Applications of Calculus to the Physical World (1 Viewer)

bored of sc · May 17, 2009

Fitzpatrick Questions I simply couldn't do.

Ex 25 (a)

7) The cross section of a trough l metre long is in the shape of an isosceles triangle of base measurement 2a metre and height b metre. Water leaks from the trough at a constant rate of c metre³/min. Find the rate at which the water level is falling when the depth of the water is b/2 metre.
Answer: c/al metres/min

22) A loading chute is in the shape of a right square pyramid of base length 10m and depth 8m. Liquid is poured in the top at a rate of 4m³/min. At what rate is the level rising when the depth is 4m?
Answer: 4/25 m/min

25) A point P moves on the curve y = x³ and the co-ordinate increases at a constant rate of 5 units/second. At what rate is:
a) the y co-ordinate of P increasing when x = 1?
b) the gradient of the curve increasing when x = 1?
Answers - (a) 15 units/s
(b) 30 units/s

27) In triangle ABC, Ab = 10cm, AC = 12cm and angle A is increasing at the rate of 0.1 radian/second. At what rate is:
a) the area of triangle ABC increasing,
b) the length of BC increasing, when angle A is pi/3 radians?
Answers - (a) 3 cm²/s
(b) 3root[3/31] cm/s

Ex 25 (e)

4) Assume Newton's Law of Cooling is T^o = P + Ae^-kt. A body, initially at room temperature 20^oC, is heated so that its temperature would rise by 5^oC/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120^oC. How long would it take to reach a temperature of 100^oC?
Answer: 32.2 minutes

6) A vessel is being filled at a variable rate and the volume of liquid in the vessel at any time 't' is given by V = A(1-e^-kt)
(a) Show that dV/dt = k(A-V)
(b) If one quarter of the vessel is filled in 5 minutes, what fraction is filled in the next five minutes?
Answer for (b): 3/16

Thank you all in advance.

gurmies · May 17, 2009

DiligentStudent · May 17, 2009

Gurmies for PM.

~caramello~ · May 17, 2009

question 22

dr/dt = dv/dt x dr/dv
dv/dt = 4
so we need to find dr/dv

B/h = 10/8
B = 5h/4

V = 1/3.B^2H
but B = 5h/4
V = 1/3.(5h/4)^2.h
= 1/3.(25h^2/16).h
= (25h^3)/48
dv/dr = (75h^2)/48

dr/dt = dv/dt x dr/dv
= 4 x 48/(75h^2)
when h =4
= 4 x 48/1200
= 48/300
= 4/25 m/min

~caramello~ · May 17, 2009

question 27

$\frac{d\theta }{dt}=0.1 r/s$

$A = \frac{1}{2}AB sin C$

$A = \frac{1}{2}(10)(12) sin C$

$A = 60 sin C$

$\frac{dA}{d\theta } = 60 cos C$

$\frac{dA}{dt}= \frac{d\theta }{dt}.\frac{dA}{d\theta }$

= $0.1 \times 60 cos C$
$when C = \frac{\pi}{3}$
= $0.1 \times 60 cos \frac{\pi}{3}$
= $0.1 \times 30$
= $3 cm^{2}/sec$

Part B
$\frac{dA}{dt}=\frac{d\theta }{dt}\times \frac{dA}{d\theta }$
$a^{2}=b^{2}+c^{2}-2bc \cos A$
$a^{2}=10^{2}+12^{2}-2(10)(12)\cos A$
$a^{2}=100+144-240 \cos A$
$a = \sqrt{244-240 \cos A}$
$a = (244-240 cos A)^{\frac{1}{2}}$
$\frac{dA}{d\theta } = \frac{1}{2}(244-240 \cos A)^{-\frac{1}{2}} \times 240 \sin A$
$\frac{240 \sin A}{2\sqrt{244-240 \cos A}}$
$\frac{dA}{dt}=\frac{d\theta }{dt}\times\frac{da}{d\theta }$
$0.1 \times \frac{240 \sin A}{2\sqrt{244-240 \cos A}}$
$when A = \frac{\pi}{3}$
simplify and you should get the answer

jet · May 17, 2009

bored of sc · May 23, 2009

Thank you all so very much. You guys are so intelligent.

Applications of Calculus to the Physical World (1 Viewer)

bored of sc

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gurmies

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~caramello~

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jet

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bored of sc

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