The radius of a sector of a circle is increasing at a rate of 3 cm/sec while the area is decreasing at a rate of 1 cm2/sec. At what rate is the angle of the sector changing when the area is 20 cm2 and the radius os 5cm?
Huh? How is this even possible? Anyway, just use the product rule to get
a = 1/2 r^2 θ
da/dt = 1/2 * 2rθ dr/dt + 1/2 r^2 dθ/dt
So, now all you ave to do is find θ when a=20 and r=5
Then plug that in to solve for dθ/dt
To find the rate at which the angle of the sector is changing, we can use related rates and the formulas for the area and central angle of a sector.
Let's denote the radius of the sector as r (in cm) and the angle of the sector as θ (in radians). We are given the following information:
dr/dt = 3 cm/sec (rate of change of radius)
dA/dt = -1 cm^2/sec (rate of change of area)
A = 20 cm^2 (area)
r = 5 cm (radius)
We know that the area of a sector is given by A = (1/2)θr^2, where θ is the central angle in radians. Taking the derivative of both sides with respect to time (t), we have:
dA/dt = (1/2)d(θr^2)/dt
Since r is changing with respect to time, we need to use the product rule on the right side:
dA/dt = (1/2)(2θr)(dθ/dt) + (1/2)(r^2)(dθ/dt)
We can simplify this to:
-1 cm^2/sec = θr(dθ/dt) + (1/2)r^2(dθ/dt)
Now, we can substitute the given values:
-1 cm^2/sec = (θ)(5 cm)(dθ/dt) + (1/2)(5 cm)^2(dθ/dt)
-1 cm^2/sec = 5θ cm(dθ/dt) + 25 cm^2(dθ/dt)
We can rearrange this equation to solve for dθ/dt:
-1 cm^2/sec - 25 cm^2(dθ/dt) = 5θ cm(dθ/dt)
Simplifying further:
-25 cm^2(dθ/dt) = 5θ cm(dθ/dt) + 1 cm^2/sec
-30θ cm(dθ/dt) = 1 cm^2/sec
dθ/dt = (1 cm^2/sec) / (-30θ cm)
Now, we can substitute the given values of θ = A / (0.5r^2) = 20 cm^2 / (0.5 * 5 cm)^2 = 20 cm^2 / 12.5 cm^2 = 8/5 radians:
dθ/dt = (1 cm^2/sec) / (-30θ cm)
dθ/dt = (1 cm^2/sec) / (-30 * 8/5 cm)
dθ/dt = (1 cm^2/sec) / (-240/5 cm)
dθ/dt = (1 cm^2/sec) / (-48 cm)
dθ/dt = -1/48 radians/sec
Therefore, the angle of the sector is changing at a rate of -1/48 radians/sec when the area is 20 cm^2 and the radius is 5 cm.