Answer:
-4.0635 cm/minute
Step-by-step explanation:
The area of a triangle is given by ...
A = 1/2bh
Then the derivative with respect to time is ...
A' = (1/2)(b'·h +b·h')
When the altitude is 10.5 cm and the area is 98 cm^2, the base is ...
b = 2A/h = 2·98/10.5 = 18 2/3
So, the rate of change of the base is ...
b' = (2A' -b·h')/h = (2·2 -(18 2/3)(2.5))/10.5 ≈ -4.0635 . . . cm/minute
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Check
0.005 minutes before the time of interest, the altitude is 10.5 -.005(2.5) = 10.4875 cm, and the area is 98 -.005(2) = 97.99 cm. Then the base is ...
2(97.99)/10.4875 ≈ 18.687008 cm
0.005 minutes after the time of interest, the altitude is 10.5+.005(2.5) = 10.5125 cm, and the area is 98 +.005(2) = 98.01 cm. Then the base is ...
2(98.01)/10.5125 ≈ 18.646373 cm
The difference in base length in that 0.01 seconds is ...
18.646373 -18.687008 ≈ -0.040635 . . . . cm
So the rate of change is about ...
-4.0635 cm/min
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A graphing calculator can also give a numerical value for the rate of change of base.
