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December 30, 2025, 09:07:03 am

Author Topic: Bucket's Questions  (Read 56777 times)  Share 

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Mao

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Re: Bucket's Questions
« Reply #165 on: June 17, 2008, 09:48:44 pm »
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I have forgotten how to derive after two weeks w/o maths :S.

, show that and specify when

last thing before bed :) [hey polky :P ]

differentiation [by product rule]:



when , it implies that the function will be "positive"

looking at , we can see three parts:

which is constant always positive

, which is a square

, the only terms that changes sign. therefore, when this part is greater than (or equals to 0), the derivative will be positive (or 0)

« Last Edit: June 17, 2008, 09:52:26 pm by Mao »
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bucket

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Re: Bucket's Questions
« Reply #166 on: June 17, 2008, 09:50:53 pm »
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hah thanks mao.
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bucket

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Re: Bucket's Questions
« Reply #167 on: June 17, 2008, 10:58:37 pm »
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Find the two positive numbers whose sum is 4 and such that the sum of the cube of the first and the square of the second is as small as possible.

The bolded part of the question is what's confusing me, what the hell do you equate these two numbers to if you don't know what 'as small as possible' would be?
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Re: Bucket's Questions
« Reply #168 on: June 18, 2008, 01:17:38 am »
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Let the two positive numbers by x and y.



You want to minimise . Let (just label what u wanna minimize)

Now, from the first equation, you have



Now, find and find the nature of 's stationary points  ;)

bucket

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Re: Bucket's Questions
« Reply #169 on: June 18, 2008, 04:35:54 pm »
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oh :S thanks.
lol i feel stupid after that one XD
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Re: Bucket's Questions
« Reply #170 on: June 18, 2008, 11:22:27 pm »
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"Find the point on the parabola that is closest to the point "
:S what the fuck do you do with this..? tangents?
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Glockmeister

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Re: Bucket's Questions
« Reply #171 on: June 18, 2008, 11:46:36 pm »
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I havent looked at the question in detail yet, but my hunch is that it involved the distance formula for some reason.
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Re: Bucket's Questions
« Reply #172 on: June 18, 2008, 11:56:00 pm »
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but (3,0) isn't a point on the graph. it's just a point. lol
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ell

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Re: Bucket's Questions
« Reply #173 on: June 19, 2008, 01:00:52 pm »
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"Find the point on the parabola that is closest to the point "
:S what the fuck do you do with this..? tangents?

Let be the points that are closest to .

As Glockmeister said we need to use the distance formula. The distance between those two points are:



Since we know we can sub this in:





We need to find when the distance is minimal, i.e. when . We can also just ignore the square root as we are going to make (since if we derive then )





We need to factorise this to find solutions. Let , then:



Hence .

We also need to ensure that this stationary point is a minimum. Find a point before and after and substitute into to verify the nature.





Drawing a gradient table:  \ _ /   hence minimum. (Also by therefore minimum).

Substitute into :



Therefore the point closest to on the parabola is .

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Re: Bucket's Questions
« Reply #174 on: June 19, 2008, 04:25:16 pm »
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hah thanks, thats good.

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Re: Bucket's Questions
« Reply #175 on: June 19, 2008, 09:47:44 pm »
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The surface area of a cube is changing at the rate of 8cm2/s. How fast is the volume changing when the surface area is 60cm?
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Re: Bucket's Questions
« Reply #176 on: June 19, 2008, 10:08:13 pm »
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Let A be surface area (), V be volume (), t be time (s).

You have

You need



To find ,







When



the volume is increasing at a rate of

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Re: Bucket's Questions
« Reply #177 on: June 19, 2008, 10:37:39 pm »
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perfect, thanks a lot man.
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Re: Bucket's Questions
« Reply #178 on: June 21, 2008, 11:19:12 pm »
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A vessel has such a shape that when the depth of the water in it is x cm, the volume is given by the equation . Water is poured in at a constant rate of . At what rate is the water rising when the depth is 8cm?
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Re: Bucket's Questions
« Reply #179 on: June 21, 2008, 11:29:52 pm »
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We are looking for (I assume you mean 30cm^3/s)

This is an application of the chain rule to relate to different rates.

This step basically differentiates your Volume formula with respect to x.

Notice how the conjugate of dV/dx is used, because we require dx/dV.

When :

« Last Edit: June 21, 2008, 11:32:46 pm by lwine »