rates of change questions (1 Viewer)

[AFGHAN22]

further rates of change question:

1. Water is pouring steadily at the rate of 1m^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
a) area of the water surface is increasing
b) 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.)

the answers for a) and b) are 2/3 m^2/min and 4/3 m^2/min respectively

more rates of change questions:

1. A weight W is attached to a rope 15m long which passes over a pulley 8m above the ground. The other end of the rope is attached to a truck at a point 2m above the ground. If the truck moves off at a speed of 2.5m/s, find how fast the weight is rising when it is 3m above the ground. (hint: at time t, let the truck be distant xm from the vertical below the weight, and let the weight be ym above the ground. then show that (7+y)^2=x^2+6^2.)

the answer to this is 2m/s

PLEASE show a diagram for this question, thats where my problem lies!!!

2. the inner and outer radii of a cylindrical tube of constant length change in such a way that the volume of the material forming the tube remains constant. Find the rate of increase of the outer radius at the instant when the radii are 3cm and 5cm, and the rate of increase of the inner radius is 3 1/3 cm/s.

the answer to this is 2cm/s

once again PLEASE show me the diagram for this as thats where my problem lies.

thanks in advance guys

 

[AFGHAN22]

according to the answers there are no pis in the solutions so yeah...maybe the working out was wrong or something

 

[Mountain.Dew]

Are you sure the answers dont have pi in them?

i got 2/3 pi m²/s for part (a)

try this chain:

dS/dt = (dS / dh)(dh / dV)(dV / dt)

S = surface area,
h= depth of water
V = volume

in this process, the pis are eliminated

 

[AFGHAN22]

can someone please try and complete the whole question...the problem isn't whethere there are pis or not in the4 solution bur rather the process of working it out

 

[SoulSearcher]

Let S be the surface are of the water
S = pi * r², where r is the radius ... (1)
dV/dt = 1 m³/min
V = 1/3 * pi * r² * h, where h is the height of the water ... (2)
We know that the semi-vertical angle is 30 degrees
.'. tan 30 = r/h
.'. 1/rt3 = r/h
.'. r = h/rt3
Therefore, substituting the value of r into (1) and (2),
S = h²pi/3
V = h³pi/9
Therefore
dS/dh = 2h*pi/3
dV/dh = h²pi/3
.'. dh/dV = 3/h²pi
dV/dt = 1 m³/min
Using dS/dt = dS/dh * dh/dV * dV/dt, when h = 3
dS/dt = 2pi * 1/3pi * 1, substituting h = 3 into dS/dh and dh/dV
= 2pi/3pi
= 2/3 m²/min

 

[alcalder]

It's been many years and a whole new syllabus since I tutored and taught Extension 2 (it was called 4 Unit back then). DO you guys do implicit differentiation? Because if you do, then this should be easy. Otherwise, it may be worth learning. eg

If V = x²

and x = f(t)

then

dV/dt = 2x dx/dt

1. A weight W is attached to a rope 15m long which passes over a pulley 8m above the ground. The other end of the rope is attached to a truck at a point 2m above the ground. If the truck moves off at a speed of 2.5m/s, find how fast the weight is rising when it is 3m above the ground. (hint: at time t, let the truck be distant xm from the vertical below the weight, and let the weight be ym above the ground. then show that (7+y)^2=x^2+6^2.)

Not sure how to get a picture here so you can see it.

Think of the rope going over a pulley. On the weight side the weight is y m above the ground. So the length of the rope from weight to pullet is (8-y)m. Thus, the length of rope connected to the truck is (15-(8-y))m = (7+y)m

The truck itself is x metres from the weight.

From the point where the rope is connected to the truck up to the pulley is (8-2)m = 6m.

Now you have a right triangle and using pythagorus:

x² +6² = (7+y)²

Using implicit differentiation with respects to time, t.

2x.dx/dt = 2y.dy/dt + 14.dy/dt

When y = 3 (in the original equation) x = 8 and dx/sdt = velocity of truck = 2.5 m/s

Thus:

2 x 8 x 2.5 = (2 x 3 + 14) . dy/dt

dy/dt = 2 m/s

 

[alcalder]

2. the inner and outer radii of a cylindrical tube of constant length change in such a way that the volume of the material forming the tube remains constant. Find the rate of increase of the outer radius at the instant when the radii are 3cm and 5cm, and the rate of increase of the inner radius is 3 1/3 cm/s.

the answer to this is 2cm/s

The diagram for this one is just a cylinder with a big hole in it of height = h.

The inner radius = r and the outer Radius = R

V = (πR² - πr²)h

volume remains constant so dV/dt = 0

dVdt = (2πR.dR/dt - 2πr.dr/dt)h

0 = (2πR.dR/dt - 2πr.dr/dt)h

therefore

R.dR/dt = r.dr/dt

When r = 3, R = 5 and dr/dt = 3 1/3

dR/dt = (3 x 10/3) / 5

= 2 cm/s

 

[SoulSearcher]

It's been many years and a whole new syllabus since I tutored and taught Extension 2 (it was called 4 Unit back then). DO you guys do implicit differentiation? Because if you do, then this should be easy. Otherwise, it may be worth learning. eg

If V = x²

and x = f(t)

then

dV/dt = 2x dx/dt

I believe implicit differentiation is still taught in the extension 2 course, it isn't in the extension 1 course.

 

[AFGHAN22]

thanks a lot everyone for all your responses

there was one part left out though....
"b) 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.)"
so yeah....does anyone have any soutions to that?
everything else was solved well and i understand it so once again thanks!

 

[SoulSearcher]

Let S be the surface are of the water
S = pi * r², where r is the radius ... (1)
dV/dt = 1 m³/min
V = 1/3 * pi * r² * h, where h is the height of the water ... (2)
We know that the semi-vertical angle is 30 degrees
.'. tan 30 = r/h
.'. 1/rt3 = r/h
.'. r = h/rt3
Therefore, substituting the value of r into (1) and (2),
S = h²pi/3
V = h³pi/9
Therefore
dS/dh = 2h*pi/3
dV/dh = h²pi/3
.'. dh/dV = 3/h²pi
dV/dt = 1 m³/min
Using dS/dt = dS/dh * dh/dV * dV/dt, when h = 3
dS/dt = 2pi * 1/3pi * 1, substituting h = 3 into dS/dh and dh/dV
= 2pi/3pi
= 2/3 m²/min

Again, we'll use from my post above,
dV/dh = h²pi/3
r = h/rt3
dV/dt = 1 m³/min
The area of the wetted surface are of the cone is given as A = pi * r * s, where r is the radius and s is the slant height.
To find s, we have to use pythagoras on the triangle, with h and r being the other 2 sides. So:
s² = h² + r²
.'. s = rt(h²+r[/sup]2[/sup])
= rt(h²+h²/3), using r = h/rt3 as the substitution
= rt(4h²/3)
= 2h/rt3
Therefore A = pi * h/rt3 * 2h/rt3
= 2h²pi/3
dA/dh = 4hpi/3
dA/dt = dA/dh * dh/dV * dV/dt
When h = 3,
dA/dt = 12pi/3 * 3/9pi * 1
= 12/9
= 4/3 m²/min