The altitude of a triangle is increasing at a rate of 2.5 centimeters/minute while the area of the triangle is increasing at a rate of 2 square centimeters/minute. At what rate is the base of the triangle changing when the altitude is 10.5 centimeters and the area is 98 square centimeters

  -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.

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answer: -10

step-by-step explanation:

s(t) = 1 - 10t

s(x) = 1 - 10x

s(x + h) = 1 - 10(x + h)

            = 1 - 10x - 10h

s(x + h) - s(x) = 1 - 10x - 10h - (1 - 10x)

                      = 1 - 10x - 10h - 1 + 10x

                      = -10h

= -10