# 2 questions.. volume and rate of change?

• Apr 7th 2009, 11:43 PM
daveee90
2 questions.. volume and rate of change?
Gday.

If anyone can show me how to do these i appreciate it muchly.
Im on a 2 week break with an exam first day back and got no idea what to do with these questions..

Question 1

Find the volume of the largest circular cone that can be cut from a solid sphere of radius 4cm.

Question 2

A boat is 2 meters below the level of a dock, is being pulled towards the dock by a rope. If the rope is pulled in at 0.1m/s how fast is the boat moving threw the water when 5 meters of rope remains between the boat and the dock

Thanks in advanced for anyone whose able to help.
Dave.
• Apr 8th 2009, 05:32 AM
stapel
Quote:

Originally Posted by daveee90
Question 1: Find the volume of the largest circular cone that can be cut from a solid sphere of radius 4cm.

Draw the cross-sectional view, being a triangle inside a circle.

Draw the vertical "height" line down the middle of the triangle.

From the center of the circle, draw radius lines to the two base vertices of the triangle. Label the three radius lines (the two you've just drawn, plus the one that coincides with part of the "height" line) as "R".

Label the cone's radius as "r", and the remaining part of the height of the cone as "h". Using Pythagorus, note that r = sqrt[R^2 - h^2].

Plug "4" in for "V" in the formula for the volume of a sphere. Solve the resulting literal equation for R.

Take the formula for the volume of a cone with radius "r" and height "h". Plug in "sqrt[R^2 - h^2]" for "r^2" in this formula. Subtitute for the R, using what you got from the previous step.

You should, I believe, now have the volume of the cone expressed only in terms of the height h. Maximize.

Quote:

Originally Posted by daveee90
Question 2: A boat is 2 meters below the level of a dock, is being pulled towards the dock by a rope. If the rope is pulled in at 0.1m/s how fast is the boat moving threw the water when 5 meters of rope remains between the boat and the dock

Draw the right triangle formed by the boat's horizontal plane of motion, the height of the rope's attachment point (through a pulley, maybe?) above the water, and the hypotenuse being the rope's length between this point and the boat.

Label the height as "h = 2", with dh/dt = 0. Label the distance between the boat and the dock as "x", with dx/dt = -0.1. Then the rope's length is clearly D = sqrt[4 + x^2]".

Differentiate the equation for the rope's length with respect to time "t". Plug in the known values, and solve for dD/dt.

If you get stuck, please reply showing how far you have gotten. Thank you! (Wink)
• Apr 8th 2009, 07:01 AM
daveee90
I can kinda see where your coming from in most of it, im just getting stuck on a few more bits..

Quote:

Originally Posted by stapel

Plug "4" in for "V" in the formula for the volume of a sphere. Solve the resulting literal equation for R.

Take the formula for the volume of a cone with radius "r" and height "h". Plug in "sqrt[R^2 - h^2]" for "r^2" in this formula. Subtitute for the R, using what you got from the previous step.

You should, I believe, now have the volume of the cone expressed only in terms of the height h. Maximize.

i got lost when you say plug 4 in for V.... Vsphere = 4/3pi(r)^3 you say solve the resulting literal equation for R, well isnt it just R=4 since it was given? i dont understand what your trying to say.

i understand making the cone formula become Vcone = 1/3pi(sqrt R^2 - h^2)^2h but subbing in 4 for R i just get 4 for H... its all to confusing lol, sorry!!!!

Quote:

Originally Posted by stapel
Draw the right triangle formed by the boat's horizontal plane of motion, the height of the rope's attachment point (through a pulley, maybe?) above the water, and the hypotenuse being the rope's length between this point and the boat.

Label the height as "h = 2", with dh/dt = 0. Label the distance between the boat and the dock as "x", with dx/dt = -0.1. Then the rope's length is clearly D = sqrt[4 + x^2]".

Differentiate the equation for the rope's length with respect to time "t". Plug in the known values, and solve for dD/dt.

If you get stuck, please reply showing how far you have gotten. Thank you! (Wink)

im sort of following a bit better on this one. we want to know dD/dt so therefore dD/dt = dx/dt x dD/dx correct? (if thats wrong them im already way off :( )

dD/dx = x/sqrt(4+x^2)
dh/dt = 0
dx/dt = 0.1

do i then just sub them in to get 0.1x/sqrt(4 + x^2)
and then make x = 5 and my answer 0.092847 m/s???

if thats correct well then good :)
but going back to the start, you said that dx/dt = 0.1 and to me, i thought that dD/dt = 0.1 because it says that the rope is being pulled in at 0.1m/s so thats where i get confused because over time, D would be changing at 0.1m/s not x.

sorry if im confusing, im not the best explainer... i just hope you can clarify things for me.

thanks heaps for taking the time to answer this!!!

dave
• Apr 8th 2009, 07:23 AM
stapel
Sorry: I'd misread the "4" in the exercise to be the fixed volume of the sphere, when actually it was the fixed radius of the sphere. So plug "4" in for "R", and go from there. (Blush)
• Apr 8th 2009, 08:18 AM
daveee90
thanks again.
• Apr 8th 2009, 12:46 PM
djc07002
Boat Problem
If u draw the triangle u get X(t) as the bottom leg and D(t) as the hypotenuse and 2 as the vertical leg....now the equation to get D(t) is pretty easy, just use pythagoreans theorem, so then differentiate the equation and solve for dD/dt...then use the original equation to get the length of X when D is 5.... then plug in all the variables into the differentiated formula and solve..

I got X = Square root of 21
and dD/dt = .49m/s

Let me know if u get stuck
• Apr 8th 2009, 07:13 PM
daveee90
obviously im a failure coz i still cant work it out haha.
what exactly am i trying to figure out? when i do it, i try and find dt/dx

i can draw the triangle fine..
i can get D = sqrt(4+x^2)
i can get dD/dx = x/sqrt(4+x^2)
i just dont know how to put the rest of it together.
i assume that dD/dt = 0.1 (or -0.1??)

so now i do dx/dt = dD/dx * dt/dD which is dt/dx = 5/sqrt(4+5^2) * 1/0.1 all that turns out that to be dx/dt = 9.285 and i want dt/dx so i flip it and get dx/dt = .1077m/s

i understand where djc07002 gets sqrt21, but i dunno where you use that to get your answer of .49???

and to stapel... i have no clue about the cone in a sphere question lol.
thanks to you both for the help!!
• Apr 9th 2009, 03:05 AM
mr fantastic
Quote:

Originally Posted by daveee90
[snip]Question 2

A boat is 2 meters below the level of a dock, is being pulled towards the dock by a rope. If the rope is pulled in at 0.1m/s how fast is the boat moving threw the water when 5 meters of rope remains between the boat and the dock

Thanks in advanced for anyone whose able to help.
Dave.

And in various trivially changed forms here:

http://www.mathhelpforum.com/math-he...es-limits.html

http://www.mathhelpforum.com/math-he...e-extrema.html

http://www.mathhelpforum.com/math-he...-problems.html

http://www.mathhelpforum.com/math-he...-problems.html

http://www.mathhelpforum.com/math-he...-problems.html

Quote:

Originally Posted by daveee90
Gday.

If anyone can show me how to do these i appreciate it muchly.
Im on a 2 week break with an exam first day back and got no idea what to do with these questions..

Question 1

Find the volume of the largest circular cone that can be cut from a solid sphere of radius 4cm.
[snip]