Answer:
[tex]\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}[/tex]
Step-by-step explanation:
The surface area of the square prism is obtained by using the following formula:
[tex]A_{s} (t) = 4\cdot l(t)\cdot h(t) + 2\cdot [l(t)]^{2}[/tex]
The rate of change of the surface area can be found by deriving the function with respect to time:
[tex]\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}[/tex]
Known variables are summarized below:
[tex]h(t) = 9\,km[/tex]
[tex]l(t) = 4\,km[/tex]
[tex]\frac{dh}{dt} = 10\,\frac{km}{min}[/tex]
[tex]\frac{dl}{dt} = -7\,\frac{km}{min}[/tex]
The rate of change is:
[tex]\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} )[/tex]
[tex]\frac{dA_{s}}{dt} = -148\,\frac{km^{2}}{min}[/tex]
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.
