A rectangle initially has width 3 meters and length 5 meters and is expanding so that the area increases at a rate of 8 square meters per hour. If the width increases by 50 centimeters per hour how quickly does the length increase initially? [hint: Using the product rule, if you know any two of the derivatives you can work out the third one. Here you know A=L*W and A' and W'.]

centimeters per hour

Using implicit differentiation, it is found that the length increases initially at a rate of 183 centimetres per hour.

The area of a rectangle of length l and width w is given by:

Applying implicit differentiation, the rate of change is given by:

In this problem:

We have to find , then:

1.83m per hour, thus 183 cm.

The length increases initially at a rate of 183 centimetres per hour.

A similar problem is given at link

Work Shown:

t = elapsed time in hours

L(t) = 5 + x*t = length at time t

L ' (t) = x = rate of change for the length.

The value of x will change depending on what t is. In this case, we will use t = 0 since we want to know how fast L(t) is increasing initially. The units for L(t) are in meters. The units for L ' (t) are in meters per hour. A similar situation happens for W(t) as well.

50 cm = 50/100 = 0.5 m

W(t) = 3 + 0.5t = width at time t

W ' (t) = 0.5 = rate of change for the width

A ' (t) = 8 = rate of change of the area

A(t) = L(t)*W(t)

A ' (t) = L ' (t)*W(t) + L(t)*W ' (t) ... product rule

8 = x*(3 + 0.5t) + (5+x*t)*0.5 substitution

8 = x*(3 + 0.5*0) + (5+x*0)*0.5 ... plug in t = 0

8 = 3x + 2.5

3x+2.5 = 8

3x = 8-2.5 subtract 2.5 from both sides

3x = 5.5

x = 5.5/3 ... divide both sides by 3

x = 1.833 ... which is approximate

L ' (t) = 1.833 is the approximate rate of change for the length. This is in meters per hour. Convert to cm per hour. To do this, multiply by 100.

1.833 m/hr = (1.833 m/hr)*(100 cm/1 m)

1.833 m/hr = (1.833*100) cm/hr

1.833 m/hr = 183.3 cm/hr

were is the equation