A rectangle has sides of length x cm and 2x - 4cm and the length x cm at time t
seconds is given by x = 2 + 3t, (t >=0) . Show that the area, Acm2 , the rectangle, in
terms of t is A =12t + 18t^2 . Hence find the rate of change of the area at the instant when
t = 2.....
Please solve it........i can't understand this question properly.
L = Length = x
W = Width = 2 x - 4
x = 2 + 3 t
L = x
L = 2 + 3 t
W = 2 x - 4 =
W = 2 ( 2 + 3 t ) - 4 =
W = 2 * 2 + 2 * 3 t - 4 =
W = 4 + 6 t - 4 =
W = 6 t
W = 6 t
L = 2 + 3 t
A = W * L =
A = 6 t * ( 2 + 3 t ) =
A = 6 t * 2 + 6 t * 3 t =
A = 12 t + 18 t ^ 2
t = 2
A = 12 t + 18 t ^ 2
A = 12 * 2 + 18 * 2 ^ 2
A = 24 + 18 * 4
A = 24 + 72
A = 96 cm ^ 2
Picking up the equation found by Bosnian at
A = 12t + 18t^2
dA/dt = 12 + 36t ---> the instantaneous rate of change for any t
when t = 2
dA/dt = 12 + 36(2) = 84 cm^2/second
To find the area of the rectangle in terms of t, we need to use the formula for the area of a rectangle: A = length * width.
Given that the length of the rectangle is x cm and the width is 2x - 4 cm, we substitute these values into the formula:
A = (x) * (2x - 4)
Now, let's substitute the given expression for x in terms of t into the equation. We know that x = 2 + 3t:
A = (2 + 3t) * (2(2 + 3t) - 4)
= (2 + 3t) * (4 + 6t - 4)
= (2 + 3t) * 6t
= 12t + 18t^2
Therefore, the area of the rectangle in terms of t is A = 12t + 18t^2.
To find the rate of change of the area at the instant when t = 2, we need to find the derivative of A with respect to t (dA/dt), and then substitute t = 2 into the derivative.
Taking the derivative of A = 12t + 18t^2 with respect to t:
dA/dt = 12 + 36t
Substituting t = 2 into the derivative:
dA/dt = 12 + 36(2)
= 12 + 72
= 84
Therefore, the rate of change of the area at the instant when t = 2 is 84 cm^2/s.
To find the area of the rectangle, we need to multiply the length by the width.
In this case, the length is given by x cm, and at time t seconds, x = 2 + 3t.
The width of the rectangle is given by 2x - 4 cm.
So, the area A of the rectangle is calculated as follows:
A = length x width
A = (2 + 3t) cm * (2(2 + 3t) - 4) cm
A = (2 + 3t) cm * (4 + 6t - 4) cm
A = (2 + 3t) cm * (6t) cm
A = 6t(2 + 3t) cm^2
A = 12t + 18t^2 cm^2
Now, let's find the rate of change of the area at the instant when t = 2.
To find the rate of change, we need to first find the derivative of the area with respect to time t.
The derivative of A (area) with respect to t (time) is given by dA/dt.
dA/dt = d(12t + 18t^2) / dt
dA/dt = 12 + 36t
Substituting t = 2 into the derivative, we get:
dA/dt (at t = 2) = 12 + 36(2)
dA/dt (at t = 2) = 12 + 72
dA/dt (at t = 2) = 84 cm^2/s
Therefore, the rate of change of the area at the instant when t = 2 is 84 cm^2/s.