Mathematics Question Thread

[jamonwindeyer]

A diagram for clarity:



The answer is \(1\text{ units}^2\)

[Rathin]

A diagram for clarity:

(Image removed from quote.)

The answer is \(1\text{ units}^2\)

Oh I see..and for q5..why is the area between the two curves the shaded area?

[jamonwindeyer]

Oh I see..and for q5..why is the area between the two curves the shaded area?

That's because in that case, they don't specify the x-axis. That's the only difference! It's a trick of the wording

[Rathin]

Is my reasoning sufficient?

[jamonwindeyer]

Is my reasoning sufficient?

This is a weird question - Given that the bounds are decided for you, I think what you actually need to prove is which curve is on top for \(1\le x\le4\). Why is it not the other way around in the integral? I'm pretty sure that is what you would need to show. Finding the point of intersection doesn't change anything, because the area is still only for \(1\le x\le4\), intersection or not

Not that this is hard to prove - Clearly for those x values one curve is always above the x-axis and one is always below, and so, one is clearly greater than the other

(You won't get this question in the HSC, it's a little awkward imo)

[Rathin]

16c pls

[jamonwindeyer]

16c pls

That is simultaneous equations, use the coordinates at C and D to create your system:



Now the fact that we have \(-60^n\) in our equations means that \(n\) needs to be an integer (we'd be beyond the bounds of the HSC course if it wasn't). We can approach this several ways, but if we consider and check some of the easy possible answers (\(n=1,2,3\)), we find that \(n=2\) and \(a=0.005\) is the solution

[Rathin]

That is simultaneous equations, use the coordinates at C and D to create your system:



Now the fact that we have \(-60^n\) in our equations means that \(n\) needs to be an integer (we'd be beyond the bounds of the HSC course if it wasn't). We can approach this several ways, but if we consider and check some of the easy possible answers (\(n=1,2,3\)), we find that \(n=2\) and \(a=0.005\) is the solution

Legend, Thanks

[katnisschung]

this one trips me up every freaking single time!

i need help differentiating the attached
pls show all working out...im bad at maths

[Syndicate]

this one trips me up every freaking single time!

i need help differentiating the attached
pls show all working out...im bad at maths

You will need to apply the product and the chain rule to work out the derivative of this expression.

First of all, using the chain rule work out the derivative of the square root part.


Now use the product rule (which states that (f(x)g(x))' = f'(x) x g(x) + g'(x) x f(x) )
Just for this moment assume f(x) = x and g(x) = \( \sqrt{x-2} \)
so f'(x) = 1 and g'(x) = \( \frac{1}{2\sqrt{x-2}} \)


You can simplify this down to \( \frac{3x-4}{2\sqrt{x-2}} \)

[katnisschung]

Thank you Syndicate!! this whole time i was using only the function rule...
idk why it was included in the function rule ex...

[RuiAce]

Thank you Syndicate!! this whole time i was using only the function rule...
idk why it was included in the function rule ex...

[jamonwindeyer]

Thank you Syndicate!! this whole time i was using only the function rule...
idk why it was included in the function rule ex...

Beyond Rui's reasoning, it is likely that it was in there because you are using the function rule within the product rule - You can be expected to use them in tandem after all a little mean if you've not learned the product rule yet though!

[Rathin]

I keep getting this wrong for some reason..
Find the area bounded by the curves y=x^2(1-x) and y=x(1-x)^2.

[jakesilove]

I keep getting this wrong for some reason..
Find the area bounded by the curves y=x^2(1-x) and y=x(1-x)^2.

That's a bloody tough one. The problem is that there are TWO areas that are bounded by these curves. Between 0 and 0.5, the first graph is on top, and between 0.5 and 1, the second graph is on top. However, the areas will be equal, so I'm going to calculate one and then double it! Make sure to sketch the graph, so you know which is on top of which for the equation to work.

So, the intercept is at x=0.5 (can you see why?)

Now, we want to find the integral between zero and 0.5 of the first graph minus the second graph.

I can't latex integrals left. The integral we need to find is



Between zero and one. This expands to



From there, just integrate between 0 and 0.5, then double the result! Let me know if you need help with that part