Trig functions (few questions) (1 Viewer)

[Makro]

1. Find the equation of the tangent to the curve y = cos(pi - x) at the point (pi/2, 0)

And

2. Find the gradient of the normal to the curve y = tan 3x at the point where x = pi/12.

 

[Makro]

Find

(needed latex)

 

[clintmyster]

1. y = -cosx
y' = sinx
y' = 1 at x = pi/2
therefore eqn of tangent is y = x - pi/2

2. y=tan3x
y' = 3sec^2(3x)
y' = 6 at x = pi/12

3. Im not quite clear with what your askin but if its the integration of (cosx)^0 then your answer is just x.

 

[Makro]

Thanks for 1, yeah 3 is a weird question. The answers show it as a degrees sign. The final one ended up being 180/pi sin x(deg) + C.

In regards to 2, can you expand on it a bit more? Just not getting it atm, pretty much from line 2 to line 3.

 

[adamcg]

Find

(needed latex)

ok, so x degrees is (pi)x/180 radiens so then your just integrating cos[(pi)x/180]

= (180/pi)sin[(pi)x/180]

where x is in radiens

 

[Uncle]

Find

(needed latex)

Is that the integral of cos⁰(x) or cos(x⁰)?

Integrating cos(x⁰) w.r.t x gives xcos(1) = 0.5403023059x
or integrating cos⁰(x) w.r.t x gives x as the other poster mentions.

 

[clintmyster]

Thanks for 1, yeah 3 is a weird question. The answers show it as a degrees sign. The final one ended up being 180/pi sin x(deg) + C.

In regards to 2, can you expand on it a bit more? Just not getting it atm, pretty much from line 2 to line 3.

um its a chain rule so...
let u = 3x, du/dx = 3
y=tanu, dy/du = sec^2u = sec^2(3x)
therefore y' = dy/du x du/dx
y' = 3sec^2(3x)

 

[jet]

That circle is a degrees sign

 

[Makro]

Is that the integral of cos⁰(x) or cos(x⁰)?

Integrating cos(x⁰) w.r.t x gives xcos(1) = 0.5403023059x
or integrating cos⁰(x) w.r.t x gives x as the other poster mentions.

it was cosx(degrees)

ok, so x degrees is (pi)x/180 radiens so then your just integrating cos[(pi)x/180]

= (180/pi)sin[(pi)x/180]

where x is in radiens

that looks right, thanks.

um its a chain rule so...
let u = 3x, du/dx = 3
y=tanu, dy/du = sec^2u = sec^2(3x)
therefore y' = dy/du x du/dx
y' = 3sec^2(3x)

Thanks.

2 more:

1. Integrate from pi to 0 cos(pi + x) dx.

2. Sketch y = tan x/2 for -2pi <= x <= 2pi

With 2. I sketched the tan graph twice, but it shows the tan graph once from -2pi to 2pi. The period is pi/b, and I got 2pi for the period, so doesn't that mean it appears twice?

Thanks in advance.

 

[smp211]

2. y=tan3x
y' = 3sec^2(3x)
y' = 6 at x = pi/12

btw for q2 it was asking for grad of normal, not tangent.. so the answer is -1/6

1. Integrate from pi to 0 cos(pi + x) dx.
2. Sketch y = tan x/2 for -2pi <= x <= 2pi

1. Well... you can take it as Integrate (cosx) between pi & 0 ...
i intregrated to sinx between pi & 0
so sin pi - sin 0 .... well that = 0 ...... :S i donno if that's right.

2. yeah the period is 2pi, so it would look like this :

 

[Makro]

Doesn't the period mean the distance it takes to complete a "cycle" of the graph? So if the period of 2pi, don't you get tan graph from -2pi to 0 and then another tan graph from 0 to 2pi? I must be missing something.

smp, for your #1 answer was 0, so you were right.

Cheers for all the help so far

 

[smp211]

yes that is correct.

as you can see on that graph, there is one full cycle between -2pi to 0, and another full cycle between 0 to 2pi.

that is because the period is 2pi, but they've asked for a graph between -2pi and 2pi, in this case meaning the tan is essentially drawn twice.

 

[Makro]

Wait so from -2pi to 0 is one full cycle of the tan graph? Cos in the textbook I see that the whole thing from -2pi to 2pi is one full cycle but it only goes from 0 to 2pi.

Blahh.