VCE Methods Question Thread!

[ecvkcuf]

Can someone please help me answer the above questions?

Also what the hell is a cusp and what functions would be considered a cusp..

[TrueTears]

Let y=ⁿ be a power function where n is a whole positive number, use calculus to find which values of n there is a local minimum of the function..

Don't really know how to approach this question.. do I just sub in a random positive number and then use calculus to find the stationary points and then the local minimum?

Please help.. please show all working out..


2)a) Consider the power function x^2/3 which is assumed to have relationship x^2/3=(x^1/3)² and this applies to all fractional indices

(i) Use calculus to show that the minimum of this function is a cusp and state its coordinates. (In this answer you expected to provide a definition of a cusp)

b) Consider y=x^4/3
i) Use calculus to investigate its minimum. Is it a local minimum or a cusp and state its coordinates.

for the first question, we require f'(x) = 0 and f''(x) >0. Apply these 2 statements and solve.

Can someone please help me answer the above questions?

Also what the hell is a cusp and what functions would be considered a cusp..

a cusp has different types of definitions, in methods, you don't really need to worry about these definitions, just remember that a cusp in VCE is a "sharp point", basically it's not a smooth joining curve at the point of a cusp.

Some formal definitions of a singular point is given here http://en.wikipedia.org/wiki/Singular_point_of_a_curve and http://en.wikipedia.org/wiki/Cusp_%28singularity%29

[pi]

"A cusp is a point at which two branches of a curve meet such that the tangents of each branch are equal. "


(wolfram mathematica definition)


Essentially, something like will have cusps at all x-intercepts


edit: beaten

[ecvkcuf]

Would a cusp point have no gradient or be undefined?

Could you give an example of two functions which have cusp points? Would |x| and |x²| be functions with cusp points?


Can you help me out with this question..

Consider a power function of the form f(x)=x^p/q where p and q are positive numbers. Find which values of p and q for the function f(x) to have a cusp point?

I don't understand this question.. do we have keep on subbing in values of p and q to see a relationship? help..

[pi]

Would a cusp point have no gradient or be undefined?

Undefined afaik

Could you give an example of two functions which have cusp points? Would |x| and |x²| be functions with cusp points?

Yes to y=|x| no to y=|x^2|, as that graph is essentially y=x^2. edit: see below post

[pi]

Actually, looking back on the question, y=|x| doesn't have a cusp either, it's more of a 'corner' as the gradients of the two tangents aren't exactly equal (or approaching equality) at x=0. I think something like y=|sin(x)| or x^3 - y^2 = 0 have more "cusp like" features.

I'll leave this nuance to kamil or TT

[Phy124]

I think something like y=|sin(x)|

I think that would be in the same boat as y=|x|, because the gradients of the line would alternate from -1 to 1 where the intercepts are (Unless I've misinterpreted)

edit: Did some reading and it appears both these functions have cusps (something like they have to undefined gradients at that point after having different gradients, which they do), so disregard the above.

edit2: Probably just best to read luffys post below, should clear things up.

[ecvkcuf]

Consider a power function of the form f(x)=xp/q where p and q are positive numbers. Find which values of p and q for the function f(x) to have a cusp point?

I don't understand this question.. do we have keep on subbing in values of p and q to see a relationship? help..

i need help with this

[Phy124]

Consider a power function of the form f(x)=x^p/q where p and q are positive numbers. Find which values of p and q for the function f(x) to have a cusp point?

I don't understand this question.. do we have keep on subbing in values of p and q to see a relationship? help..

i need help with this

If p is an even number and q is an odd number and p<q, then the equation should have a cusp. (I think )

[luffy]

Would a cusp point have no gradient or be undefined?

Could you give an example of two functions which have cusp points? Would |x| and |x²| be functions with cusp points?


Can you help me out with this question..

Consider a power function of the form f(x)=x^p/q where p and q are positive numbers. Find which values of p and q for the function f(x) to have a cusp point?

I don't understand this question.. do we have keep on subbing in values of p and q to see a relationship? help..

1. Would a cusp point have no gradient or be undefined?

A cusp point is really just a point that isn't "smooth" on the graph and the tangents are equal on both sides (i.e. the two branches are somewhat mirror images of each other). After reading quite a few resources, the definition seems more complicated than this, but a cusp is defined for a function, and has "no gradient" at that point. The function certainly isn't undefined at the cusp point e.g. By definition, it seems y = |x| has a cusp point at x = 0, but its still defined at x=0.


