Author Topic: Differentiation (Read 8897 times) Tweet Share

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

dekoyl · « **Reply #1 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.

d0minicz · « **Reply #2 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...

TrueTears · « **Reply #3 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 $\frav{dV}{dh}$

$\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$

d0minicz · « **Reply #4 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

Flaming_Arrow · « **Reply #5 on:** June 14, 2009, 05:29:14 pm »

differentiate the two fuctions and make them equal each other and solve for x

d0minicz · « **Reply #6 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 =]

Flaming_Arrow · « **Reply #7 on:** June 20, 2009, 04:59:08 pm »

what part do u need help with?

d0minicz · « **Reply #8 on:** June 20, 2009, 05:04:05 pm »

everything cept a)

dcc · « **Reply #9 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.

dekoyl · « **Reply #10 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).

d0minicz · « **Reply #11 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

dcc · « **Reply #12 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.

ilovevce · « **Reply #13 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 $\frac {dm}{dt}$

dcc · « **Reply #14 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 $\frac{dm}{dt}$

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