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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: d0minicz on June 14, 2009, 03:21:36 pm

Title: Differentiation
Post by: d0minicz on June 14, 2009, 03:21:36 pm: the function y=ax^2 -bx has a zero gradient for x=2 only. The x-axis intercepts of the graph of this function are ...
thanks
Title: Re: Differentiation
Post by: dekoyl on June 14, 2009, 03:27:25 pm: $\frac{dy}{dx}|_{x=2} = 4a-b = 0$
$\implies 4a=b$
$\therefore y = ax^{2}-bx$
$=x(ax-b)$
$=x(ax-4a)$
$=ax(x-4)$
At x-ints, y = 0
$0=ax(x-4)$
$\therefore (4,0), (0,0)$

I think.
Title: Re: Differentiation
Post by: d0minicz on June 14, 2009, 04:06:20 pm: Water is draining from a cone-shaped funnel at a rate of 500cm^3/min. The cone has a base radius of 20cm and a height of 100cm. Let h cm be the depth of water in the funnel at time t minutes. The rate of decrease of h in cm/min is...
thanks...
Title: Re: Differentiation
Post by: TrueTears on June 14, 2009, 04:48:51 pm: $\frac{dV}{dt} = 500$

$\frac{dh}{dt} = ?$

We also know the volume of a cone with height h is $V = \frac{1}{3} \pi r^2h$

From this you can work out $\frav{dV}{dh}$

$\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$
Title: Re: Differentiation
Post by: d0minicz on June 14, 2009, 05:26:29 pm: For what value(s) of x do the graphs of $y= x^3$ and $y= x^3 +x^2 +x -2$ have the same gradient?
thanks
Title: Re: Differentiation
Post by: Flaming_Arrow on June 14, 2009, 05:29:14 pm: differentiate the two fuctions and make them equal each other and solve for x
Title: Re: Differentiation
Post by: d0minicz on June 20, 2009, 04:49:09 pm: the mass m kg of radioactive lead remaining in a sample t hours after observations began is given by $m= 2e^{-0.2t}$
a) Find the mass left after 12 hours.
b) Find how long it takes to fall to half of its value at t=0
c) find out how long it takes for the mass to fall to
i) one quarter
ii) one eighth
of its value at t=0.
d) express the rate of decay as a function of m
thanks =]
Title: Re: Differentiation
Post by: Flaming_Arrow on June 20, 2009, 04:59:08 pm: what part do u need help with?
Title: Re: Differentiation
Post by: d0minicz on June 20, 2009, 05:04:05 pm: everything cept a)
Title: Re: Differentiation
Post by: dcc on June 20, 2009, 05:15:57 pm: Can you tell us what you've tried doing for each part?

Much more satisfying then us just giving you answers.
Title: Re: Differentiation
Post by: dekoyl on June 20, 2009, 05:23:14 pm: Quote from: d0minicz on June 20, 2009, 04:49:09 pm
the mass m kg of radioactive lead remaining in a sample t hours after observations began is given by $m= 2e^{-0.2t}$
a) Find the mass left after 12 hours.
b) Find how long it takes to fall to half of its value at t=0
c) find out how long it takes for the mass to fall to
i) one quarter
ii) one eighth
of its value at t=0.
d) express the rate of decay as a function of m
thanks =]
a) let t= 12.
b) find value at t=0. put that as m. solve for t.
c) do the same as part b except divide it by 4 or 8. solve for t.
d) make t the subject... after finding the rate.

Edit: Thanks to dcc. Changed my (d).
Title: Re: Differentiation
Post by: d0minicz on June 20, 2009, 05:31:03 pm: thanks for that dekoyl
i need help with d) now dunno how they get their answer
ty verymuch
Title: Re: Differentiation
Post by: dcc on June 20, 2009, 05:33:38 pm: Quote from: d0minicz on June 20, 2009, 05:31:03 pm
thanks for that dekoyl
i need help with d) now dunno how they get their answer
ty verymuch

First of you, you have to analyse what they want. The rate of decay of the radioactive lead? They are asking for some rate, and we all know what this requires.