2. Could you give an example of two functions which have cusp points? Would |x| and |x2| be functions with cusp points?
So, as I just explained, cusp points are when the gradient is undefined. The best way to see this is simply graphing the two equations on your calculator. You will see that |x| has a sudden bend ("pointy end") at x = 0. This is a cusp point.
|x^2| has no cusp point? Why? Because x^2 is always >0 and hence, the absolute value signs have NO impact on the expression. If you graph y=|x^2|, you will note that it looks exactly the same as y = x^2.

3. Consider a power function of the form f(x)=xp/q where p and q are positive numbers. Find which values of p and q for the function f(x) to have a cusp point?

This is a good question and to be frank, I think the answer is far more complicated than the one I will give. If you test out a few values, you will find that you have a cusp for x^(2/3) or x^(4/7), but no cusp for graphs like y = x^(4/3)  -- oddly enough. I think this question is outside the methods course =.=.

Having said that, my logic was as follows:
- If q is even, then there is no cusp point (as the graph has an end point at x = 0). Hence, q must be odd.
- When q is odd, the cusp only occurs when p is even.
- When you differentiate the function, you get f'(x) = (p/q)x^(p/q - 1). We require p/q - 1 < 0 so that at x = 0, you are "dividing by 0" and hence, showing it has no derivative at that point. Hence, p/q <1. As you said both are positive, 0<p/q<1. (as x^0 clearly has no cusp)

Thus, I came to the conclusion that q must be odd (i.e. q = 2n - 1, where n is an element of the Natural Number set). For p/q to be less than 1 AND even, we require p = 2k, where k is an element of the natural number set AND p<2n - 1. i.e. p is even and less than q.

Assumptions I made:
- p/q must be the most simplest form of the fraction.
- The natural number set does not include 0. I forgot which definition VCAA methods uses, but the definition of the "natural number set" is debatable as some believe it to include 0.
- p and q are positive WHOLE numbers.

Hope the stuff I said is correct. I'm not 100% sure on question 3, so it might be best to ask someone else for confirmation....

Hope I helped.

[ecvkcuf]

Thanks I figured it out. It turns out that if p<q then the f'(x)=0 will be undefined and therefore will have a cusp.

while if q>p then f'(x)=0, would be 0 and therefore will have a local minimum

These questions are outside of the methods course, yes. My teacher is making us do them because she hates us

[ecvkcuf]

I need help with this question..

(1) Function of the form f(x)= xⁿ+1/x^m where n and m are positive numbers.

(i) Use calculus to show that the coordinates of the local minimum are:

x_min=(m/n)^(1/n+m)

y_min=(m/n) x (n/n+m) + n/m x m/n+m

[luffy]

Thanks I figured it out. It turns out that if p<q then the f'(x)=0 will be undefined and therefore will have a cusp.

while if q>p then f'(x)=0, would be 0 and therefore will have a local minimum

These questions are outside of the methods course, yes. My teacher is making us do them because she hates us

Yeah - thats essentially my answer. Except, p/q must also be between 0 and 1 for what you said to be correct.

I need help with this question..

(1) Function of the form f(x)= xⁿ+1/x^m where n and m are positive numbers.

(i) Use calculus to show that the coordinates of the local minimum are:

x_min=(m/n)^(1/n+m)

y_min=(m/n) x (n/n+m) + n/m x m/n+m

1) i) f'(x) = nx^(n-1) -mx^(-m-1) = 0 for min.
nx^(n-1) = mx^(-m-1)
n/m = x^(-m-1) / x^(n-1)
n/m = x^(-m-1-n+1)
n/m = x^(-m-n)
m/n = x^(m+n)
x = (m/n)^(1/(m+n)), as required

f[(m/n)^(1/(m+n))] = [(m/n)^(1/(m+n)]^n + [(m/n)^(1/(m+n))]^(-m)
= (m/n)^(n/(n+m)) + (m/n)^(-m/(m+n))
=(m/n)^(n/(n+m)) + (n/m)^(m/(m+n)), as required.

Should have used latex xD. Oh well. Hope I helped.

[IndefatigableLover]

Find the equation of the tangent to the curve with the equation
y=3x³-4x²+2x-10
at the point of intersection with the 'y' axis.

 

[TrueTears]

Find dy/dx first, then the intersection with the y axis is when x = 0.

Thus sub x = 0 into dy/dx to yield the gradient.

Then you know that (0, -10) lies on this cubic curve.

So you have y-y_1 = m(x-x_1) where m = dy/dx_|x=0 , x_1 = 0 and y_1 = -10