At what rate is the area of the square increasing?

Question

Each side of a square is increasing at a rate of 4 cm/s. At what rate is the area of the square increasing when the area of the square is 81 cm²?

Answer 1

The square's Area as a function of time = A(t) = x(t)²

where x(t) is the value of x @ time t.

When A(t) = 81 sq cm, the square's side length (x) is equal to 9 cm.

At any given time, each side of the square is increasing at the rate of 4 cm / s (= dx / dt).

Apply the Calculus to determine the rate of change in area when A(t) = 81 and dx/dt = 4 cm / s:

Recall, A(t) = x(t)²

Area rate of change = dA / dt = d[x(t)²] / dt = 2 * x(t) * dx / dt (The Chain Rule)

At time t, A = 81 and x = 9; therefore,

dA / dt = 2 * 9(cm) * 4(cm / s) = 72 cm² / s

I hope this helps you.

Gary

Answer 2

Hi Ely, this looks like a related rates problem! Lets start with are givens and solve it from there.

First I will define everything so we know what we are talking about:

X = side length of the square

A(X) = area of the square with side lengths X

X' = 4cm/s

A'(X) = ?

if the Area = 81 cm², then this implies that are side lengths X are = to 9 cm i.e,we just need to take the square root of 81 to get X!

A(X) = X², we will differentiate this using implicit differentiation, remember that X is a function of time.

A'(X) = 2X * X'

X' = 4cm/s

X= 9cm

Plug in our numbers into A'(X)

A'(9) = 2*9cm * 4cm/s = 72 cm²/s

The area of the square is increasing at a rate of 72 cm²/s

Answer 3

Hi,

when problem tells you the rate of side is 4cm/s it means:

you assume side is S so derivative of side respect to t will be dS/dt = 4

Now you need to find of rate of area please write the formula of area and make a function

area = side ^2

y = S^2

Now, (S^2)' = 2 S S' and you Know area = 81= S^2 ---> side = 9 now make a derivative of area

y' =dy/dt= 2S dS/dt

dy/dt= 2 * 9 * 4 =72

I hope it is useful, please let me know if you have any question.

Minoo