Question

The length ℓ, width w, and height h of a box change with time. At a certain instant the dimensions are ℓ = 1 m and w = h = 9 m, and ℓ and w are increasing at a rate of 4 m/s while h is decreasing at a rate of 5 m/s. At that instant find the rates at which the following quantities are changing. (a) The volume. m3/s (b) The surface area. m2/s (c) The length of a diagonal. (Round your answer to two decimal places.)

Answer 1

Answer:

a.)$\frac{∂V}{∂t} = 315 m^{3}s^{-1}$

b.)$\frac{∂S}{∂t} = 124 m^{2}s^{-1}$

c.)$\frac{∂D}{∂t} = -10ms^{-1}$

Explanation:

a.)

    We know that

               Volume = Length × Width × height

                     V = l × w × h

As volume is a dependent variable and its depends upon the length, width and height of the box. These three variable further dependent on time. So, to calculate the rate of change of volume, we need to take the partial derivative of volume with respect to time.

$\frac{∂V}{∂t} = \frac{∂V}{∂l}\frac{∂l}{∂t} + \frac{∂V}{∂w}\frac{∂w}{∂t} + \frac{∂V}{∂h} \frac{∂h}{∂t}$

$\frac{∂V}{∂t} = wh\frac{∂l}{∂t} + hl\frac{∂w}{∂t} + wl \frac{∂h}{∂t}$

$\frac{∂V}{∂t} = (9)(9)(4) + (9)(1)4) + (9)(1)(-5)$

$\frac{∂V}{∂t} = 315 m^{3}s^{-1}$

b.)

We know that

    Surface area = 2(length × width + length × height + width × heigth)

            S    =  2( l × w + l × h + w × h )

Using the same technique as in part a

$\frac{∂S}{∂t} = \frac{∂S}{∂l}\frac{∂l}{∂t} + \frac{∂S}{∂w}\frac{∂w}{∂t} + \frac{∂S}{∂h} \frac{∂h}{∂t}$

$\frac{∂S}{∂t} = 2[(w\frac{∂l}{∂t} +l\frac{∂w}{∂t})+( l\frac{∂h}{∂t}+ h\frac{∂l}{∂t}) + (w \frac{∂h}{∂t}+ h \frac{∂w}{∂t})]$

$\frac{∂S}{∂t} = 2[(9.4+1.4)+( 1.-5+ 9.4) + (9.-5+ 9.4)]$

$\frac{∂S}{∂t} = 124 m^{2}s^{-1}$

c.)

We know that

    Length of diagonal = (length)² × (height)² × (width)²

            D    =  (l)²× (w)² × (h)²

Using the same technique as in part a

$\frac{∂D}{∂t} = \frac{∂D}{∂l}\frac{∂l}{∂t} + \frac{∂D}{∂w}\frac{∂w}{∂t} + \frac{∂D}{∂h} \frac{∂h}{∂t}$

$\frac{∂D}{∂t} = [2(l\frac{∂l}{∂t})+2( h\frac{∂h}{∂t})+ 2(w\frac{∂w}{∂t})]$

$\frac{∂D}{∂t} = [2(4)+2(-45)+ 2(36)]$

$\frac{∂D}{∂t} = -10ms^{-1}$

Minus sign indicate that length of diagonal decreases.