A box is partially filled with sand. The length of the box is 8 mm. The width of the box is 17 mm. If the volume of the sand is decreasing at a rate of 562 cubic mm per minute, what is the rate, in mm per minute, at which the height of the sand is changing when the height of the sand is 2 mm?

Question

Emery Wilkins

Physics

28 March, 02:06

A box is partially filled with sand. The length of the box is 8 mm. The width of the box is 17 mm. If the volume of the sand is decreasing at a rate of 562 cubic mm per minute, what is the rate, in mm per minute, at which the height of the sand is changing when the height of the sand is 2 mm?

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Answers (1)

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Kale Huff · Answer 1

- 4.13 mm/min

Explanation:

length of box, l = 8 mm

width of box, w = 17 mm

dV/dt = - 562 cubic mm per minute

let the height is h.

dh / dt = ? when h = 2 mm

The volume of the box is given by

V = l x w x h

Differentiate with respect to h

here, l and w be the constants as the length and width are not changing

dV/dt = l x w x dh/dt

- 562 = 8 x 17 x dh/dt

dh/dt = - 4.13 mm/min

negative sign sows that the height is decreasing.