VCE Specialist 3/4 Question Thread!

[Aelru]

Find the volume of a truncated cone of height 10cm , a base radius of 5cm and a top radius of 2cm.

This falls in the integration chapter, but I used a different method to calculate the answer (overall cone volume - imaginary small cone volume at the top)

Would anyone have any idea how to solve this integration wise?

[Ancora_Imparo]

Think about volumes of solids of revolution.

What graph, when rotated around the x-axis, gives a cone? A straight line starting at the origin. However, this cone is truncated, so the difference is that the line will not start at the origin.

We know that the top radius is 2cm, so it would start at (0,2). Since the height is 10cm and the the bottom radius is 5cm, the line ends at (10, 5).

Now, you can find the equation of the line:




The truncated cone is formed by rotating this line around the x-axis from x=0 to x=10:


Evaluating this gives you the answer.

[Aelru]

We know that the top radius is 2cm, so it would start at (0,2). Since the height is 10cm and the the bottom radius is 5cm, the line

Everything else seems to fall in place, however I'm unsure as to why it would start at (0,2). The top radius is 2cm, so wouldn't it be (2, 10) , or (-2,10)?

[Phy124]

Ancora_Imparo worked along the x axis so that the smaller end (that with radius of 2) is along the line x = 0 and the larger end (that with radius of 5) is along the line x = 10. He then rotated it around the x-axis to form the cone.

You could alternatively take the points (2,0) and (5,10) and rotate it around the y-axis.

edit: changed a word for clarity's sake.

[Ancora_Imparo]

The first graph in this image might be useful for visualising it: http://ars.els-cdn.com/content/image/1-s2.0-S1090780710001540-gr2.jpg

[Aelru]

Ah, the wonders of maths never cease to amaze me. 3 different ways, all visually appealing. Thanks for your help.

MODIFY: Another Q: Find the volume of this circle about the x axis, with the equation
The answer gives , but that seems rather odd.

[lzxnl]

Ah...the case of the delightfully devilish donut...

If you look carefully, the volume of the solid formed isn't given by , even though it's a revolution around the x-axis. The actual volume is a difference of volumes; take the upper semicircle , rotate this around the x axis and find the volume, then take the lower semicircle , rotate this and find the volume.
You should get

Note that the integral is a multiple of times the area of a semicircle of radius 4, which is merely , so the final integral is the product of and , or as the answers suggest.

[saba.ay]

Could someone please help with the following:
Find y in terms of x:
1) dy/dx = (1/4) + y^2, Given (pi/2, 1/2)
2) dy/dx = -(2^(1/2))-y^2 , given (0,1)

Also, how do you find the domains of the above differential equations?

I also have the two questions attached. My answers don't match with BOB so any/all help is appreciated.
For Q1, I cannot express the equation in terms of N correctly.
For Q8, I'm off entirely. :/

Sorry for the bombardment of questions, but thought to post them collectively.

Please and thanks in advance.

[Deleted User]

If dy/dx = f(y),

dx/dy = 1/f(y)

Integrate that to get x in terms of y and sub in the point to find the constant. Then rearrange to get y in terms of x.

[Aelru]

Q1: Water is leaking from a triangular prism tank, at a constant rate of 0.03 m^3/min.
Find the expression for the volume in terms of the depth, h.
Answer was:. I'm not sure as to how they obtained that answer.

Q2: A boy slides down a slide that is in the shape of a parabola. The bottom of the slide is at the origin, and the top of the slide is 5m above, and 2m out horizontally from the base. When his height above the ground is 2.5m, he is descending at 0.8m per second. Find the rate at which he is moving horizontally forward at that instant.

I am unsure if the method I used was correct , as I obtained a different answer (actual answer was ) . Therefore, I was wondering if you guys could check if I used the right process, and if there is a more simple process to this question.

METHOD USED:
I assumed 'y' as the vertical displacement, while 'x' is the horizontal.
dy/dx=0.8
dx/dt = dx/dy x dy/dt

dx/dy process:
one solution (1 x intercept at origin), so discriminant =0.
. But since bottom of parabola is at 0, c=0 (very unsure about this part, as I think only applies to pwr fns)


ie, parabola eq , or

At (2,5),
5=4a
a= 5/4
ergo,

From there on, I simply found the inverse of that, and multiplied it to 0.8, while subbing in the x=2.5.

[lzxnl]

I'll look at question 2 first.
Defining positive to be up and from the origin to the base:
Top of the slide is 5m above, so if , then c = 5
Two metres out horizontally => b=2
so
Also, the bottom of the slide is at the origin. Therefore we can say that the curve passes through (0,0)
4a+5=0
a=-5/4

y=2.5,

as x>0


Now

[SocialRhubarb]

Is the top of the slide necessarily a turning point? I think it's needed to do the problem but I don't think it's stated.

Also, congratulations on learning latex.

[lzxnl]

Question one surely doesn't have enough info.

Is the top of the slide necessarily a turning point? I think it's needed to do the problem but I don't think it's stated.

Yeah technically it's not really there, but they mentioned slide was a parabolic shape, and highest point of slide = local maximum. At least by VCE maths logic.

[Aelru]

I'll look at question 2 first.
Defining positive to be up and from the origin to the base:
Top of the slide is 5m above, so if , then c = 5
Two metres out horizontally => b=2
so
Also, the bottom of the slide is at the origin. Therefore we can say that the curve passes through (0,0)
4a+5=0
a=-5/4

y=2.5,

as x>0


Now

Brilliant insight, as always. Thanks for Q2~
For Q1, I'll post a picture.

 

[brightsky]

V = 1/2*2*2*3 - 1/2*(2-h)*(2-h)*3= 6 - 3/2*(2-h)^2 [hint: use similar triangles]