VCE Methods Question Thread!

[Sine]

is related rates still in the methods study design?

nope it isn not longer on the study design but may turn up on school sacs

[michaeljacksonftw]

nope it isn not longer on the study design but may turn up on school sacs

Could it turn up on the exam by any chance?

[Sine]

Could it turn up on the exam by any chance?

it "shouldn't" be on the exam as per the study design cahnges unless in the exam they explain the concept. However related rates is just an application of the chain rule so you never know.

EDIT: sorry I can't give a definitive answer

[michaeljacksonftw]

it "shouldn't" be on the exam as per the study design cahnges unless in the exam they explain the concept. However related rates is just an application of the chain rule so you never know.

EDIT: sorry I can't give a definitive answer

Thanks
Also, since probability is the hardest part of methods, what's the best way to understand it?
How would you sketch (1/4x)+x without a calculator?

[michaeljacksonftw]

Thanks
Also, since probability is the hardest part of methods, what's the best way to understand it?
How would you sketch (1/4x)+x without a calculator?

Bump
Got another question
y = x^2(sin(2pix))
find dy/dx
i first derived sin(2pix) using the chain rule and got cos(2pix)(2pi)
then i derived x^2 to get 2x
to my final answer was 2x (cos(2pix)(2pi)), which was wrong
Can someone please explain what's wrong with my working out and how the corredt answer should be worked out?
Thanks

[RuiAce]

Bump
Got another question
y = x^2(sin(2pix))
find dy/dx
i first derived sin(2pix) using the chain rule and got cos(2pix)(2pi)
then i derived x^2 to get 2x
to my final answer was 2x (cos(2pix)(2pi)), which was wrong
Can someone please explain what's wrong with my working out and how the corredt answer should be worked out?
Thanks

\[ \frac{d}{dx} x^2 \sin 2\pi x = x^2 (2\pi \cos 2\pi x) + (2x) \sin 2\pi x \]

[michaeljacksonftw]

\[ \frac{d}{dx} x^2 \sin 2\pi x = x^2 (2\pi \cos 2\pi x) + (2x) \sin 2\pi x \]

Don't really understand your explanation - could you please explain it another way perhaps?

[darkz]

Don't really understand your explanation - could you please explain it another way perhaps?

Your equation is made up of two functions of x, x^2 and sin(2pix), so you have to use the product rule when differentiating

[michaeljacksonftw]

Your equation is made up of two functions of x, x^2 and sin(2pix), so you have to use the product rule when differentiating

But wouldn't be sin(2pix) be differentiated using the chain rule?

[darkz]

But wouldn't be sin(2pix) be differentiated using the chain rule?

Indeed, however the overall picture is still differentiation using the product rule. The chain rule is only used when differentiating the sin(2pix) component.

Lets think about it this way,
f(x) = x^2 * sin(2pix)
Let x^2 be a(x)
Let sin(2pix) be b(x)

f'(x) = a'(x)*b(x) + a(x)*b'(x)
a'(x) = 2x
b'(x) = 2pi * cos(2pix)

Now just sub in the values of a'(x) & b'(x) into f'(x) and you've got your answer

[RuiAce]

But wouldn't be sin(2pix) be differentiated using the chain rule?

The idea is that this is a question, in which you must use both rules.

In fact, you must use the chain rule inside the product rule, as demonstrated above. You should always be alert for the possibility of requiring two derivative rules at the same time.

[michaeljacksonftw]

I think this question requires both the chain rule and the product rule
y = x^2sin^2(3x)
y = x^2(sin(3x)^2

let u = x^2
let v = (sin(3x)^2

du/dx = 2x
dv/dx
let u = sin(3x)
du/dx = 3cos(3x)
y = u^2
dy/du = 2u
dy/dx = (2u)*(3cos(3x)
= (2)(sin(3x)(3cos(3x)

so dv/dx = (2)(sin(3x)(3cos(3x)
and du/dx = 2x

so (x^2)((2)(sin(3x)(3cos(3x))+((sin(3x)^2)(2x)
Is this correct? If not, would the correct answer be worked out?

[RuiAce]

I think this question requires both the chain rule and the product rule
y = x^2sin^2(3x)
y = x^2(sin(3x)^2

let u = x^2
let v = (sin(3x)^2

du/dx = 2x
dv/dx
let u = sin(3x)
du/dx = 3cos(3x)
y = u^2
dy/du = 2u
dy/dx = (2u)*(3cos(3x)
= (2)(sin(3x)(3cos(3x)

so dv/dx = (2)(sin(3x)(3cos(3x)
and du/dx = 2x

so (x^2)((2)(sin(3x)(3cos(3x))+((sin(3x)^2)(2x)
Is this correct? If not, would the correct answer be worked out?

Looks correct.

[michaeljacksonftw]

Looks correct.

Is my answer all simplified?

[TheBigC]

Bump
Got another question
y = x^2(sin(2pix))
find dy/dx
i first derived sin(2pix) using the chain rule and got cos(2pix)(2pi)
then i derived x^2 to get 2x
to my final answer was 2x (cos(2pix)(2pi)), which was wrong
Can someone please explain what's wrong with my working out and how the corredt answer should be worked out?
Thanks

$$ \mbox{Let y = }x^{2}\sin(\frac{2\pi}{x}) $$
Product rule:
$$ \frac{dy}{dx} = \frac{d}{dx}[x^{2}]\sin(\frac{2\pi}{x}) + x^{2}\frac{d}{dx}[\sin(\frac{2\pi}{x})] $$
Chain rule:
$$  = 2x\sin(\frac{2\pi}{x}) + x^{2}(\cos(\frac{2\pi}{x})\frac{d}{dx}[\frac{2\pi}{x}])  $$
$$  = 2x\sin(\frac{2\pi}{x}) + x^{2}(\cos(\frac{2\pi}{x})\frac{-2\pi}{x^{2}})  $$
$$  = 2x\sin(\frac{2\pi}{x}) + (-2\pi\cos(\frac{2\pi}{x}))  $$

EDIT: Sorry for the delayed post, I presumed page 1101 was the most recent page and that this question hadn't been answered yet.