Rates of Change; Help required!!! (1 Viewer)

NizDiz · Jun 18, 2013

I've tried doing 38 in so many ways but don't seem to get anywhere near the answer. Can some please help me?? You're free to try 39 and 40 Thanks

Makematics · Jun 18, 2013

just cause i'm bored.
$\\\mathbf{(i)}~$V$=\pi\int_{0}^{h}x^2~dy\\\\~~~~~~~=\pi\int_{0}^{h}(20y-y^2)~dy\\\\~~~~~~~=\pi[10y^2-\frac{y^3}{3}]_{0}^{h}\\\\~~~~~~~=\pi(10h^2-\frac{h^3}{3})\\\\~~~~~~~=10\pi h^2-\frac{\pi h^3}{3}\\\\~~~~~~~=\frac{\pi h^2}{3}(30-h)~$cm$^3\\\\$In order to find the surface area,~we need to use the formula$~~A=\pi r^2\\\\$When$~y=h~$,~let$~x=r\\\\r^2+h^2-20h=0\\\\r^2=20h-h^2\\\\S=\pi r^2\\\\~~~=\pi(20h-h^2)\\\\~~~=\pi h(20-h)~$cm$^2$

$\\\mathbf{(ii)(\alpha )}~$We are given that$~\frac{dV}{dt}=50\\\\V=10\pi h^2-\frac{\pi h^3}{3}\\\\\frac{dV}{dh}=20\pi h-\pi h^2\\\\h=5,~\frac{dV}{dh}=100\pi-25\pi\\\\~~~~~~~~~~~~~~~=75\pi\\\\\frac{dh}{dV}= \frac{1}{75\pi}\\\\\frac{dh}{dt}=\frac{dV}{dt} \times \frac{dh}{dV}\\\\~~~~~=50\times\frac{1}{75\pi}\\\\~~~~~=\frac{2}{3\pi}\\\\~~~~~\approx 0.21~$cm$/$sec.$$

$\\\mathbf{(\beta )}~S=20\pi h-\pi h^2\\\\\frac{dS}{dh}=20\pi-2\pi h\\\\h=5,~\frac{dS}{dh}=20\pi-10\pi\\\\~~~~~~~~~~~~~~~=10\pi\\\\$From$~\mathbf{(\alpha )}$,we know that$~\frac{dh}{dt}=\frac{2}{3\pi}\\\\\frac{dS}{dt}=\frac{dS}{dh} \times \frac{dh}{dt}\\\\~~~~~=10\pi\times\frac{2}{3\pi}\\\\~~~~~=\frac{20}{3}\\\\~~~~~\approx 6.7~$cm$^2/$sec.$$

HeroicPandas · Jun 18, 2013

40)

(i) Alternate angles somewhere
Then apply a specific trig ratio that involves x and 200

(ii)Differntiate theta with respect to x (utilising equation found in part (i))
$\frac{d\theta}{dx} = \frac{\frac{-200}{x}}{1+\left ( \frac{200}{x} \right )^2}$

We want TIME, and to get rid of x...

Apply chain rule:

$\frac{d\theta}{dt}.\frac{dt}{dx} = \frac{\frac{-200}{x}}{1+\left ( \frac{200}{x} \right )^2}$

We know that the aeroplane is approaching a speed of 80m/s

Since its apporaching in the negative direction (left) (always think mathematically)

$\frac{dx}{dt} = -80$

Sub in,

$\frac{d\theta}{dt}.\frac{1}{80} = \frac{\frac{-200}{x}}{1+\left ( \frac{200}{x} \right )^2}$

Simplfiy and u will get the answer if u make d(theta)/dt the subject!

(iii)
"Find the rate in which THETA is CHANGING when x = pi/4 "

Simply means, change of theta in terms of time

$\frac{d\theta}{dt} ($ at \theta = \frac{\pi}{4}) = ?$

This is hard to find... and we did part (ii) for a specific reason

$\frac{d\theta}{dt} =\frac{16000}{x^2 +40000}$

All we have to do is sub in theta = pi/4 and we will complete this question

But there is no theta........, there is a DIRTY x.........

Let us try and eliminate it!

So we must find a relationship of THETA and X

*Looks at part (i)*, oh there it is...

$\theta = tan^{-1} \frac{200}{x}$

Make 'x' the subject of this relation, square it, sub into $\frac{d\theta}{dt}$

Then finally sub in theta

OR u can sub in theta as pi/4 into $\theta = tan^{-1} \frac{200}{x}$ and this value of x will correspond to that value of theta (pi/4)

Therefore when u sub this 'x' into ur $\frac{d\theta}{dt}$ , its like subbing in theta = pi/4

Makematics · Jun 18, 2013

looks like you too are bored Heroic

haha i like how you guide him through

HeroicPandas · Jun 18, 2013

Makematics said:
looks like you too are bored Heroic haha i like how you guide him through

lol

Rates of Change; Help required!!! (1 Viewer)

NizDiz

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Makematics

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HeroicPandas

Heroic!

Makematics

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HeroicPandas

Heroic!

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