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October 04, 2025, 05:33:10 am
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Author Topic: TyErd's questions  (Read 42899 times)  Share 

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TyErd

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Re: TyErd's questions
« Reply #240 on: June 13, 2010, 07:36:15 pm »
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Also help with this: , where n is a positive even integer.

show that


Also show that for all values of x. How do I go about doing this? draw a graph?
« Last Edit: June 13, 2010, 07:43:21 pm by TyErd »
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TyErd

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Re: TyErd's questions
« Reply #241 on: June 13, 2010, 07:37:20 pm »
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Quote from: kenhung123 on June 13, 2010, 05:32:39 pm
Quote from: TyErd on June 13, 2010, 05:06:08 pm
use the chain rule to prove , where n is a negative number. I dont get it.
Hmm not sure about the "where n is a negative number" but I would just do it normally like lol thats basically it I guess...

but it says using chain rule
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kenhung123

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Re: TyErd's questions
« Reply #242 on: June 13, 2010, 07:48:41 pm »
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Yea thats chain rule, it looks like its very simple but you can see that u=x
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TyErd

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Re: TyErd's questions
« Reply #243 on: June 13, 2010, 07:57:48 pm »
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Also what is the equation of the tangent of
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kenhung123

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Re: TyErd's questions
« Reply #244 on: June 13, 2010, 08:35:04 pm »
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find dy/dx then sub in x=0
This would give the gradient of the point x=0
Now sub the gradient (m) into y=mx+c
You are given a point also (x,y) when x=0, find y using original equation
Sub the point into the linear equation and find c
Then sub c into the y=mx+c equation.
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kenhung123

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Re: TyErd's questions
« Reply #245 on: June 13, 2010, 08:37:21 pm »
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Quote from: brightsky on June 13, 2010, 01:54:30 pm
Quote from: kenhung123 on June 13, 2010, 01:36:59 pm
Is there at all many tricks to find the gradient graph only given the original graph? This is extremely tricky!
Just find the equation of the original graph and then derive it to get gradient graph as far as I know.
But sometimes they don't even give you 1 point basically they want you to draw an approximate gradient graph but it has to be somewhat reasonable like...positive gradient at a particular point must have f'(x) above x axis at least..
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TyErd

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Re: TyErd's questions
« Reply #246 on: June 13, 2010, 08:51:34 pm »
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Quote from: kenhung123 on June 13, 2010, 08:35:04 pm
find dy/dx then sub in x=0
This would give the gradient of the point x=0
Now sub the gradient (m) into y=mx+c
You are given a point also (x,y) when x=0, find y using original equation
Sub the point into the linear equation and find c
Then sub c into the y=mx+c equation.

Yeah I got that but at x=0 there is a cusp, does that mean there's many tangents?
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/0

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Re: TyErd's questions
« Reply #247 on: June 13, 2010, 09:22:03 pm »
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Quote from: TyErd on June 13, 2010, 07:37:20 pm
Quote from: kenhung123 on June 13, 2010, 05:32:39 pm
Quote from: TyErd on June 13, 2010, 05:06:08 pm
use the chain rule to prove , where n is a negative number. I dont get it.
Hmm not sure about the "where n is a negative number" but I would just do it normally like lol thats basically it I guess...

but it says using chain rule

I'm not sure how the chain rule plays into it, but here's how I would do it. Since is a negative number, it's usually convenient to 'bring the negative out'. To do this, set , then is a positive number.

Then , so

Differentiating implicitly,





So we have proved differentiation works for negative exponents. Notice that nowhere in the proof did we actually differentiate a negative exponent, since is positive.
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TyErd

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Re: TyErd's questions
« Reply #248 on: June 13, 2010, 09:51:18 pm »
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Thanks /0!
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TyErd

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Re: TyErd's questions
« Reply #249 on: June 13, 2010, 09:54:18 pm »
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can someone help me out with post number 240.
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TyErd

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Re: TyErd's questions
« Reply #250 on: June 13, 2010, 10:30:45 pm »
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Im stumped on this one: One side of a rectangle is three times the other. If the perimeter inceases by 2%, what is the percentage increase in the area.
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the.watchman

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Re: TyErd's questions
« Reply #251 on: June 14, 2010, 09:33:10 am »
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Let the length of the shorter side of the rectangle be x
Let the perimeter be P(x)
Let the area be A(x)

Then the other side length is 3x

So



From the linear approximation formula, the change in perimeter is

So if the change in perimeter is 2%, then





Because

Therefore the change in area is





So the change is 4%

I hope that made some sense :)
« Last Edit: June 14, 2010, 12:21:15 pm by the.watchman »
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TyErd

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Re: TyErd's questions
« Reply #252 on: June 14, 2010, 11:05:53 am »
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Wow you are a legend mate, question though, what did you sub into the x in the last step to get 4?
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TyErd

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Re: TyErd's questions
« Reply #253 on: June 14, 2010, 11:25:25 am »
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I cant seem to get this one either: A 2% error is made in measuring the radius of a sphere. Find the percentage error in the surface area. (The surface area of a sphere is given by
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the.watchman

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Re: TyErd's questions
« Reply #254 on: June 14, 2010, 12:27:06 pm »
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Quote from: TyErd on June 14, 2010, 11:05:53 am
Wow you are a legend mate, question though, what did you sub into the x in the last step to get 4?

Sorry, my LaTeX didnt come out how I wanted, it should be there now :)

Quote from: TyErd on June 14, 2010, 11:25:25 am
I cant seem to get this one either: A 2% error is made in measuring the radius of a sphere. Find the percentage error in the surface area. (The surface area of a sphere is given by

From the linear approximation formula:



In this instance, h is 2% of r, so:









So the percentage error is 4% :)
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Remember, remember the 5th of November

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