Trigonometric Values (1 Viewer)

[FDownes]

I've got yet another maths problem that I'm stuck on. Lets see if you guys can help me out of this one, although I'm sure it's a pretty simple question. Here it is;

Find the derivative of 3 cos^3 5x

And that's not cos to the power of 35x, if anyone was wondering.

 

[Mark576]

d/dx (3cos³5x) = -15sin5x.3cos²5x = -45sin5x.cos²5x, we can verify this by using the substitution u = cos5x to find the integral, and it turns out to be correct.

 

[munchiecrunchie]

I've got yet another maths problem that I'm stuck on. Lets see if you guys can help me out of this one, although I'm sure it's a pretty simple question. Here it is;

Find the derivative of 3 cos^3 5x

And that's not cos to the power of 35x, if anyone was wondering.

I'm a bit rusty on this stuff, but I'm pretty sure this is right:

y = 3 cos ^3 5x
= 3 (cos 5x)^3

now it looks a bit like the function of a function chain rule thing.

dy/dx = 3 x 3 x (cos 5x)^2 x -5 x sin5x
= -45 sin 5x (cos 2x) ^2

so basically its the derivative of the whole thing, then the derivative of wats inside the brackets.

Correct me if I'm wrong.

 

[me121]

FDownes, you can put superscript and subscript in your post using the tags,

[noparse]

_{insert text here}
and
^{insert text here}

[/noparse]

 

[FDownes]

Thanks guys, it makes more sense now.

And I'll keep that in mind, me121, in case I have any further questions. Thanks.

 

[FDownes]

Speaking of which... I've got me another problem;

Find the equation of the tangent to the curve y = sin(pi - x) at the point (pi/6, 1/2), in exact form.

I'm fairly familiar with this type of question, but I'm a little shaky on exactly how to differentiate equation, and exactly how the result will fit in to the y - y1 = m(x - x1) format.

 

[tommykins]

hey finlay, its tommy =p.

y = sin(pi-x)
dy/dx = -cos(pi-x) since -x is in brackets.

Mp = -cos(pi-[pi/6]) = -cos 5pi/6 => sqrt3/2

y - [1/2] = sqrt3/2 (x - pi/6)

2y -1 = sqrt3x - sqrt3pi/6

12y - 6 = 6sqrt3x - sqrt3pi

6sqrt3x -12y = 6+sqrt3pi

18x - 12sqrt3y = 6sqrt 3 + 3pi
.:. 6x-4sqrt3y = 2sqrt3 + pi should be the equation .

 

[YannY]

Speaking of which... I've got me another problem;

Find the equation of the tangent to the curve y = sin(pi - x) at the point (pi/6, 1/2), in exact form.

I'm fairly familiar with this type of question, but I'm a little shaky on exactly how to differentiate equation, and exactly how the result will fit in to the y - y1 = m(x - x1) format.

y=sin(pi-x)=sinx
expand: sin(a-b)=sinacosb-sinbcosa
sin(pi-x)=sinpisinx-sinxcospi =sinx
sinpi=0 cospi=-1

y'=cosx

at x=pi/6

y'=sqrt3/2 Where sqrt: square root. i.e root sign

Tangent:
y-1/2=sqrt3/2(x-pi/6)

sqrt3/2x-y-(pi.sqrt3+6)/12=0

 

[tommykins]

y=sin(pi-x)=sinx
expand: sin(a-b)=sinacosb-sinbcosa
sin(pi-x)=sinpisinx-sinxcospi =sinx
sinpi=0 cospi=-1

y'=cosx

at x=pi/6

y'=sqrt3/2 Where sqrt: square root. i.e root sign

Tangent:
y-pi/6=sqrt3/2(x-1/2)

sqrt3/2x-y+(2pi-3sqrt3)/12=0

Isn't it the other way around mate?

 

[YannY]

Isn't it the other way around mate?

yeah sorry man i must've been on drugs

 

[rooeys2]

yeah sorry man i must've been on drugs

lol yea most definitely

 

[FDownes]

Gots me some more problems of the mathematical kind. Lets see who can dig me out of this hole;

Find ∫e ^{tan x} sec²x dx using the substitution u = tan x

 

[tommykins]

haha mostapha just asked me that on msn.

u = tan x
du/dx = sec²x
.:. dx = du/sec²x

∫e tan x sec2x

= ∫e^u du => since the sec²x cancels out
= e^u
= [e^tan x] + C

 

[FDownes]

Awesome, thanks.

 

[FDownes]

Here we go again;

Evaluate ⁵₀∫(25 - x²)^1/2 dx using the substitution x = 5 sin u.

 

[SpinCobra]

₀∫⁵ (25 - x²)^1/2 dx

-----------------------
Let..

x = 5 sin u
dx = 5 cos u du

when x = 0, u = 0
when x = 5, u = pi/2

-----------------------

Therefore

₀∫⁵ (25 - x²)^1/2 dx

= ₀∫^pi/2 (25 - 25 sin²u)^1/2 (5 cos u) du

= ₀∫^pi/2 root(25 - 25 sin²u) (5 cos u) du

= ₀∫^pi/2 5 root(1 - sin²u) (5 cos u) du

(Take the 25 out)

= 25 ₀∫^pi/2 root(1 - sin²u) (cos u) du

(1 - sin²u = cos²u)

= 25 ₀∫^pi/2 root(cos²u) (cos u) du

= 25 ₀∫^pi/2 (cos²u) du

-----------------------

cos 2u = 2 cos²u - 1
2 cos²u = cos 2u + 1
cos²u = 1/2 (cos 2u + 1)

(Shove 1/2 outside integral)

-----------------------

= 25/2 ₀∫^pi/2 cos 2u + 1 du

Theres your easy integral.

 

[SpinCobra]

I think thats right. Assuming i didnt miss something while typing it up.

In the end I got 25pi/4

Whats the answer meant to be?

 

[FDownes]

Yup, that's correct. Thanks muchly.

 

[FDownes]

I thought I had this problem figured, but my answer is way off. I don't suppose someone could walk me through this one too?

Evaluate ₀∫^4√3 dt / (16 + t²)^3/2 using the substitution t = 4tan[FONT=&quot]Ө[/FONT].

 

[SpinCobra]

Whoo. This one is messy. xD