Then, they ask for you to express it as a function of m. So, once you've found this decaying rate, can you then express it using only m (and some constants)? Look to see if there are any similarities between your m function and your rate function, hopefully it'll be obvious then.
Title: Re: Differentiation
Post by: ilovevce on June 20, 2009, 06:33:13 pm: Another way to go about part (d) is:

1. Transpose the equation to make t the subject

2. Find $\frac {dt}{dm}$

3. Flip to give $\frac {dm}{dt}$
Title: Re: Differentiation
Post by: dcc on June 20, 2009, 06:43:51 pm: Quote from: ilovevce on June 20, 2009, 06:33:13 pm
Another way to go about part (d) is:

1. Transpose the equation to make t the subject

2. Find $\frac {dt}{dm}$

3. Flip to give $\frac {dm}{dt}$

That's making it unnecessarily difficult, it'd be much easier just to do it by looking at $\frac{dm}{dt}$ :P
Title: Re: Differentiation
Post by: ilovevce on June 20, 2009, 07:04:06 pm: Quote from: dcc on June 20, 2009, 06:43:51 pm
Quote from: ilovevce on June 20, 2009, 06:33:13 pm
Another way to go about part (d) is:

1. Transpose the equation to make t the subject

2. Find $\frac {dt}{dm}$

3. Flip to give $\frac {dm}{dt}$

That's making it unnecessarily difficult, it'd be much easier just to do it by looking at $\frac{dm}{dt}$ :P

Well, for this particular question, it might be easier to do it by recognition, but not in every situation. I was just giving the general rule. :P
Title: Re: Differentiation
Post by: d0minicz on June 22, 2009, 09:20:56 pm: If h=6, $V= 6\pi r^2 + \frac {2}{3} \pi r^3$ . For r=4:
i)Show that a small increase of p cm in the radius results in an increase of $80\pi p$ in the volume. Done
ii) Show that a small increase of q% in the radius will cause an approximate increase of 2.3%q in the volume. Need help
need to see workings not using Liebniz notation, thanks =]
Title: Re: Differentiation
Post by: d0minicz on June 22, 2009, 09:52:23 pm: Also, how do you do these q's:
A right circular cylinder is placed inside a sphere of radius 5
cm. Find the largest possible volume of the cylinder.
thanks
Title: Re: Differentiation
Post by: kamil9876 on June 22, 2009, 10:11:30 pm: first post:

ii.) $<br />p=r * \frac{q}{100}$
$=\frac{qr}{100}$

$\Delta V=\frac{4\pi qr}{5}$ (are you satisfied with that avoidance of Liebniz notation?)

Now the percentage increase of Volume is:

$\frac{\Delta V}{V(4)}*100$
Title: Re: Differentiation
Post by: ilovevce on June 22, 2009, 10:24:53 pm: Quote from: d0minicz on June 22, 2009, 09:52:23 pm
Also, how do you do these q's:
A right circular cylinder is placed inside a sphere of radius 5
cm. Find the largest possible volume of the cylinder.
thanks

Find the equation for, and graph, a circle with radius 5 centred on the origin.
Now, choose any point (x,y). The x-coordinate will be the radius of the flat surfaces of your cylinder, which will have area $A = \pi r^2$ . The y-coordinate will be half the height of you cylinder.

Now construct a formula for the volume of your cylinder in terms of x only. You can differentiate this to find the maximum value.
Title: Re: Differentiation
Post by: d0minicz on June 23, 2009, 08:18:45 pm: The tangent to the curve with equation y=tan2x at the point where $x= \frac {\pi}{8}$ meets the y=axis at point A. Find the distance OA where O is the origin.
need to check answer, thanks.
Title: Re: Differentiation
Post by: TrueTears on June 23, 2009, 08:23:27 pm: $\frac{dy}{dx} = 2sec^2(2x)$

when $x = \frac{\pi}{8}$

$\frac{dy}{dx} = 2sec^2(2(\frac{\pi}{8})) = 4$

When $x = \frac{\pi}{8}$ $y = 1$

$y - y1 = m(x-x1)$

$y - 1 = 4(x-\frac{\pi}{8})$

The distance OA is just the $|$ y intercept $|$ .
Title: Re: Differentiation
Post by: d0minicz on June 23, 2009, 08:31:05 pm: mmm the book got $\frac {\pi}{2} -1$ but i got the same as you...
thanks

