the area of a circle increases at a rate of 1 cm²/s.\na. how fast is the radius changing when the radius is…

Explanation:

Step1: Recall the formula for the area of a circle

The area of a circle is (A=\pi r^{2}). Differentiate both sides with respect to time (t) using the chain - rule. (\frac{dA}{dt}=2\pi r\frac{dr}{dt}).

Step2: Solve for (\frac{dr}{dt}) in part (a)

We know that (\frac{dA}{dt} = 1\space cm^{2}/s) and (r = 5\space cm). From (\frac{dA}{dt}=2\pi r\frac{dr}{dt}), we can solve for (\frac{dr}{dt}). [ \frac{dr}{dt}=\frac{\frac{dA}{dt}}{2\pi r} ] Substitute (\frac{dA}{dt}=1) and (r = 5) into the formula: (\frac{dr}{dt}=\frac{1}{2\pi\times5}=\frac{1}{10\pi}\space cm/s)

Step3: Recall the formula for the circumference of a circle

The circumference of a circle is (C = 2\pi r), so (r=\frac{C}{2\pi}). The area formula (A=\pi r^{2}=\pi(\frac{C}{2\pi})^{2}=\frac{C^{2}}{4\pi}) Differentiate (A) with respect to (t): (\frac{dA}{dt}=\frac{2C}{4\pi}\frac{dC}{dt}). But we also know from the original area - radius relation (\frac{dA}{dt}=2\pi r\frac{dr}{dt}), and since (C = 2\pi r) (so (r=\frac{C}{2\pi})), (\frac{dA}{dt}=C\frac{dr}{dt})

Step4: Solve for (\frac{dr}{dt}) in part (b)

We know that (\frac{dA}{dt}=1\space cm^{2}/s) and (C = 3\space cm). From (\frac{dA}{dt}=C\frac{dr}{dt}), we solve for (\frac{dr}{dt}) [ \frac{dr}{dt}=\frac{\frac{dA}{dt}}{C} ] Substitute (\frac{dA}{dt}=1) and (C = 3) into the formula: (\frac{dr}{dt}=\frac{1}{3}\space cm/s)

Answer:

a. (\frac{1}{10\pi}\space cm/s) b. (\frac{1}{3}\space cm/s)