Cambridge Prelim MX1 Textbook Marathon/Q&A (1 Viewer)

wiliaamnguuyen · Aug 1, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Can anyone do this?
Find the exact gradient, with rational denominator, of the normal to the parabola y^2 = 12x at the point where x=4 in the first quadrant.

wiliaamnguuyen · Aug 1, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I used implicit differentiation and got the answer, but is there any way I can square root both sides and take the positive answer (y=root12x instead of y=-root12 because it states "at the point x=4 in first quadrant".)

wiliaamnguuyen · Aug 1, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

LOL nevermind, I got it thanks

appleibeats · Aug 3, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

8b from 11G

Graph of y = 5x and y = 5

Shaded area is the area within the y axis, y = 5x and y = 5

Find the volume of the solid generated by rotating the region about the x axis ..

I get the answer 25 pi / 3

However the answer in the book say 50 pi / 3

What is the correct answer??

VBN2470 · Aug 3, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Take the difference volumes and you should get $V={\pi}\int_{0}^{1} 5^2-(5x)^2 \text{d}x=\frac{50\pi}{3}$

appleibeats · Aug 3, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Why can't you just take the integral as usual , ie V = pi integral 0 to 1 25x^2 dx

Why do you need to take the differences of volume??

VBN2470 · Aug 3, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Because the region you are rotating is bounded by the y-axis (not x-axis), so you are taking the the difference of volumes when rotating around the x-axis. In this case you are findig the volume of solid of revolution of the rectangle bounded by $[0,1] \times [0,5]$ (which is a cylinder with volume $25{\pi}$ ) and subtracting the volume of the solid of revolution bounded by the x-axis, x = 1 and y = 5x. A diagram will greatly help you visualise what's going on.

InteGrand · Aug 3, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

We can also do it without using integration by subtracting the volume of a cone (using the cone volume formula) from the cylinder volume.

appleibeats · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Also similarly for 8c

ii) volume of the solid generated by rotating region about the y axis

region is the area between x axis, x = y^2 and x = 4

My working:

V = pi integral from 0 to 2 8 - y^4 dy

then integrate normally.

What is wrong with my working??

InteGrand · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
Also similarly for 8c

ii) volume of the solid generated by rotating region about the y axis

region is the area between x axis, x = y^2 and x = 4

My working:

V = pi integral from 0 to 2 8 - y^4 dy

then integrate normally.

What is wrong with my working??

What is the reason for your working?

appleibeats · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I am trying to find the volume. Not sure where I have gone wrong. The answer is 128 pi / 5 u^3 and I am getting way off.

InteGrand · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
I am trying to find the volume. Not sure where I have gone wrong. The answer is 128 pi / 5 u^3 and I am getting way off.

What's the reasoning for your working, as in, why do you think you've done it right in the first place?

appleibeats · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

I realised my mistake. Really stupid of me. I finally got the answer.

appleibeats · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

c)i ) A sphere of radius r is generated by rotating the semicircle y = (r^2 - x^2)^ 1/2 about the x asix. Show that the volume of the sphere is given by 4/3 pi r^3

I can do this first part

Not sure about part ii)

ii) A spherical cap of radium r and heigh h is formed by rotating the semicircle y = (r^2 - x^2)^1/2 between x = r and x = r - h about the x - axis . Show that the volume of the cap is geven by 1/3 pi h^2( 3r - h)

I am quite certain it has something to do with part i) but am not sure how to use part i) .

integral95 · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
c)i ) A sphere of radius r is generated by rotating the semicircle y = (r^2 - x^2)^ 1/2 about the x asix. Show that the volume of the sphere is given by 4/3 pi r^3

I can do this first part

Not sure about part ii)

ii) A spherical cap of radium r and heigh h is formed by rotating the semicircle y = (r^2 - x^2)^1/2 between x = r and x = r - h about the x - axis . Show that the volume of the cap is geven by 1/3 pi h^2( 3r - h)