Edit; dw got it now lol
Title: Re: Differentiation
Post by: julianpeiriez on June 24, 2009, 06:12:34 pm: methods cas 3 and 4 cambridge - exercise 11 C
q17

The volume V cm3, of water in a dish when the depth is "h" cm is given by the rule V = pi/2(e^2h -1). the depth of the dish is 2.5cm. If water is being poured in at 5cm^3/s, find:

a) The rate at which the depth of the water is increasing when the depth is 2cm?
plz reply asap thamx :)
Title: Re: Differentiation
Post by: d0minicz on June 24, 2009, 06:44:55 pm: well you find $\frac {dV}{dh}$ from the formula they give you
and we have $\frac {dV}{dt} = 5 cm/s$ and we want to find $\frac {dh}{dt}$
so, $\frac {dh}{dt} = \frac {dh}{dV}. \frac {dV}{dt}$
Title: Re: Differentiation
Post by: d0minicz on June 24, 2009, 07:31:37 pm: An aeroplane is flying horizontally at a constant height of 1000m. At a certain instant the angle of elevation is $30^o$ and decreasing and the speed of the aeroplane is 480 km/h.
a) How fast is $\theta$ decreasing at this instant? Answer is degrees/s.
b) How fast is the distance between the aeroplane and the observation point changing at this instant?
thanks :)
Title: Re: Differentiation
Post by: Mao on June 24, 2009, 08:56:45 pm: let x be the horizontal distance of the aeroplane from observer.

construct a triangle, through geometry, you can see that angle of elevation is opposite to vertical distance and adjacent to horizontal distance

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

Using related rates, $\frac{d\theta}{dt} = \frac{d\theta}{dx} \cdot \frac{dx}{dt}$ , where $\frac{dx}{dt} = 480 km/hr = \frac{400}{3} m/s$

Hence, $\frac{d\theta}{dt} \approx -\frac{1}{30} radians \approx -1.91^o s^{-1}$

let y be the hypotenuse of the above triangle, this is its distance from the observer.

then by trig ratio, $y = \frac{1000}{\sin \theta}\implies \frac{dy}{d\theta} = -\frac{1000 \cos \theta}{\sin^2 \theta}$ , when $\theta = 30^o, \; \frac{dy}{d\theta} = -2000\sqrt{3}$ (be aware that $\theta$ here is in radians)

using related rates, $\frac{dy}{dt} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dt} = -2000\sqrt{3} \cdot -\frac{1}{30} = \frac{200\sqrt{3}}{3} \approx 115.5 ms^{-1}$

that was an extremely tough question at methods level =\
Title: Re: Differentiation
Post by: Mao on June 24, 2009, 08:58:57 pm: Also, note to people in reply to julianpeiriez's misplacement of posts, whilst it is good to point new member to the right place, I do expect it to be done with respect and good intention. Antagonizing new members is not acceptable. If you feel it is misplaced or thread-hijacking, report the post to moderators so it can be deleted/split.
Title: Re: Differentiation
Post by: ilovevce on June 26, 2009, 12:57:50 pm: Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.
Title: Re: Differentiation
Post by: kamil9876 on June 26, 2009, 01:29:11 pm: Quote from: ilovevce on June 26, 2009, 12:57:50 pm
Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

I think chain rule would be more efficient :P
Title: Re: Differentiation
Post by: ilovevce on June 26, 2009, 04:35:59 pm: Quote from: kamil9876 on June 26, 2009, 01:29:11 pm
Quote from: ilovevce on June 26, 2009, 12:57:50 pm
Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

I think chain rule would be more efficient :P

This is just to find the term $\frac {d\theta}{dx}$ which is needed to apply the chain rule. :P
Title: Re: Differentiation
Post by: Mao on June 27, 2009, 02:01:06 am: Quote from: ilovevce on June 26, 2009, 12:57:50 pm
Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

you expect a methods student to differentiate arctan? :P
Title: Re: Differentiation
Post by: TrueTears on June 27, 2009, 02:02:25 am: Quote from: Mao on June 27, 2009, 02:01:06 am
Quote from: ilovevce on June 26, 2009, 12:57:50 pm
Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

you expect a methods student to differentiate arctan? :P
d0minicz is quite pr0 at spesh :P
Title: Re: Differentiation
Post by: d0minicz on June 27, 2009, 02:05:33 am: quite shit :P
thx for help guys
Title: Re: Differentiation
Post by: ilovevce on June 29, 2009, 07:23:34 pm: Quote from: Mao on June 27, 2009, 02:01:06 am
Quote from: ilovevce on June 26, 2009, 12:57:50 pm
Quote from: Mao on June 24, 2009, 08:56:45 pm

