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Author Topic: VCE Methods Question Thread!  (Read 4802397 times)  Share 

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Reus

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Re: VCE Methods Question Thread!
« Reply #4785 on: May 16, 2014, 11:19:46 pm »
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Can someone showing me the working out of the derivative of (loge e^x)? I know it equals 1, but how?
Much appreciated.
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b^3

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Re: VCE Methods Question Thread!
« Reply #4786 on: May 16, 2014, 11:26:11 pm »
+5
Using log rules, you can bring the power down, then log_a(a) is 1.


It might be interesting to point out that if you had it the other way around, that is the log in the power of the exponential it would be similar, but not exactly the same. You'd then have a domain restriction, as the function wouldn't be defined for zero and negative values of .

« Last Edit: May 16, 2014, 11:28:47 pm by b^3 »
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Reus

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Re: VCE Methods Question Thread!
« Reply #4787 on: May 16, 2014, 11:29:09 pm »
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Using log rules, you can bring the power down, then log_a(a) is 1.
Wow thanks! Could you do one more?
3 loge e^2x
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b^3

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Re: VCE Methods Question Thread!
« Reply #4788 on: May 16, 2014, 11:33:10 pm »
+4
Remember, if we have something in the form we can bring the power down to the front to get . Also anything that has the base being the same as the inside of the bracket will give you one. . You might want to review your log rules a little and then try it.

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lzxnl

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Re: VCE Methods Question Thread!
« Reply #4789 on: May 17, 2014, 12:16:54 pm »
+1
3 ln e^(2x)?

Well you can always do it the dumb way without knowing log laws :P

  Use chain rule on this





So for y=3 ln e^2x, you can either simplify it first to 6x and differentiate to get 6, or do a similar thing to what I did above by letting u = e^2x
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Reus

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Re: VCE Methods Question Thread!
« Reply #4790 on: May 17, 2014, 01:02:46 pm »
0
Differentiate with the respect of x,

x3 √1–5x
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swagsxcboi

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Re: VCE Methods Question Thread!
« Reply #4791 on: May 17, 2014, 01:11:37 pm »
+1
Differentiate with the respect of x,

x3 √1–5x
use product rule

x3 √1–5x

x3  ->  3x2             √1–5x  -> -5/2√1-5x

3x2 times √1–5x   +  x3  times -5/2√1-5x

3x2√1–5x  +   -5x3/2√1-5x

common denominator, final answer is (-35x3 + 6x2)/2√1–5x
(rationalise if you need to)
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Reus

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Re: VCE Methods Question Thread!
« Reply #4792 on: May 17, 2014, 01:23:30 pm »
0

3x2√1–5x  +   -5x3/2√1-5x

common denominator, final answer is (-35x3 + 6x2)/2√1–5x
(rationalise if you need to)

Thats exactly what I got, however I do not understand the step from 3x2√1–5x  +   -5x3/2√1-5x

to

(-35x3 + 6x2)/2√1–5x
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b^3

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Re: VCE Methods Question Thread!
« Reply #4793 on: May 17, 2014, 01:31:33 pm »
+4
Thats exactly what I got, however I do not understand the step from 3x2√1–5x  +   -5x3/2√1-5x

to

(-35x3 + 6x2)/2√1–5x

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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #4794 on: May 17, 2014, 03:47:53 pm »
0
I don't understand rates of related change. Really struggling :(

Could someone do the steps to see where I went wrong?

Grain is being poured at a rate of 6m^3 per minute into a pile and forms a right circular cone which has a height equal to twice its radius. Find the exact rate at which the heigh of the pile is increasing when the height is 3m.

Help would be very much appreciated. Thanks!

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Re: VCE Methods Question Thread!
« Reply #4795 on: May 17, 2014, 04:03:29 pm »
+7
"Grain is being poured at a rate of 6m^3 per minute"

"a right circular cone which has a height equal to twice its radius."

Also we'll use the formula for the volume of a cone.
.
"Find the exact rate at which the heigh of the pile is increasing "
So we want to find . When .

So we'll need to use the chain rule.
.
We already have so we just need to find .

Then we can substitute this back into our expression from the chain rule above, along with .


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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #4796 on: May 17, 2014, 04:59:07 pm »
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Only thing I don't understand is how 1/3 pi (h/2)^2 x h gets to 1/12 pi h ^3

But otherwise thanks so much :)

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Re: VCE Methods Question Thread!
« Reply #4797 on: May 17, 2014, 05:13:13 pm »
+1
Only thing I don't understand is how 1/3 pi (h/2)^2 x h gets to 1/12 pi h ^3

But otherwise thanks so much :)

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Cort

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Re: VCE Methods Question Thread!
« Reply #4798 on: May 18, 2014, 12:16:32 pm »
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Hi, just wanted to ask a quick concept question in
2011 Exam 2 Q4 b)i) "to find the coordinates..from the derivative)
 Q: when it states to find the coordinates of the point; why is it we have to make dy/dx=0 then find m? I thought the concept of dy/dx was to find the gradient function and when it's =0, that means it will be a turning point. Does that mean, (m,n) would be a turning point? I'm rather confused or I forgot about something.
I actually have no idea what I'm saying or talking about.

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Re: VCE Methods Question Thread!
« Reply #4799 on: May 18, 2014, 01:06:10 pm »
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Hi, just wanted to ask a quick concept question in
2011 Exam 2 Q4 b)i) "to find the coordinates..from the derivative)
 Q: when it states to find the coordinates of the point; why is it we have to make dy/dx=0 then find m? I thought the concept of dy/dx was to find the gradient function and when it's =0, that means it will be a turning point. Does that mean, (m,n) would be a turning point? I'm rather confused or I forgot about something.

finding the derivative of the function (dy/dx) gives us the rule where if we sub an x-value in, it tells us the gradient at that point. Now, if we equalise dy/dx =0 and solve for x, we find the x-value(s) (of the original function) at which the gradient equals 0.
well, since the gradient at turning points is 0, we can find the x-values at which these occur (solve dy/dx =0). This only provides us with the x-coordinate of the turning point(s). To find the corresponding y-value -  just sub it into the original function.
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