Related Rates -- Fitzpatrick 25(a) (1 Viewer)

[Pyramid]

Hiya there, any worked solutions to any of these would be appreciated



Q26:
Grain is ejected froma chute at the rate of 0.1m^3/min and is forming a heap on a flat horizontal floor. THe heap is in the form of a circular cone of semi-vertical angle 45'. Find the rate, in metres per minute, at which the height of the cone is increasing, at the instant 3 minutes after the opening of the chute.

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Q27:
In the triangle ABC, AB = 10cm, AC = 12cm, and angle A is increasing at the rate of 0.1 radian/second.

At what rate is:

1) the area of triangle ABC increasing.
2) the length of BC increasing, when angle A is pi/3 radians.

Thanks very much

 

[Timothy.Siu]

Hiya there, any worked solutions to any of these would be appreciated



Q26:
Grain is ejected froma chute at the rate of 0.1m^3/min and is forming a heap on a flat horizontal floor. THe heap is in the form of a circular cone of semi-vertical angle 45'. Find the rate, in metres per minute, at which the height of the cone is increasing, at the instant 3 minutes after the opening of the chute.

* * * * * * * * * * *

well, we want to find dh/dt, where h is the height in metres and t is time in minutes.
dh/dt=dG/dt x dH/dG where G is the grain
dG/dt is given to be 0.1m^3/min.
we know G=1/3*Pi*r^2*h=1/3*Pi*h^3 since its a 45' cone
dG/dh=Pi*h^2
dh/dG=1/(Pi*h^2)
In 3 minutes there is 0.3m^3 of grain...
so when g=0.3, 0.3=1/3*Pi*h^3
h=0.317 (3s.f)

Therefore, dh/dt=dG/dt x dH/dG=0.1x1/(Pi*(0.317)^2)=0.317m/min

edit:

Area of triangle=A=0.5 x 12 x 10 x sin @
differentiating both sides with respect to time.
dA/dt=60cos @ x d@/dt=60cos @ x 0.1=6cos @

similarly,
BC^2=AB^2+AC^2 - 2xABxACxcos @
BC^2=100+144-240cos@
differentiating both sides w.r.t time,
2BC x BC/dt=240sin@ x d@/dt
BC/dt=24sin@/2BC
BC/dt=24sin@/(2x sqrt(100+144-240cos@))
BC/dt=0.933cm/s