The length of rectangle is increasing at a rate of 8cm/s and its width is increasing at a rate of 3cm/s. When the length is 20cm and the width is 10cm, how fast is the area of the rectangle increasing?
a = lw
da/dt = w dl/dt + l dw/dt
da/dt = 10*8 + 20*3 = 140 cm^2/s
To find how fast the area of the rectangle is increasing, we need to use the formula for the area of a rectangle: A = length * width.
Given:
- The rate of change of length (dl/dt) = 8 cm/s
- The rate of change of width (dw/dt) = 3 cm/s
- The length (L) = 20 cm
- The width (W) = 10 cm
We want to find:
- The rate of change of area (dA/dt)
Using the product rule of differentiation, we can differentiate the area formula with respect to time:
dA/dt = d/dt (L * W)
Since L and W are functions of time, we can apply the chain rule:
dA/dt = dL/dt * W + L * dW/dt
Substituting the given values:
dA/dt = (8 cm/s) * (10 cm) + (20 cm) * (3 cm/s)
Simplifying:
dA/dt = 80 cm²/s + 60 cm²/s
dA/dt = 140 cm²/s
Therefore, the area of the rectangle is increasing at a rate of 140 cm²/s when the length is 20 cm and the width is 10 cm.