Author Topic: 4U Maths Question Thread (Read 665377 times) Tweet Share

itssona · « **Reply #1275 on:** June 25, 2017, 03:43:31 pm »

Heey how would I graph
y^2=x^4 (4 + x)

Thank you

RuiAce · « **Reply #1276 on:** June 25, 2017, 03:57:46 pm »

Quote from: MisterNeo on June 25, 2017, 03:56:10 pm

If you're looking for what a graph would look like, you can do it online I recommend Desmos Graphing Calculator

It would look like this...
(Image removed from quote.)

Which really is not obvious at all.

I'll explain where it comes from later; was having a nap. Desmos only shows what it looks like, not where to actully get it which is important for 4U.

jakesilove · « **Reply #1277 on:** June 25, 2017, 04:08:56 pm »

Quote from: MisterNeo on June 25, 2017, 03:56:10 pm

If you're looking for what a graph would look like, you can do it online I recommend Desmos Graphing Calculator

It would look like this...
(Image removed from quote.)

But thanks for the post! Always a good idea to have a graphing calculator on hand

RuiAce · « **Reply #1278 on:** June 25, 2017, 04:16:17 pm »

Quote from: sssona09 on June 25, 2017, 03:43:31 pm

Heey how would I graph
y^2=x^4 (4 + x)

Thank you

$\text{Let }f(x)=x^4(4+x)\\ \text{The graph of }y=f(x)\text{ should be standard. This is just a degree 5 polynomial with intercepts}\\x=-4,0\text{ noting 0 is a quadruple root.}\\ \text{This is a 3U method.}$
$\text{Using the graph of }y=f(x),\text{ sketch }y=\sqrt{f(x)}$
By now you should be familiar with techniques on drawing square root curves, e.g. Block out what's below the x-axis
$\text{Lastly, you should also be aware that going from}y=\sqrt{f(x)}\text{ to }y^2=f(x)\\\text{ is a matter of reflection about the }x\text{-axis.}$

itssona · « **Reply #1279 on:** June 25, 2017, 04:47:06 pm »

Quote from: RuiAce on June 25, 2017, 04:16:17 pm

$\text{Let }f(x)=x^4(4+x)\\ \text{The graph of }y=f(x)\text{ should be standard. This is just a degree 5 polynomial with intercepts}\\x=-4,0\text{ noting 0 is a quadruple root.}\\ \text{This is a 3U method.}$
$\text{Using the graph of }y=f(x),\text{ sketch }y=\sqrt{f(x)}$
By now you should be familiar with techniques on drawing square root curves, e.g. Block out what's below the x-axis
$\text{Lastly, you should also be aware that going from}y=\sqrt{f(x)}\text{ to }y^2=f(x)\text{ is a matter of reflection about the }x\text{-axis.}$

thank you Rui!!

beau77bro · « **Reply #1280 on:** June 25, 2017, 05:43:24 pm »

Hey guys q2, the answer is:
(g x cot(theta))/4(n x Pi)^2

But I'm getting tan? Just wondering if you guys could check it - can't see where I'm getting it wrong

RuiAce · « **Reply #1281 on:** June 25, 2017, 05:59:32 pm »

Quote from: beau77bro on June 25, 2017, 05:43:24 pm

(Image removed from quote.)

Hey guys q2, the answer is:
(g x cot(theta))/4(n x Pi)^2

But I'm getting tan? Just wondering if you guys could check it - can't see where I'm getting it wrong

Check that you didn't misplace your theta

itssona · « **Reply #1282 on:** June 25, 2017, 07:06:09 pm »

Heey alsoo

Find the volume of rotation when the region bounded by the x and y axes, x=2 and the curve y= 1/(x^2 - 4x +13) is rotated by the y axis

I attempted it and got an answer of 4pi/3 arctan 2/3 + piln (9/13) but tbh got no idea

Thank you

bluecookie · « **Reply #1283 on:** June 25, 2017, 07:11:24 pm »

For the circular motion around a circle of x^2+y^2=r^2, with x=rcostheta, y=rsintheta, x*=r(-sintheta)dtheta/dt. y*=rcosthetadtheta/dt. I don't get how the last two equations were derived.

RuiAce · « **Reply #1284 on:** June 25, 2017, 07:11:33 pm »

Quote from: sssona09 on June 25, 2017, 07:06:09 pm

Heey alsoo

Find the volume of rotation when the region bounded by the x and y axes, x=2 and the curve y= 1/(x^2 - 4x +13) is rotated by the y axis

I attempted it and got an answer of 4pi/3 arctan 2/3 + piln (9/13) but tbh got no idea

Thank you

$\text{From the cylindrical shells method, the integral you should've ended up with is}\\ V = 2\pi \int_0^2 \frac{x}{x^2-4x+13}dx$
This is what WolframAlpha tells me.

RuiAce · « **Reply #1285 on:** June 25, 2017, 07:12:50 pm »

Quote from: bluecookie on June 25, 2017, 07:11:24 pm

For the circular motion around a circle of x^2+y^2=r^2, with x=rcostheta, y=rsintheta, x*=r(-sintheta)dtheta/dt. y*=rcosthetadtheta/dt. I don't get how the last two equations were derived.

$x = r\cos \theta\\ \text{and we assume that }\theta\text{ is a function of the time, }t$
$\text{Using implicit differentiation}\\ \begin{align*}\ddot{x} &= \frac{dx}{dt}\\ &= \frac{d}{dt} (r\cos \theta)\\ &= r\frac{d}{dt}\cos \theta\\ &= r \times -\sin \theta \frac{d\theta}{dt}\end{align*}$

bluecookie · « **Reply #1286 on:** June 25, 2017, 07:15:55 pm »

Quote from: RuiAce on June 25, 2017, 07:12:50 pm

$x = r\cos \theta\\ \text{and we assume that }\theta\text{ is a function of the time, }t$
$\text{Using implicit differentiation}\\ \begin{align*}\ddot{x} &= \frac{dx}{dt}\\ &= \frac{d}{dt} (r\cos \theta)\\ &= r\frac{d}{dt}\cos \theta\\ &= r \times -\sin \theta \frac{d\theta}{dt}\end{align*}$

I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?

kiwiberry · « **Reply #1287 on:** June 25, 2017, 07:21:58 pm »

Quote from: bluecookie on June 25, 2017, 07:15:55 pm

I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?

$r \frac{d}{dt} \cos \theta = r \frac{d \theta }{dt} \times \frac{d}{d\theta} \cos \theta= r \times -\sin \theta \frac{d \theta }{dt}$

RuiAce · « **Reply #1288 on:** June 25, 2017, 07:23:24 pm »

Quote from: bluecookie on June 25, 2017, 07:15:55 pm

I don't get the second last step to the last step, why does the cos change to a -sintheta and why does the theta appear at the top of the derivative?

Like I said, implicit differentiation. This is a very standard 4U technique that you're expected to know.

Thanks for stepping it out kiwiberry

bluecookie · « **Reply #1289 on:** June 25, 2017, 07:29:44 pm »

Thanks! Can I've some help with this question?

A particle P is attached by a light string of length l to a fixed point A, and P describes a horizontal circle with uniform angular velocity w around the vertical AC. Prove that if theta is the inclination of the string to the vertical, then h does not depend on l, and find a expression for h in terms of w.

Find the tension in the string when the particle P is rotating at 1 rev/s, the mass of P is 1kg and the length of the string is 35cm. Take pi^2=10.

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