Respuesta :
Answer: approximately 183.3 cm per hour
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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
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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
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:
[tex]A = lw[/tex]
Applying implicit differentiation, the rate of change is given by:
[tex]\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]
In this problem:
- Width of 3m, length of 5m, thus [tex]w = 3, l = 5[/tex].
- Area increases at a rate of 8 m² per hour, thus [tex]\frac{dA}{dt} = 8[/tex].
- Width at a rate of 50 cm = 0.5m per hour, thus [tex]\frac{dw}{dt} = 0.5[/tex].
We have to find [tex]\frac{dl}{dt}[/tex], then:
[tex]\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}[/tex]
[tex]8 = 5(0.5) + 3\frac{dl}{dt}[/tex]
[tex]3\frac{dl}{dt} = 5.5[/tex]
[tex]\frac{dl}{dt} = \frac{5.5}{3}[/tex]
[tex]\frac{dl}{dt} = 1.83[/tex]
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 https://brainly.com/question/9543179