I am quite certain it has something to do with part i) but am not sure how to use part i) .

doesn't need part i)

rand_althor · Aug 4, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Just in case you were still unsure about part ii):

$\begin{align*}y&=\sqrt{r^2-x^2} \to y^2=r^2-x^2 \\V&=\pi\int^r_{r-h}r^2-x^2\,\mathrm{d}x \\&=\pi \left [r^2x-\frac{x^3}{3} \right ]^r_{r-h} \\&=\pi \left [\left (r^3-\frac{r^3}{3} \right ) -\left (r^2(r-h)-\frac{(r-h)^3}{3} \right ) \right ] \\&=\pi \left [ \frac{2r^3}{3} -\frac{1}{3}\left (3r^3-3r^2h -(r^3-3r^2h+3rh^2-h^3) \right ) \right ] \\&=\frac{\pi}{3} \left [ 2r^3 -(3r^3-3r^2h -r^3+3r^2h-3rh^2+h^3)\right ] \\&=\frac{\pi}{3} \left [2r^3 - (2r^3-3rh^2+h^3) \right ] \\&=\frac{\pi}{3} \left [2r^3-2r^3 +3rh^2-h^3 \right ] \\&=\frac{\pi}{3}\left [h^2(3r-h) \right ] \textup{units}^3\end{align*}$

appleibeats · Aug 5, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Don't know how to sketch y = x^1/2 (1 - x)

Indicating stationary points and intercepts with the axes.

And Hence sketch a graph of the function y^2 = x(1 - x)^2

laters · Aug 6, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

appleibeats said:
sketch y = x^1/2 (1 - x)

And Hence sketch a graph of the function y^2 = x(1 - x)^2

The domain of $y=\sqrt{x} (1-x)$ is $x\leq 0$
The intercepts are at x=0 and x=1
Expanding and then differentiating, the derivative function is $y'=\frac{1}{2\sqrt{x}}-\frac{3}{2}\sqrt{x}$
(And kind of unnecessary here, but $y''=\frac{-1}{4x\sqrt{x}} -\frac{3}{4\sqrt{x}}$ )

The stationary points are found by letting y'=0 as this is when the tangent to the curve is horizontal. Solving y'=0 you get x= 1/3 (and so $y=\frac{2}{3\sqrt{3}}$

Using the first derivative test you will find that to the left of x=1/3 the derivative is positive (so increasing curve) and to the right the derivative is negative (so decreasing curve), meaning you have a maximum turning point. (Also, as this is the only turning point and there are no asymptotes etc. you know it is also the 'global' maximum). In addition, as y' is undefined at x=0, a point at which y exists, we have a vertical tangent at x=0.

Letting y''=0 for possible inflexion points we get x=-1/3 but that is not in our domain so it can be disregarded. We therefore can say the curve is concave down because we already know we have a maximum turning point and that the curve cannot change in concavity.

We can then draw the curve.

Also to draw the y^2=... function we can square root both sides. that brings us to the $\pm$ version of the original curve. So we simply draw the curve we just drew and then draw the 'flip' of it around the x-axis... but we KEEP BOTH CURVES.

appleibeats · Aug 7, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

Consider the function x^2 = 8y . Tangents are drawn at the poitns A (4,2) and B (-4,2) and intersect on the y - axis. Find teh area bounded by the curve and the tangents.

rand_althor · Aug 7, 2015

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread

$$To find the area calculate double the blue area on the graph. The tangent we consider for the blue area$: y=x-2$

$\begin{align*}2\int^4_0 \frac{x^2}{8}-(x-2) \mathrm{d}x &=2 \left [\frac{x^3}{24}-\frac{x^2}{2}+2x \right ]^4_0 \\&=2 \left[\frac{4^3}{24}-\frac{4^2}{2}+2(4) \right ] \\&=2\times\frac{8}{3} \\&=\frac{16}{3} \textup{ units}^2\end{align*}$

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