The radius of a circle is increasing at the rate of 0.1 cm/sec. At what rate is the area increasing at the instance when r=5cm?​

3.1416

Step-by-step explanation:

A=pi*r^2, differentiate with respect to t both sides

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

dA/dt=2*pi*5*(0.1)

dA/dt=pi=3.1416 cm^2/sec

Step-by-step explanation:

since the user is listed as beginner, I was wondering, if (while correct) the answer should be based on differentiation (rather advanced topic).

I thought originally this would be about sequences.

and I wondered about the start value.

in any case, here a different view.

the area of a circle is

Ac old = pi × r²

now, r is increasing by 0.1

Ac new = pi×(r+0.1)² = pi×(r² + 0.2r + 0.01) =

= pi×r² + pi×0.2r + pi×0.01 =

= Ac old + pi×0.2r + pi×0.01

so, the increase of the area is

pi×0.2r + pi×0.01

for r=5

pi×0.2×5 + pi×0.01

pi×1 + pi×0.01 = pi + p×0.01 = pi×(1 + 0.01) =

= pi×(1 + (radius change)²)

now, it depends on what your teacher wants to see here.

a "digital stair case" 0.1 by 0.1 increase/sequence approach ?

in this case you might also want to calculate the above with r=4.9 (as only with the last 0.1 step r reaches 5).

and either the r=4.9 (result a tiny bit less than pi) or r=5 (result a tiny bit larger than pi) is correct, of simply the value in the middle (practically pi).

or it was meant to be a continuous increase (not step by step).

in which case we need then to calculate the limit with "radius change" going to 0. which delivers pi as rate result (as with the differentiation).