Question

The length s(t)s(t)s, (, t, )of the side of the base of a square prism is decreasing at a rate of 777 kilometers per minute and the height h(t)h(t)h, (, t, )of the prism is increasing at a rate of 101010 kilometers per minute. At a certain instant t_0t 0 ​ t, start subscript, 0, end subscript, the base's side is 444 kilometers and the height is 999 kilometers. What is the rate of change of the surface area A(t)A(t)A, (, t, )of the prism at that instant?

Answer 1

Answer:

Step-by-step explanation:

Given:

Base length, l = 4 km

Height, h = 9 km

dl/dt = -7 km/min

dh/dt = 10 km/min

Surface area of a prism, A = 2 × (lw + lh + wh)

For a square prism,

Length, l = width, w

= 4km

A = 2 × (l^2 + lh + lh)

= 2(l^2) + 4lh

A = 2(l^2) + 4 lh

= 2 × (l^2 + 2lh)

dA/dt = 2 × (2I × dI/dt + 0 × dh/dt + 2 × 2h × dl/dt + 2l × dh/dt)

dA/dt = 2 × (2l × dl/dt + 0 + 2 × 2h × dl/dt + 2l × dh/dt)

= 2 × (2(4) × (-7) + 2(9) × (-7) + 2(4) × (10))

= 2 × (-56 + (-126) + (80))

= 2 × -102

= -204 km/min

The rate of change of the surface area, A is decreasing by 204 km/min.

Answer 2

Answer:

$\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}$

Step-by-step explanation:

The surface area of the square prism is obtained by using the following formula:

$A_{s} (t) = 4\cdot l(t)\cdot h(t) + 2\cdot [l(t)]^{2}$

The rate of change of the surface area can be found by deriving the function with respect to time:

$\frac{dA_{s}}{dt} = 4\cdot [h(t)\cdot \frac{dl}{dt} + l(t)\cdot \frac{dh}{dt}] + 2\cdot l(t)\cdot \frac{dl}{dt}$

Known variables are summarized below:

$h(t) = 9\,km$

$l(t) = 4\,km$

$\frac{dh}{dt} = 10\,\frac{km}{min}$

$\frac{dl}{dt} = -7\,\frac{km}{min}$

The rate of change is:

$\frac{dA_{s}}{dt} = 4\cdot [(9\,km)\cdot (-7\,\frac{km}{min} )+(4\,km)\cdot (10\,\frac{km}{min} )] + 2\cdot (4\,km)\cdot (-7\,\frac{km}{min} )$

$\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}$