Author Topic: VCE Methods Question Thread! (Read 6156760 times) Tweet Share

StupidProdigy · « **Reply #10080 on:** May 04, 2015, 07:54:50 pm »

Quote from: Shinkaze on May 04, 2015, 07:12:32 pm

Find two quadratic functions f and g such that f(1)=0,g(1)=0 and f(0)=10, g(0) = 10 and both have a maximum value of 18

Can anyone help me? Thank you so much ^_^

This is just a simultaneous equations question, easily done with cas but a fair bit more trivial by hand. write each equation in the form y=a*(x-b)+c so you'll have g(x)=a*(x-b)+c and f(x)=a*(x-b)+c. However we already know one of the unknown variables because of the 'maximum value info'. However I don't know if that value of 18 is for b or c since I don't know if its the x or y coordinate. Hope that gets you started

Callum@1373 · « **Reply #10081 on:** May 04, 2015, 07:58:50 pm »

so it's f(x) = a(x-h)^2 + 18

sub in the x values

a(1-h)^2 + 18 = 0

ah^2 + 18 = 10

use cas to solve these

TheMereCat · « **Reply #10082 on:** May 04, 2015, 08:47:57 pm »

Yeah, some questions are much more convenient by CAS, although it is not too difficult by hand when you grasp the concept.

Shinkaze · « **Reply #10083 on:** May 04, 2015, 08:49:52 pm »

Quote from: TheMereCat on May 04, 2015, 08:47:57 pm

Yeah, some questions are much more convenient by CAS, although it is not too difficult by hand when you grasp the concept.

Hmm if you don't mind can you help me? I really have no idea how to do it without a CAS ahahah >.<

Redoxify · « **Reply #10084 on:** May 04, 2015, 08:56:03 pm »

can someone show me how to get the derivative of
20e^-0.02t x (e^0.62t-1)

cosine · « **Reply #10085 on:** May 04, 2015, 09:08:22 pm »

Quote from: Redoxify on May 04, 2015, 08:56:03 pm

can someone show me how to get the derivative of
20e^-0.02t x (e^0.62t-1)

$20e^{-0.02t} * (e^{0.62t}-1)$

With the use of the product rule, because we have the product of two functions:

$\frac{dy}{dx}=\frac{udv}{dt}+\frac{vdu}{dt}$

$Let u = 20e^{-0.02t}$

$Let v = e^{0.62t}-1$

$\therefore y=uv$

$\frac{du}{dt}= -0.02*20e^{-0.02t} = -0.4e^{-0.02t}$

$\frac{dv}{dt}= 0.62e^{0.62t}$

$\therefore \frac{dy}{dt} = 20e^{-0.02t}*0.62e^{0.62t} + (e^{0.62t}-1)*(-0.4e^{-0.02t})$

Now clean it up:

$12.4e^{0.6t} - 0.4e^{0.6t} + 0.4e^{-0.02t}$

StupidProdigy · « **Reply #10086 on:** May 04, 2015, 09:17:18 pm »

Quote from: Shinkaze on May 04, 2015, 08:49:52 pm

Hmm if you don't mind can you help me? I really have no idea how to do it without a CAS ahahah >.<

define g(x) as g(x)=a*(x-h)^2+18
define f(x) as f(x)=a*(x-h)^2+18
go menu, 3 (algebra), 7 (simul eqs)
it will ask for how many equations, make it 4 since that's what you have, and call the unknowns whatever (a and h in this case)
It will come it with a template, type in f(1)=0 and so forth for the rest, then hit enter. As for casio...I have no idea

keltingmeith · « **Reply #10087 on:** May 04, 2015, 09:32:09 pm »

Quote from: Redoxify on May 04, 2015, 08:56:03 pm

can someone show me how to get the derivative of
20e^-0.02t x (e^0.62t-1)

An arguably simpler method, expand the brackets:

$20e^{-0.02t}\left(e^{0.62t}-1\right)=20e^{-0.02t}e^{0.62t}-20e^{-0.02t} \\ =20e^{0.62t-0.02t}-20e^{-0.02t}=20e^{0.6t}-20e^{-0.02t} \\ \implies \frac{d}{dx}\left[20e^{-0.02t}\left(e^{0.62t}-1\right)\right] \\ =\frac{d}{dx}\left[20e^{0.6t}-20e^{-0.02t}\right] \\ =20(0.6)e^{0.6t}-20(-0.02)e^{-0.02t} \\ =12e^{0.6t}+0.4e^{-0.02t}$

which, of course, matches cosine's answer.

keltingmeith · « **Reply #10088 on:** May 04, 2015, 09:45:27 pm »

Quote from: Callum@1373 on May 04, 2015, 07:58:50 pm

so it's f(x) = a(x-h)^2 + 18

sub in the x values

a(1-h)^2 + 18 = 0

ah^2 + 18 = 10

use cas to solve these

These can be solved by hand - worth knowing how to do, in case you're asked this on a calc-free paper:

$a(1-h)^2+18=0\implies a=\frac{-18}{(1-h)^2} \\ ah^2+18=10\implies a=\frac{-8}{h^2}* \\ \implies \frac{-18}{(1-h)^2}=\frac{-8}{h^2} \\ 18h^2=8(1-h)^2 \\ 9h^2=4(1-2h+h^2) \\ 9h^2-4h^2+8h-4=0 \\ 5h^2+8h-4=0 \\ (5h-2)(h+2)=0 \\ \implies h=-2,\frac{2}{5}$

Subbing both of these into *:

$a=\frac{-8}{(-2)^2}=\frac{-8}{4}=-2 \\ a=\frac{-18}{(2/5)^2}=-18/\frac{4}{25}=\frac{-18\times 25}{4}=-\frac{225}{2}$

So, your two functions should be:

$f_1(x)=-2(x+2)^2+18 \\ f_2(x)=\frac{-225}{2}\left(x-\frac{2}{5}\right)^2+18$

qwerty101 · « **Reply #10089 on:** May 05, 2015, 06:22:19 pm »

doing some rev questions.

for "when simplified" of log4(32). i get log10(32)/log10(4) = log10(32/4) = log10(

, this gets me an evaluation of 0.903, which differs from the answer of 2.5 (5/2), how did they get this?

i see that 32 = 4x
2^5 = 2^2x, where x = 5/2, but how can i do so with change of base? that when i put it in my calc will get me 5/2
thanks

dankfrank420 · « **Reply #10090 on:** May 05, 2015, 06:23:25 pm »

(x^2 + 1)(2x-4)^1/2

I can do it until I get stuck when I have to simplify it.

Edit: Forgot to say differentiate it lol

Second edit: nvm got it

kinslayer · « **Reply #10091 on:** May 05, 2015, 06:34:07 pm »

Quote from: qwerty101 on May 05, 2015, 06:22:19 pm

doing some rev questions.

for "when simplified" of log4(32). i get log10(32)/log10(4) = log10(32/4) = log10(, this gets me an evaluation of 0.903, which differs from the answer of 2.5 (5/2), how did they get this?

i see that 32 = 4x
2^5 = 2^2x, where x = 5/2, but how can i do so with change of base? that when i put it in my calc will get me 5/2
thanks

32 = 2^5 = (4^(1/2))^5 = 4^(5/2).

So $\log_4 (32) = \log_4 4^{\frac 5 2} = \frac 52 \log_4 4 = \frac 52$ .

edit: for the change of base you can do $\frac {\log_{10}(32)}{\log_{10} 4}$

RKTR · « **Reply #10092 on:** May 05, 2015, 06:57:14 pm »

Quote from: qwerty101 on May 05, 2015, 06:22:19 pm

doing some rev questions.

for "when simplified" of log4(32). i get log10(32)/log10(4) = log10(32/4) = log10(, this gets me an evaluation of 0.903, which differs from the answer of 2.5 (5/2), how did they get this?

i see that 32 = 4x
2^5 = 2^2x, where x = 5/2, but how can i do so with change of base? that when i put it in my calc will get me 5/2
thanks

Log(32)/log(4) does not equal log(32/4)

TheAspiringDoc · « **Reply #10093 on:** May 05, 2015, 07:55:10 pm »

Quote from: RKTR on May 05, 2015, 06:57:14 pm

Log(32)/log(4) does not equal log(32/4)

No, quite right, log(32) - log(4) = log(32/4) so what is log (a) / log (b)?

keltingmeith · « **Reply #10094 on:** May 05, 2015, 08:02:08 pm »

Quote from: TheAspiringDoc on May 05, 2015, 07:55:10 pm

No, quite right, log(32) - log(4) = log(32/4) so what is log (a) / log (b)?

It's log(a)/log(b). You could simplify it by a change of base rule, like so:

$\frac{\log_c(a)}{\log_c(b)}=\frac{\log_b(a)}{\log_b(b)}=\log_b(a)$

but this isn't something that you're ever expected to do unless it's mentioned specifically.

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