VCE Methods Question Thread!

[Rob16]

Can anyone give me a complete summary of the new study design changes for methods 3/4?

So far i know that:
- Modulus functions are being removed
- Matrices are being removed
- Markov chains are being removed from probability

- Statistical Inference is the new topic

am i missing anything?

[keltingmeith]

Can anyone give me a complete summary of the new study design changes for methods 3/4?

So far i know that:
- Modulus functions are being removed
- Matrices are being removed
- Markov chains are being removed from probability

- Statistical Inference is the new topic

am i missing anything?

Nah, that seems to have it all. Also, normals are out, but that's pretty tiny, so.

[Photon]

Can someone please help me with this question:

Find the value of c such that the line with equation y = 2x + c is tangent to the parabola
with equation y = x2 + 3x. (Answer:  c= -1/4)

I'm stuck on this question and it's really annoying.

Any help would be greatly appreciated.

[MightyBeh]

Can someone please help me with this question:

Find the value of c such that the line with equation y = 2x + c is tangent to the parabola
with equation y = x2 + 3x. (Answer:  c= -1/4)

I'm stuck on this question and it's really annoying.

Any help would be greatly appreciated.

Gotcha covered. Use the derivative of y to find the value of x that has a gradient of two, sub that back into y to find some coordinates and then sub those into your tangent equation. If that didn't make sense, working's in the spoiler.

Spoiler

Derivative (gradient function) of y is:


Because the gradient of the tangent is also the gradient of where it touches the curve, we can solve for dy/dx = 2




However there's three unknowns in the tangent equation so we also need to find a y value. Subbing the x value back into the parabola will give us one.




Which gives us the ordered pair

Subbing that into the tangent gives us:




Hope that's clear

[TheAspiringDoc]

Gotcha covered. Use the derivative of y to find the value of x that has a gradient of two, sub that back into y to find some coordinates and then sub those into your tangent equation. If that didn't make sense, working's in the spoiler.

Spoiler

Derivative (gradient function) of y is:


Because the gradient of the tangent is also the gradient of where it touches the curve, we can solve for dy/dx = 2




However there's three unknowns in the tangent equation so we also need to find a y value. Subbing the x value back into the parabola will give us one.




Which gives us the ordered pair

Subbing that into the tangent gives us:




Hope that's clear

But doesn't that produce a tangent line with a slope of -2?

[MightyBeh]

But doesn't that produce a tangent line with a slope of -2?

Nope, -2 would be on the other side of the parabola, unless I'm misunderstanding what you're asking. 

Here's a graph (which on paper could also be used to approximate the tangent)

[babushka818]

I heard related rates was going to be taken out, is that true?

[Orb]

I heard related rates was going to be taken out, is that true?

Apparently so.

Can anyone give me a complete summary of the new study design changes for methods 3/4?

So far i know that:
- Modulus functions are being removed
- Matrices are being removed
- Markov chains are being removed from probability

- Statistical Inference is the new topic

am i missing anything?

Matrices haven't been removed, it's still on the new Essentials textbook.

[Photon]

Gotcha covered. Use the derivative of y to find the value of x that has a gradient of two, sub that back into y to find some coordinates and then sub those into your tangent equation. If that didn't make sense, working's in the spoiler.

Spoiler

Derivative (gradient function) of y is:


Because the gradient of the tangent is also the gradient of where it touches the curve, we can solve for dy/dx = 2




However there's three unknowns in the tangent equation so we also need to find a y value. Subbing the x value back into the parabola will give us one.




Which gives us the ordered pair

Subbing that into the tangent gives us:




Hope that's clear

Thank you so much!

I understand you used 'dy/dx (x^2+3x)' to represent the derivative of x^2+36 but shouldn't it be 'd/dx (x^2+3x)' because y = x^2+3x?

[zsteve]

Thank you so much!

I understand you used 'dy/dx (x^2+3x)' to represent the derivative of x^2+36 but shouldn't it be 'd/dx (x^2+3x)' because y = x^2+3x?

You're correct. d/dx[expression] is operator notation so y is subsituted into [expression]

[MightyBeh]

Thank you so much!

I understand you used 'dy/dx (x^2+3x)' to represent the derivative of x^2+36 but shouldn't it be 'd/dx (x^2+3x)' because y = x^2+3x?

Oh yeah you're right there - dy/dx is different to d/dx. Nice catch 

Edit: Damn zsteve always beats me to it

[Photon]

Oh okay 

Find the value of c such that the line with equation y = 2x + c is tangent to the parabola
with equation y = x^2 + 3x. (Answer:  c= -1/4)

But what if y= 2x + c was a secant? Would it be possible to find c?^^^
 

[MightyBeh]

Oh okay 

But what if y= 2x + c was a secant? Would it be possible to find c?^^^

There's probably a better way, but every possible secant with a gradient of two will lie between the corresponding tangent and a line of the same slope that passes through the turning point of the parabola (I think? ).

So the 'answer' would look a bit like this: , where .

Edit: Actually, after thinking about it is probably better. I was thinking of a secant as a way to approximate the gradient but obviously we know the gradient so there's not really much point in restricting c.
_{someone correct me if I'm wrong pls}

[keltingmeith]

Oh okay 

But what if y= 2x + c was a secant? Would it be possible to find c?^^^

Very possible, particularly with the Mean Value Theorem. I wouldn't expect it from VCE, though. Also, what sort of secant? Is it parallel to the tangent? Does it have a higher gradient? Lower gradient?

[brightsky]

There's probably a better way, but every possible secant with a gradient of two will lie between the corresponding tangent and a line of the same slope that passes through the turning point of the parabola (I think? ).

So the 'answer' would look a bit like this: , where .

Edit: Actually, after thinking about it is probably better. I was thinking of a secant as a way to approximate the gradient but obviously we know the gradient so there's not really much point in restricting c.
_{someone correct me if I'm wrong pls}

MightyBeh is pretty much on the money here. A secant is defined simply as a line that cuts a curve at two or more points, which means that any line that has two or more points of intersection with the parabola can be considered a secant.