Rate of Change Question5

In this page rate of change question5 we are going to see solution of some practice question of the worksheet.

Question 5:

The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2 cm²/min. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm².

Solution:

Let "b" and "h" be the base and height of the triangle ABC

Area of triangle ABC

(A)= (1/2) b h

dh/dt = 1 cm/min

 dA/dt = cm²/min

 dh/dt = ?

  h = 10      area = 100

    100 = (1/2) x b x 10

       b = (100 x 2)/10

       b = 20

      dA/dt = (1/2) [b (dh/dt) + h (db/dt)]

   h (db/dt) = 2 (dA/dt) - b (dh/dt)

      (db/dt) = (2/h) (dA/dt) - (b/h) (dh/dt)

      (db/dt) = (2/10) (2) - (20/10) (1)

      (db/dt) = (4/10) - (20/10)

      (db/dt) = (4 - 20)/10

      (db/dt) = - 16/10

      (db/dt) = - 1.6 cm/min