Author Topic: Differentiation (Read 8919 times) Tweet Share

ilovevce · « **Reply #30 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

This is just to find the term $\frac {d\theta}{dx}$ which is needed to apply the chain rule.

Mao · « **Reply #31 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?

TrueTears · « **Reply #32 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?

d0minicz is quite pr0 at spesh

d0minicz · « **Reply #33 on:** June 27, 2009, 02:05:33 am »

quite shit

thx for help guys

ilovevce · « **Reply #34 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?

No need to!

$\frac {d}{d\theta}(\frac {1000}{\tan \theta}) = \frac {-1000 \sec^2 \theta}{\tan^2 \theta} = \frac {-1000}{\sin^2 \theta} \therefore \frac {d\theta}{dx} = \frac{-sin^2\theta}{1000}$

d0minicz · « **Reply #35 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.

d0minicz · « **Reply #36 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 !

ilovevce · « **Reply #37 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.

d0minicz · « **Reply #38 on:** July 02, 2009, 11:45:11 pm »

sorry fixed it
wtf am i tripping on =]...

ilovevce · « **Reply #39 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).

d0minicz · « **Reply #40 on:** July 03, 2009, 12:08:52 am »

ohh okay thanks
are 2nd derivatives required for methods aswell?

ilovevce · « **Reply #41 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.

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