Question

The side of a square sheet is increasing at the rate of 4 cm per minute. At what rate is the area increasing when the side is 8 cm long ?

Answer 1

Answer:

64 cm / min

Step-by-step explanation:

Given --->

-----------

Rate of increasing of the side of a square sheet = 4 cm / minute

To find --->

------------

Rate of increasing of area when the side is 8 cm long

Solution--->

--------------

Let side of square sheet be x

ATQ

Rate of increasing of side of a square

sheet = 4 cm / minute

dx

----- = 4 cm / minute

dt

Area of square sheet = Side²

A = x²

Differentiating with respect to t

dA d

------- = -------- ( x² )

dt dt

We have a formula

d / dx ( xⁿ ) = n x ⁿ⁻¹ Applying it here

dA dx

---------- = 2 x -------

dt dt

Putting dx / dt = 4 cm / min

dA

---------- = 2 x ( 4 )

dt

dA

--------- = 8 x

dt

Rate of increasing of area when x = 8cm

dA

------ = 8 ( 8 ) = 64 cm / min

d t

Answer 2

Given:

The side of a square sheet is increasing at the rate of 4 cm per minute.

To find:

At what rate is the area increasing when the side is 8 cm long ?

Explanation:

$\sf let\: the \:side\: of \:the \:square=x$

According to the question

$\bullet\sf \dfrac{dx}{dt}=4cm/min$

We know that

$\boxed{\sf Area \:of \:square =x^2sq.units}$

Differentiating both sides wrt time

$\longrightarrow \sf \dfrac{dA}{dt}=\dfrac{d}{dt}(x^2)$

$\longrightarrow \sf \dfrac{dA}{dt} = 2 {x}^{1} (\dfrac{dx}{dt} ) = 2x( \dfrac{dx}{dt} )$

$\boxed{\tt \dfrac{d(x^n)}{dt}=nx^{n-1} \dfrac{dx}{dt}}$

$\longrightarrow \sf \dfrac{dA}{dt} = 2x(4) = 8x$

At x=8 cm

$\longrightarrow \sf \dfrac{dA}{dt} = 8(8) = 64$

$\huge\boxed{\tt \dfrac{dA}{dt}=64 cm/min}$

Area increases at the rate of 64 cm/min when the side is 8cm long