related rates questions (1 Viewer)

[5647382910]

1) A conical tank whose axis is vertical has a semi-vertical angle of 45 degree. water initially at a depth of 5 metres leaks out through a hole at the bottom at the rate of 0.2sqrth metres^3/min when the depth is h metres. find the rate at which the depth is decreasing when the depth is 4 metres Solution: 1/40pi m/min

2) The altitude of a right angled triangle is 6cm and the base is increasing at a constant rate of 2cm/s. At what rate is the hypotenuse increasing when its length is 10cm? solution: 1.6cm/sec

3) In triangle ABC, AB = 10cm, AC = 12cm and angle A is increasing at the rate of 0.1 radians/second. At what rate is the area of triangle ABC increasing? Solution: 3cm^2/sec I keep getting 6cosA

4) it is given that the volume of a cone with semi vertical angle a is V = (1/3).pi.tan^2(a).h^3. Water is flowing out through a hole at the vertex of an inverted cone whose vertex angle is 60 degrees at a rate pi times the square root of the depth of the water at any time. At what rate, in cm/s, would water be flowing out when the depth is 9cm Solution: 1/9 cm/s


These q's are questions 20,21,27,31 respectively of excercise 25 a in fitzpatrick BTW

I know its alot of work, but any help would be very much appreciated

 

[jet]

1) For this one, you say, let the radius of the water be r. Hence, tan45 = r/h
Therefore r = h.
Now, the volume of the water is V = (1/3)πr²h
=(1/3)πh³

Differentiating, V' = (dV/dh) x h' (I am using the ' instead of a dot to indicate with respect to time)

0.2sqrt(h) = πh² h'
Therefore h' = sqrt(h)/5πh²

When h = 4, h' = 2/80π m/min
= 1/(40π) m/min. They've added extra info to trick you.

2) Ok, for this, the altitude is the vertical height. Let the base be b and the hypotenuse be x.

Then the sides are related by x = sqrt(36 + b²)
x' = (dx/db) b'

So, x' = b/sqrt(36 + b²)

So, when the hypotenuse is 10 and the altitude is 6, then the base is 8cm. Therefore when b = 8, x' = 16/(10) = 1.6cm/sec

3)Let angle A = θ, therefore the area of the triangle, A = 0.5 x 10 x 12 sinθ
= 60sinθ

Therefore dA/dt = dA/dθ x dθ/dt
= 60cosθ dθ/dt.

Now, unless you know the triangle is right angled
or something I don't really think you can do anything else with it.

4) Here, with the information, V = (1/3)π(tan30)²h³ as the semi vertex angle is 30.
= (1/9)πh³.

Now, V' = dV/dh h'
Therefore π sqrt(h) = (1/3)πh² h'

When h = 9, 3π = 27πh'
h' = 1/9 cm/s.

Ill have a think about the third one. If you want diagrams, tell me.
Therefore h' =

 

[5647382910]

1) For this one, you say, let the radius of the water be r. Hence, tan45 = r/h
Therefore r = h.
Now, the volume of the water is V = (1/3)πr²h
=(1/3)πh³

Differentiating, V' = (dV/dh) x h' (I am using the ' instead of a dot to indicate with respect to time)

0.2sqrt(h) = πh² h'
Therefore h' = sqrt(h)/5πh²

When h = 4, h' = 2/80π m/min
= 1/(40π) m/min. They've added extra info to trick you.

2) Ok, for this, the altitude is the vertical height. Let the base be b and the hypotenuse be x.

Then the sides are related by x = sqrt(36 + b²)
x' = (dx/db) b'

So, x' = b/sqrt(36 + b²)

So, when the hypotenuse is 10 and the altitude is 6, then the base is 8cm. Therefore when b = 8, x' = 16/(10) = 1.6cm/sec

3)Let angle A = θ, therefore the area of the triangle, A = 0.5 x 10 x 12 sinθ
= 60sinθ

Therefore dA/dt = dA/dθ x dθ/dt
= 60cosθ dθ/dt.

Now, unless you know the triangle is right angled
or something I don't really think you can do anything else with it.

4) Here, with the information, V = (1/3)π(tan30)²h³ as the semi vertex angle is 30.
= (1/9)πh³.

Now, V' = dV/dh h'
Therefore π sqrt(h) = (1/3)πh² h'

When h = 9, 3π = 27πh'
h' = 1/9 cm/s.

Ill have a think about the third one. If you want diagrams, tell me.
Therefore h' =

with 1) i understood it as inverted, did u?
2) isnt an altitude from an angle to a side?
3) im sure its the books mistake

 

[jet]

1) I don't get what you really mean.

2) Well, if you really think about it, one of the sides adjacent the the right angle really is an altitude from the corner to the base, as it is perpendicular. While it does go to the other corner, it still meets the base.

3) Okay. You should have been given a value for the angle then, which would have been π/3.

 

[5647382910]

1) I don't get what you really mean.

2) Well, if you really think about it, one of the sides adjacent the the right angle really is an altitude from the corner to the base, as it is perpendicular. While it does go to the other corner, it still meets the base.

3) Okay. You should have been given a value for the angle then, which would have been π/3.

WIth 1) it just says its in the shape of a cone (apex facing upwards), but with the way u did it it seems as though the cone was inverted, i.e standing on its apex; is that how u did it?

 

[jet]

yeah. Because it called it a tank, its fine to assume that the cone is inverted, as it wouldnt make sense the other way.