so, 6 = 2 * pi * r * (dr/dt) ; [dA/dt = 6 & dA/dr = 2 * pi * r]
from this equation we find,
dr/dt = 3 / (pi * r)
dr/dt = 3/ (pi * 1.69)
dr/dt = .56 (ans)

Answer 2

Let A be the area, and r be the radius.

You want dr/dt, given that dA/dt = 6.

You have A=pi*r^2. Differentiate both sides with respect to t and get

dA/dt = pi*2r*dr/dt.

Plug in dA/dt=6. You also need r, but since A=pi*r^2, you can plug in A=9 and solve for r. Then plug r into

6 = pi*2r*dr/dt

and solve for dr/dt.

Answer 3

Find an equation for the radius.
Area = 6t
r = (A/pi)^.5

r = (6t/pi)^.5
take the derivitive with respect to t
dr/dt = .5(6/pi)^.5 * t^-.5
plug in t where area == 9. (t=1.5)

dr/dt = .5(6/(1.5pi)), which should be your final answer.
(calculations may be off.)

Answer 4

Area of circle = pi r^2

r = sqrt(A/PI) = (A/pI)^(1/2)

dr/dA = (A/pi)^(-1/2)/2
delta r = 2/(A/pi)^(1/2) delta A

A = 9
delta A = 6
so delta r = 2*6/(9/PI)^(1/2) = 12/sqrt(9/PI) 7.08 mi/h

Answer 5

3.14*r*r=area of a circle=9 m^2
r=sqrt(9/3.14)=1.69m

rate of increase=6 m^2/hr
r=sqrt(6/3.14)=1.38 m/hr

so the radius increasing by,
1.69-1.38=0.31m

Answer 6

take the derivative and find d(9)/dx