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Question 1

Page history last edited by Rence 11 years, 6 months ago

 

 Calculator MAY NOT be used to help answer this question.

 

Water is being poured into a hemispherical bowl of radius 6 inches at the rate of .

 

 

(a) Given that the volume of the water in the spherical segment shown above is;

 

 

where R is the radius of the sphere, find the rate that the water is rising when the water is 2 inches deep.

 

(b) Find an expression for r, the radius of the surface of the spherical segment of water in terms of h.

 

(c) How fast is the circular area of the surface of the spherical segment of water growing (in in2/sec) when the water is 2 inches deep?

 

Solution


Quick Find Code For This Problem.

 

Legend: [QUES1LGND]

Solution For Part a.): [QUES1PA]

Solution For Part b.): [QUES1PB]

Solution For Part c.): [QUES1PC]

 

How To Use Quick Find.

 

-To Use Quick Find, copy (Ctrl+C/Command+C) the code (All the stuff INSIDE the square brackets in the code region above) for which ever section you'd like to go to, then, hit in this order;

 

Ctrl+F

Ctrl+V

Enter/Return

 

or if your on a Mac You'd hit

 

Command+F

Command+V

Enter/Return.

 

Then, your webbrowser should bring you directly to the beginning of the solution indicated by the code, making navigation through longish solutions much easier. Enjoy :]

 

 

[QUES1LGND]

Legend:

 

Explaination on How to Solve It (mmm, grape)

 

The Actual Solution in Numbers and Formulas (mmm, orange)

 

Author's Notes (mmm... electric)

 

 

 

 

 

 

 

 

 

In this question it appears R would be the radius for the bowl itself, while r is the radius for the surface of the water.

[QUES1PA]

Part (a) This question is asking for the value of the rate at time where h = 2 (h being height). 

 

To find the rate, we want to find the derivative of

 

 

Using differentiation rules:  

 

 

 

 

 

 

 

 

 

In general the rate that the water is rising when the water is 2 inches deep is

 


[QUES1PB]

Part (b)The way to solve this question is relatively simple, since V also equals (4/3)πr^3 and we have the value of R. Rearrange algebraically.

 

 

therefore

 

 

Divide out π

 

 

 

 

A little rearraging with algebra...

 

 

 

 

there for the expression for r is

, when the radius of the surface of the spherical segment of water in terms of h


[QUES1PC]

(c)  Find the rate at which the surface area of your water is increasing. This simply means find the derivative of the equation you found in (b).

 

To demonstrate the use of a differentiation rule, we'll create two sub-functions...

 

 

 

 

 

A small sub problem...

 

9 * 72 = (9*70) + (9*2) 

 

(9*7)10 + (18)

 

63(10) + 18

 

630 + 18

 

=648 

 

So, using the rules for differentiation, we know that...

 

So we input 2 as h and....

 

 

 

And so because you cannot use a calculator to solve this question, this is the farthest you can go!

 

Or you can simplifies it more, which -36/81 will be come -4/9, and 648/81 will be 8.

 

there the answer will be come:                    Formula

 

 

In conclusion the  circular area of the surface of the spherical segment of water growing (in in2/sec) is

Formula

when the water is 2 inches deep?

Comments (1)

Benchmen said

at 7:28 pm on Apr 18, 2009

Haha mmmm...electric

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