Let A(t) be the area of a circle with radius r(t), at time t in min. Suppose the radius is changing at the rate of drdt=6 ft/min. Find the rate of change of the area at the moment in time when r=9 ft

The rate of change is ft^(2)/min

Step-by-step explanation:

The area of a circle is given by the following equation:

To solve this question, we have to realize the implicit differentiation in function of t. We have two variables, A and r. So

We have that:

.

We want to find

So

Since the area is in square feet, the rate of change is in ft^(2)/min.

So the rate of change is ft^(2)/min

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.