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acmilan

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I was wondering if someone can help me out with some related rates questions. They are from Fitzpatrick Exercise 25 (a). I am either doing them wrong or the answers are wrong.

9. Sand is poured into a heap in the shape of a right circular cone whose semi-vertex angle is a, where tan a = 3/4. When the height of the cone is 16cm, the height is increasing at the rate of 2 cm/min. At what rate is the volume increasing at that instant? (Answer = 288pi cm^3/min)

17. A straight railway track and a straight road intersect at right angles. At a given instant a motor car, at 40 km/h, and a train, at 50 km/h, are moving away from the intersection and are 40 and 30 km respectively from the intersection. At what rate is the distance between them changing one hour later? At what rate would the distance between them be changing at that instant if they were both traveling towards the intersection? (Answer = 45root2 km/h, 50 km/h)

22. A loading chute is in the shape of a right square pyramid of base length 10 m and depth 8 m. Liquid is poured in the top at a rate of 4 m^3/min. At what rate is the level rising when the depth is 4m? (Answer = 4/25 m/min, I keep getting 4/125)

31. For the first part of this question I proved that V = pi/3 (tan a)^2 h^3
The second part asks: Water is flowing out through a hole at the vertex of an inverted cone whose vertex angle is 60 degrees at a rate equal to pi times the square root of the depth of the water at any time. At what rate, in cm/s, would the water be flowing out when the depth is 9cm? (Answer = 1/9 cm/s)
 

Xayma

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Originally posted by acmilan1987
9. Sand is poured into a heap in the shape of a right circular cone whose semi-vertex angle is a, where tan a = 3/4. When the height of the cone is 16cm, the height is increasing at the rate of 2 cm/min. At what rate is the volume increasing at that instant? (Answer = 288pi cm^3/min)
V=1/3*pi*r<sup>2</sup>h

We are trying to find dV/dt and have dh/dt, so we need to find dV/dh.

We have a relationship bewteen height and radius (from the angle, (ie radius=3/4h)

so V=1/3*pi*(3/4*h)<sup>2</sup>h
=1/3*pi*9/16*h<sup>3</sup>
=3/16*pi*h<sup>3</sup>

.: dV/dh=9/16*pi*h<sup>2</sup>
when h=16cm, dh/dt=2cm min<sup>-1</sup>
dV/dt=9/16*pi*16<sup>2</sup>*2
=288pi cm<sup>3</sup> min<sup>-1</sup>
 

acmilan

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woops sorry i didnt realise i posted it in the wrong forum. Its ext 1 stuff

How do you get that r = 3/4 h from tan a = 3/4?
 
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