$\tan \theta = \frac{1000}{x} \implies \theta = \tan^{-1} \frac{1000}{x}$ , plot this on the graphics calculator

Using trig knowledge, when $\theta = 30^o$ , $x=1000\sqrt{3}$ , use the graphics calculator's derivative function to find the gradient at this point, this will give you $\frac{d\theta}{dx}$

You can also work out $\frac{d \theta}{dx}$ without a calculator. Make $x = \frac{1000}{\tan \theta}$ and use the quotient rule to find $\frac {dx}{d \theta}$ and then flip that to get $\frac{d \theta}{dx}$ .The numbers are not too large to handle.

you expect a methods student to differentiate arctan? :P

No need to! :P $\frac {d}{d\theta}(\frac {1000}{\tan \theta}) = \frac {-1000 \sec^2 \theta}{\tan^2 \theta} = \frac {-1000}{\sin^2 \theta} \therefore \frac {d\theta}{dx} = \frac{-sin^2\theta}{1000}$
Title: Re: Differentiation
Post by: d0minicz on July 02, 2009, 10:02:48 pm: Just a question i dont have answers to, so im curious.
And this isnt an assignment or SAC im tryna cheat by the way.

During a Melbourne summer day, the temperature after 5am can be modelled by the function $T= -6cos(\frac {\pi t}{12}) + 28 , 0\le t \ge 19$ where t is the time in hours after 5am and T is the temperature in degrees Celsius. Using this model:
a) find the maximum temperature and the time of day when this will occur
b) calculate the temperature at 7am, correct to the nearest degree
c) calculate, to the nearest minute, the length of time that the temperature remains over 30 degrees celcius
d) find $\frac {dT}{dt}$ , hence find the exact rate of change of temperature at 9am
e) find the greatest rate of increase in temperature and the time of day this occurs.

thanks.
Title: Re: Differentiation
Post by: d0minicz on July 02, 2009, 10:57:40 pm: ok can someone help me with part e) please. jjust need to be pointed in the right direction =)
thanks !
Title: Re: Differentiation
Post by: ilovevce on July 02, 2009, 11:25:25 pm: I think it must mean 'find the greatest rate of increase in temperature after 9am'. The question doesn't make sense otherwise.

To find this, you simply need to find the maximum value of the derivative function, the same way you would find the maximum of any other function.
Title: Re: Differentiation
Post by: d0minicz on July 02, 2009, 11:45:11 pm: sorry fixed it
wtf am i tripping on =]...
Title: Re: Differentiation
Post by: ilovevce on July 03, 2009, 12:05:02 am: So you get $\frac{dT}{dt} = \frac{\pi}{2}\sin(\frac{\pi t}{12})$

To find when this reaches a maximum, you have to differentiate again (find the second derivative):

$\frac{d^2T}{dt^2} = \frac{\pi^2}{24}\cos(\frac{\pi t}{12}) = 0$

$\implies \frac {\pi t}{12} = \frac{\pi}{2}, \frac{3\pi}{2}$

$\implies t = 6, t = 18$

t = 6 is a maximum stationary point, t = 18 is a minimum stationary point (can be confirmed by looking at the graph).
Title: Re: Differentiation
Post by: d0minicz on July 03, 2009, 12:08:52 am: ohh okay thanks
are 2nd derivatives required for methods aswell?
Title: Re: Differentiation
Post by: ilovevce on July 03, 2009, 12:12:11 am: Quote from: d0minicz on July 03, 2009, 12:08:52 am
ohh okay thanks
are 2nd derivatives required for methods aswell?

Technically they're not, you won't get any questions on a VCAA exam on them. However, I don't see why you shouldn't learn them. After all, it's not a new concept - just a very easy and logical extension on what you already know about derivatives.