Question

A rectangle is growing such that the length of a rectangle is 5t+4 and its height is t4, where t is time in seconds and the dimensions are in inches. Find the rate of change of area, A, with respect to time. g

Answer 1

Answer:

Therefore the rate of change of area $25t^4+16t^3$ square inches/ s

Step-by-step explanation:

Given that,

A rectangle is growing such that the length of rectangle is(5t+4) and its height is t⁴.

Where t is in second and dimensions are in inches.

The area of a rectangle is = length× height

Therefore the area of the rectangle is

A(t) = (5t+4) t⁴

⇒A(t) = t⁴(5t+4)

To find the rate change of area we need to find out the first order derivative of the area.

Rules:

$(1)\frac{dx^n}{dx} = nx^{n-1}$

$(2) \frac{d}{dx}(f(x) .g(x))= f'(x)g(x)+f(x)g'(x)$

A(t) = t⁴(5t+4)

Differentiate with respect to t

$\frac{d}{dt} A(t)=\frac{d}{dt} [t^4(5t+4]$

$\Rightarrow \frac{dA(t)}{dt} =(5t+4)\frac{dt^4}{dt} +t^4\frac{d}{dt} (5t+4)$

$\Rightarrow \frac{dA(t)}{dt} = (5t+4)4t^{4-1}+t^4.5$

$\Rightarrow \frac{dA(t)}{dt} =4t^3(5t+4)+5t^4$

$\Rightarrow \frac{dA(t)}{dt} = 20t^4+16t^3+5t^4$

$\Rightarrow \frac{dA(t)}{dt} = 25t^4+16t^3$

Therefore the rate of change of area $25t^4+16t^3$ square inches/ s