Rates question :-\ (1 Viewer)

[Grey Council]

blah

Water is pouring at the rate of 1 m^3 / min into a conical reservoir whose semi-vertical angle is 30 degrees. When the water is 3m deep. Find the rate at which the
i) Water level is rising
ii) Area of the water surface is increasing
iii) Wetted surface area of the reservoir is increasing. (the curved surface area of a cone is pi . r . s, where r is the base radius and s the slant height)

I've got the first bit. Am stuck on the second bit.

Let W be area of water surface.
w = pi . s^2
where s is slope of conical thingo.
h/s = rt3 / 2
so s = 2h / rt3

dW/dt = dW/dh . dh/dt
from first part
dh/dt = 3 / (pi . h^2)
and
dW/dh = (8pi.h) / 3

so dW/dt SHOULD be 8/3
but it isn't.

:-\

 

[Grey Council]

I got a bad feeling about this.

I think I got my surface area formula wrong.

...

*smacks himself on head*

 

[CrashOveride]

Hmm this is harder than anything in my 3u book chapter on rates =p Is this from Fitzpatrick?

What's the significance of the 30 degrees ?

 

[Estel]

After attempting and then deleting I now know that I am hopeless at typing stuff up without having first done it on paper

The 30 degrees is needed to establish the relationship between h and r.

My solution (please note- error ridden, I'm haven't done calc- just know chain rule and reduce index rule so i'm just doin for the fun of it)

r/h = tan30
r = h/rt3
V = (1/3)(pi)r^2h
= (1/9)(pi)h^3
Given dV/dt = 1

1)
dh/dt
= dh/dV x dV/dt
= 1/(dV/dh) x 1
= 3/[(pi)h^2]

A = (pi)r^2
= (pi)h^2/3

2)
dA/dt
= dA/dh x dh/dt
= (2/3)(pi)h x 3/((pi)h^2)
= 2/h

3)
EDIT: (wrong again)
ANOTHER EDIT:
dW/dt
= dW/dh x dh/dt
Since
W= (pi)rs
= (pi)sh/rt3
dW/dt
= (pi)s/rt3 x 3/(pi)h^2
= 3s/(h^s[rt3])

I'm bound to get it eventually

 

[Grey Council]

can you please put them under like a heading?

i think the first two are right.
last is wrong

and 30 degrees is what Estel said it is.
Get a conical shape. Stand it on its point, exactly vertically. Now imagine a line that splits it in half. That line and the slope of the cone make a 30 degree angle.

You need it to establish the relationship between the radius of the circle of the base, and the height of the cone, cause the volume of a cone is like 1/3 pi r^2h or whatever.

btw, Estel, how you get:
A = (pi)r^2
= (pi)h^2/3

?! Thats where I'm stuck on.
your using the r from the first question, but r in this case means the slope, doesn't it? and the slope is of a different length than the radius of the circle of the BASE of the cone.



EDIT:
AND Its still wrong.

 

[CrashOveride]

Is the answer to ii) 2/3 m²/min

Oh sorry, Estel has already done this

 

[Estel]

Orange Council: don't forget my disclaimer sayin I don't know calculus- I'm in Yr 11 and we are up to coord geometry now...

Anyhow, since we know

r/h = tan30
r = h/rt3

The surface area of the water is a circle:
A = (pi)r^2
subbing in h/rt3
A = (pi)(h/rt3)^2
= (pi)h^2/3

When I say r, I mean r as the radius of the circle. As r decreases, h decreases, and they decrease in proportion so that since the tan ratio is always the same, you can still sub in that earlier result.

 

[CrashOveride]

I got your result Estel for dW/dt.

To eliminate the s, i said sin 30 = r/s

Yields s = 2h/ rt(3)

then wehn i sub that back in, i end up getting 2/h..again.

 

[CM_Tutor]

Originally posted by Orange Council
Water is pouring at the rate of 1 m^3 / min into a conical reservoir whose semi-vertical angle is 30 degrees. When the water is 3m deep. Find the rate at which the
i) Water level is rising
ii) Area of the water surface is increasing
iii) Wetted surface area of the reservoir is increasing. (the curved surface area of a cone is pi . r . s, where r is the base radius and s the slant height)

Let's start by defining some variables. Let V m³ be the volume of water in the reservoir at time t min. At time t min, the height of water is h m, the radius of the surface is r m, the slant height is s m, the area of the water surface is A m², and the wetted surface area is W m².

Now, we have a semi-vertical angle of 30 degrees, so tan 30 = r / h, and hence r = h / sqrt(3) _____ (1)

We are also told that dV/dt = 1 m³min^-1

V = (1 / 3) * pi * r² h = (1 / 3) * pi * (h / sqrt(3))² * h, using (1)
So, V = pi * h³ / 9
and so dV/dh = 3 * pi * h² / 9 = pi * h² / 3

(i) We seek dh/dt = dh/dV * dV/dt (Chain Rule)
So, dh/dt = 3 / (pi * h²) * 1 = 3 / (pi * h²)

When h = 3 m, dh/dt = 1 / (3 * pi) mmin^-1

(ii) We seek dA/dt. Since A is a circle, we know that A = pi * r² = pi * (h / sqrt(3))², using (1)
So, A = pi * h² / 3
and so dA/dh = 2 * pi * h / 3

dA/dt = dA/dh * dh/dt (Chain Rule)
So, dA/dt = (2 * pi * h / 3) * 3 / (pi * h²) = 2 / h

When h = 3 m, dA/dt = 2 / 3 m²min^-1

(iii) We seek dW/dt. Since W is the curved surface of a cone, we know that W = pi * r * s.
Furthermore, r² + h² = s² (Pyth)
So, W = pi * r * sqrt(r² + h²) = pi * (h / sqrt(3)) * sqrt[(h / sqrt(3))² + h²], using (1)
So, W = pi * (h / sqrt(3)) * sqrt(4h² / 3) = 2 * pi * h² / 3
and so dW/dh = 4 * pi * h / 3

dW/dt = dW/dh * dh/dt (Chain Rule)
So, dW/dt = (4 * pi * h / 3) * 3 / (pi * h²) = 4 / h

When h = 3 m, dW/dt = 4 / 3 m²min^-1

 

[Grey Council]

argh! I thought area of water surface meant all the parts where the water touches the conical thing.

DAMN, how the HELL didn't I get that?!?

blah

Thanks CM_Tutor. Tis good to have you back. ^_^

 

[CrashOveride]

Doh, i substitued value for s after having differentiated :/