Mathematics Question Thread

[anotherworld2b]

I was working on this question but i ended up with 0 as an answer 

Edit: I was working on this question but when i differentiated it i ended up 60? 

[jamonwindeyer]

I was working on this question but i ended up with 0 as an answer 

Edit: I was working on this question but when i differentiated it i ended up 60? 

Just a little error,



You divided instead of subtracting!

[jamonwindeyer]

ah I see
thank you very much for your help 
I was wondering would you have any tips on dealing with questions on the application of differentiation? We were just taught this so I am still relatively a newbie that is slow to understand what to do.

Oh, and on this, just lots of practice unfortunately. The process is always the same:

1 - Rearrange to get a single variable
2 - Use the first derivative to find critical point

It's just about doing these over and over until it becomes second nature, and you get used to all the different things that can get thrown at you

[Shadowxo]

Just remember the derivative is the gradient, the max/min is where the gradient equals zero
Put things in terms of one thing (eg surface area in terms of L), if you can, use the easiest option - in this case putting it in terms of L so you don't have to use a square root
Make sure not to make any little mistakes. I saw with your attempt you made a little mistake saying 1000^1/2*6 = 6000^^1/2, instead of 6√1000 = 60√10. I like to check over my working frequently to minimise mistakes
I think you basically have the hang of it. Can't think of much else to say, but if you want any advice or questions answered feel free to ask

[anotherworld2b]

I now see what I did wrong. Thank you
I found r= 40/pi
I'm not sure how to find y

Just a little error,



You divided instead of subtracting!

[jamonwindeyer]

I now see what I did wrong. Thank you
I found r= 40/pi
I'm not sure how to find y

Great work! Go back to the equation, \(y=120-2\pi r\), with that new value of \(r\)

[anotherworld2b]

thank you for your help 

Great work! Go back to the equation, \(y=120-2\pi r\), with that new value of \(r\)

[anotherworld2b]

I am not sure what to do for this question 

[jakesilove]

I am not sure what to do for this question 

We'll have to do this without a graphing calculator. First, we need to find a function for the perimeter of the isosceles triangle. We know that the triangle has a fixed perimeter, so



Where x is the two equal sides, and y is the base (unequal) side. c is a constant. Now, we care about the area of the triangle. This can be calculated by



The base is going to be equal to y, and we can find the height using pythag.





So, our area function is going to be



We can rewrite our perimeter function like this



and sub every y for the function above.



Great! Now, we're looking for a maximum area. So, let's differentiate the function. Since you can use a graphic calculator, I'm going to use wolfram alpha.



Clearly, the turning point will occur when



I'll leave you to show that this is a maximum. Well, if two of the sides (ie. x) are equal to c/3, then the third side must be equal to c/3 (as the perimeter must add up to c). Therefore, the triangle is equilateral

[anotherworld2b]

Like this?

We'll have to do this without a graphing calculator. First, we need to find a function for the perimeter of the isosceles triangle. We know that the triangle has a fixed perimeter, so



Where x is the two equal sides, and y is the base (unequal) side. c is a constant. Now, we care about the area of the triangle. This can be calculated by



The base is going to be equal to y, and we can find the height using pythag.





So, our area function is going to be



We can rewrite our perimeter function like this



and sub every y for the function above.



Great! Now, we're looking for a maximum area. So, let's differentiate the function. Since you can use a graphic calculator, I'm going to use wolfram alpha.



Clearly, the turning point will occur when



I'll leave you to show that this is a maximum. Well, if two of the sides (ie. x) are equal to c/3, then the third side must be equal to c/3 (as the perimeter must add up to c). Therefore, the triangle is equilateral

[jakesilove]

Like this?

Almost! Instead of 1/3 (which assumes that the perimeter of the triangle is 1), you should be writing c/3 (which assumes that the perimeter of the triangle is some length, c).

[anotherworld2b]

Ah okay
For these two questions i dont understand how to use the small changes rule?

I was able to find the radius for q11.
R= 6.180 but I'm not sure where to go from here

Almost! Instead of 1/3 (which assumes that the perimeter of the triangle is 1), you should be writing c/3 (which assumes that the perimeter of the triangle is some length, c).

[jakesilove]

Ah okay
For these two questions i dont understand how to use the small changes rule?

I was able to find the radius for q11.
R= 6.180 but I'm not sure where to go from here

Unfortunately, this isn't in the HSC! Sorry

[anotherworld2b]

oh okay 

Unfortunately, this isn't in the HSC! Sorry

[jamonwindeyer]

oh okay 

Since you've started the 1st one, I'll show you the full working of the 2nd and hopefully it will help you finish the first!

Remember, the formula we use is:



This comes from rearranging the visually simpler approximation, \(\frac{\delta C}{\delta n}\approx\frac{dC}{dn}\) - Remember that the small change (on the left) is roughly equal to the derivative (on the right), when that derivative is evaluated at the correct point. Geometrically, this is the same as using a tangent to approximate a curve (as shadow said):



We also know that \(\delta n=1\), so:



So the additional cost will be $200, roughly. If you compare that to the actual cost found by substitution, it is $216, so we're reasonably close

So the steps are to find the derivative, and evaluate it at our base point (what we're increasing from), then use the formula