Mathematics, 14.02.2020 04:12 DangerDash

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

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Answer from: sodiumvlogsYT

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:

A = lw

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

$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

In this problem:

Width of 3m, length of 5m, thus w = 3, l = 5

.Area increases at a rate of 8 m² per hour, thus $\frac{dA}{dt} = 8$ .Width at a rate of 50 cm = 0.5m per hour, thus $\frac{dw}{dt} = 0.5$ .

We have to find $\frac{dl}{dt}$ , then:

$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

$8 = 5(0.5) + 3\frac{dl}{dt}$

$3\frac{dl}{dt} = 5.5$

$\frac{dl}{dt} = \frac{5.5}{3}$

$\frac{dl}{dt} = 1.83$

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

Answer from: negritobull78

approximately 183.3 cm per hour

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

Answer from: Quest

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

$\frac{1}{2}$ (0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

$\frac{1}{2}$ (4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)

Answer from: Quest

answer: yard ramps with a level off platform are typically 36 feet long. truck, rather than at an angle, enabling access to those pallets at the very back of the truck. if you are using your yard ramp to convert a loading dock into a drive-in door, we even at 60 inches (5 feet) high and a 30 foot ramp, the slope of the ramp will be less

step-by-step explanation: