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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: xD_aQt on April 09, 2011, 10:25:06 pm

Title: Maths Help
Post by: xD_aQt on April 09, 2011, 10:25:06 pm: Hi,

I've got two questions that I'm unsure of.
1) If cosh(x)=2.5, what is sinh(x) exactly?
2) If an angle is measured to be 0.4 radians, what is its sine?

Any help would be appreciated.
Thanks :)
Title: Re: Maths Help
Post by: Gloamglozer on April 09, 2011, 10:33:14 pm: With part a, I think you can use the trig identity:

$cosh^{2}(x) - sinh^{2}(x) = 1$

Though I'm not sure.
Title: Re: Maths Help
Post by: xD_aQt on April 10, 2011, 10:42:42 am: I'm not sure about that formula either, would it work for that particular question?
Title: Re: Maths Help
Post by: m@tty on April 10, 2011, 11:09:40 am: Yes, you can use the formula. Also for 2) you have an angle of 0.4^c, to find the sine just evaluate sin(0.4)
Title: Re: Maths Help
Post by: xD_aQt on April 10, 2011, 11:31:12 am: Oh okay, thanks guys!
Title: Re: Maths Help
Post by: xD_aQt on April 10, 2011, 03:22:30 pm: What's the difference between sinh(x), cosh(x), tanh(x) and sin(x), cos(x), tan(x) ?
Title: Re: Maths Help
Post by: m@tty on April 10, 2011, 03:24:31 pm: They're the hyperbolic functions - http://en.wikipedia.org/wiki/Hyperbolic_function

They're based off hyperbolae rather than a circle, as with sine, cosine and tangent
Title: Re: Maths Help
Post by: taiga on April 10, 2011, 04:01:47 pm: You can use trig identities to turn them into hyperbolic identities, generally involves swapping a minus sign somewhere :P

The rule for it is here:
http://mathworld.wolfram.com/OsbornesRule.html

There are some equations (that wiki page has them) that relate the two types of functions together.

like sinh(iz) = i sinz
Title: Re: Maths Help
Post by: xD_aQt on April 14, 2011, 08:40:31 pm: need help on a questions :P

find dx/dt when x = (5+2t)^6t
I got 12t(5+2t)^(6t-1) but the calculator says (6ln(2x+5)+(12x/2x+5)).(2x+5)^6t

could someone please explain?
Title: Re: Maths Help
Post by: moekamo on April 14, 2011, 09:27:53 pm: if you had x^x, to differentiate you would have to log both sides and do it implicitly, here you do the same thing since you cant differentiate with respect t t since it is both a power and in the bracket:

$x=(5+2t)^{6t} \implies \ln x = 6t \ln (5+2t)$
$\implies \frac{d}{dt} \ln x = \frac{d}{dt} 6t \ln (5+2t)$
$\implies \frac{1}{x} \frac{dx}{dt} = \left ( \frac{12t}{5+2t}\right ) + 6 \ln (5+2t)$
$\implies \frac{dx}{dt} = x \left ( \left ( \frac{12t}{5+2t}\right ) + 6 \ln (5+2t)\right ) = (5+2t)^{6t} \left ( \left ( \frac{12t}{5+2t}\right ) + 6 \ln (5+2t)\right )$
Title: Re: Maths Help
Post by: xD_aQt on April 14, 2011, 10:01:56 pm: I see, cheers mate! :)
Title: Re: Maths Help
Post by: xD_aQt on April 20, 2011, 09:32:40 pm: I've got a differentiation question that I'm unsure of as I'm not sure how to solve it.
The question asks to use implicit differentiation to find dy/dx of 3(x^2)y+4x(y^2)=5x-7
Title: Re: Maths Help
Post by: brightsky on April 20, 2011, 09:35:29 pm: 3x^2y + 4xy^2 = 5x - 7
Use chain rule:
y*6x + y' * 3x^2 + 4 * y^2 + 4x * 2yy' = 5
6xy + 3x^2y' + 4y^2 + 8xyy' = 5
y'(3x^2 + 8xy) = 5 - 6xy - 4y^2
y' = (5 - 6xy - 4y^2)/(3x^2 + 8xy)
hope there's no errors.

edit: arithmetic error
Title: Re: Maths Help
Post by: xD_aQt on April 20, 2011, 10:39:17 pm: Just out of curiousity also, how would you solve it on the T.I. NSpire? :P
Title: Re: Maths Help
Post by: xD_aQt on April 22, 2011, 07:24:41 pm: A red car is travelling east towards an intersection at a speed of 40km/hr while a blue car is simultaneously travelling north towards the intersection at a speed of 60km/hr if the red car is 6km from the intersection and the blue car is 8km from the intersection at what rate is the distance between the cards changing?
Title: Re: Maths Help
Post by: brightsky on April 22, 2011, 07:42:24 pm: So we want to find dD/dt, where D is the distance.
D = sqrt((6 - 40t)^2 + (8 - 60t)^2)
Find dD/dt.
then sub t = 0.
Title: Re: Maths Help
Post by: xD_aQt on April 22, 2011, 09:03:14 pm: Quote from: brightsky on April 22, 2011, 07:42:24 pm
So we want to find dD/dt, where D is the distance.
D = sqrt((6 - 40t)^2 + (8 - 60t)^2)
Find dD/dt.
then sub t = 0.
I got -72, but I'm unsure if thats the right answer because it doesn't look right. Also what would the unit be? Would it be Km/Hr? Thanks
Title: Re: Maths Help
Post by: brightsky on April 22, 2011, 09:13:49 pm: Quote from: xD_aQt on April 22, 2011, 09:03:14 pm
Quote from: brightsky on April 22, 2011, 07:42:24 pm
So we want to find dD/dt, where D is the distance.
D = sqrt((6 - 40t)^2 + (8 - 60t)^2)
Find dD/dt.
then sub t = 0.
I got -72, but I'm unsure if thats the right answer because it doesn't look right. Also what would the unit be? Would it be Km/Hr? Thanks

Yeah, that's what I got. And I'm pretty sure the units are km/hr.
Title: Re: Maths Help
Post by: xD_aQt on April 22, 2011, 09:31:15 pm: Alright I'll take your word for it, cheers!!
Title: Re: Maths Help
Post by: xD_aQt on May 15, 2011, 09:41:48 pm: I've been struggling to answer these kind of questions, any help? :P
$\int$ 4 cos 3 x - 12 x sin 3 x dx = k ( x cos 3 x ) + C
Title: Re: Maths Help
Post by: moekamo on May 15, 2011, 09:52:36 pm: i assume you want to solve for k?

differentiate both sides:

$\frac{d}{dx} \int 4 \cos (3x)-12x \sin (3x) dx = 4 \cos (3x)-12x\ sin(3x)$

and $k \frac{d}{dx} \left ( x \cos (3x) + C \right ) = k(-3x \sin (3x) + \cos (3x))$

so $4 \cos (3x)-12x\ sin(3x)=k(-3x \sin (3x) + \cos (3x))$

i.e $k=4$
Title: Re: Maths Help
Post by: xD_aQt on May 15, 2011, 10:08:05 pm: haha Yeah finding values of k, I think that makes sense, cheers!
Title: Re: Maths Help
Post by: xD_aQt on May 15, 2011, 10:54:01 pm: $\int$ $\frac{x^3}{\sqrt{3x^4+5}}$ dx
Apparently the answer is on the calc
$\frac{\sqrt{3x^4+5}}{6}$
but I got some unusual answer that is no where near it :-\
Title: Re: Maths Help
Post by: brightsky on May 15, 2011, 11:00:01 pm: let u = 3x^4 + 5, then du = 12x^3 dx
so after substitution the integral becomes:
1/12 int 1/sqrt(u) du
= 1/12 * 2sqrt(u)
= sqrt(u)/6
= sqrt(3x^4 + 5)/6
Title: Re: Maths Help
Post by: xD_aQt on May 16, 2011, 09:02:49 pm: Oh alright, I see cheers!
Title: Re: Maths Help
Post by: xD_aQt on May 16, 2011, 10:24:38 pm: $\int$ sin²x cos²x dx

I got as far as $\int$ ( $\frac{1}{2}$ sin 2x)² dx

Help? :)
Title: Re: Maths Help
Post by: brightsky on May 16, 2011, 10:43:11 pm: sin^2(x) cos^2(x) = (sin(x) cos(x))^2 = (1/2 sin(2x))^2
so the integral becomes:
1/4 int sin^2(2x) dx
let u = 2x, du = 2 dx
integral becomes:
1/8 int sin^2(u) du
now sin^2(u) = 1/2(1 - cos(2u))
so 1/16 int (1-cos(2u)) du
= u/16 - sin(2u)/32
= x/8 - sin(4x)/32
Title: Re: Maths Help
Post by: xD_aQt on May 18, 2011, 08:16:56 pm: Thanks mate!

Just one last question for now
$\int$ sin 3 x cos 4 x dx
Title: Re: Maths Help
Post by: brightsky on May 18, 2011, 08:34:00 pm: $\sin(3x) \cos(4x) = \frac{e^{3ix} - e^{-3ix}}{2i} \frac{e^{4ix} + e^{-4ix}}{2} = \frac{-(e^{ix} - e^{-ix}) + (e^{7ix} - e^{-7ix})}{4i} = \frac{1}{2} \sin(7x) - \frac{1}{2} \sin(x)$

Using this the int becomes:

$\frac{1}{2} \int (\sin(7x) - \sin(x)) dx = \frac{1}{2} \cos(x) - \frac{1}{14} \cos(7x) + C$
Title: Re: Maths Help
Post by: xD_aQt on May 18, 2011, 08:45:39 pm: Just a tad rusty on integration at the moment, but seriously appreciate it, thanks! :)
Title: Re: Maths Help
Post by: xD_aQt on May 25, 2011, 10:20:02 pm: Find the volume obtained when the region above y = 0 and below y = 4-(x-2)² is revolved around the x-axis

is this question basically asking for the area? do we integrate then let y=0? help?
Title: Re: Maths Help
Post by: jasoN- on May 25, 2011, 10:26:11 pm: this is asking for the volume of revolution, which is in specialist maths.
but the formula is $V_y = \int_a^b \pi x^2 dy$
Title: Re: Maths Help
Post by: xD_aQt on May 25, 2011, 10:39:06 pm: Oh alright, never did spesh, cheers ...