onur369's Methods Question Thread :)

[Respect]

I would go



As required.

i got no idea how u did that

the second step.. how did u simplfy into  3(x+5)^2 ... .. .. .

 im stuffed for the test tomrrow.

[onur369]

The test were going to have is going to be handwritten so we are all dead.

[jasoN-]

I would go



As required.

i got no idea how u did that

the second step.. how did u simplfy into  3(x+5)^2 ... .. .. .

 im stuffed for the test tomrrow.

[xZero]

i got no idea how u did that

the second step.. how did u simplfy into  3(x+5)^2 ... .. .. .

 im stuffed for the test tomrrow.

use product rule,









take out the common factor

[kefoo]

Hey guys I'm back unfortunately =_=
need help on this question::

1)a.Sketch graph of  y = cos(x) and y = sqrt(3)sin(x) [Sketched] (0,2pi)
   b. Find the coordinates of the points of intersection [Equated them to get tanx = 1/sqrt3, got pi/6 and 7pi/6 correct, just don't know how to get the y values :S

[jasoN-]

 just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)

[kefoo]

just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)

but wouldn't sin(pi/6) be different to cos(pi/6)?

[luffy]

just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)

but wouldn't sin(pi/6) be different to cos(pi/6)?

Yes. However,

The equation you had was not . It was

Hope I helped.

[kefoo]

just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)

but wouldn't sin(pi/6) be different to cos(pi/6)?

Yes. However,

The equation you had was not . It was

Hope I helped.

oh right, forgot about sqrt3... x_x careless mistakes.

Got another q..

tan(2x-pi/4) = sqrt(3) solve for 0< x < 2pi

[luffy]

just sub pi/6 and 7pi/6 into either of the equations to get the y value (cos both graphs intersect at the same coordinate)

but wouldn't sin(pi/6) be different to cos(pi/6)?

Yes. However,

The equation you had was not . It was

Hope I helped.

oh right, forgot about sqrt3... x_x careless mistakes.

Got another q..

tan(2x-pi/4) = sqrt(3) solve for 0< x < 2pi

Below is how I would do it:






The same principle applies to this question, however, you want all the solutions of between and







Tell me if I didn't explain anything clear enough (sorry if I made any errors).

Hope I helped.

[panicatthelunchbar]

Can someone please help, mainly just a) and b):

11) A patient suffering with hypothermia after being lost in the snow for 2 days is covered in
thermal blankets and warmed so that his body temperature can return to normal levels.The change in body temperature (T°C) from the initial temperature that occurs during thermal treatment is modelled by the equation T = A loge (t − b) where t represents the time in minutes. The initial body temperature is 35.7°C and it reaches 36.1°C ten minutes after treatment is initiated.
(a)   What is the value ofT when t = 0?
(b)   Find the value of b.
(c)   Find the value of A.
(d)   What will be the body temperature after 20 minutes?
(e)   How long will it take for the body temperature to reach 36.8°C? Write your answer
in hours and minutes.

thanks

[onur369]

Hey guys Im having trouble with some worded questions in Apps of Diff, Ive got one question how would I analyse it.

Question 6:
A cylinder is expanding in such a way that is length h is always double the radius r of its
base. Find the rate of change of its volume V with respect to:
a) r.
b) h.

[brightsky]

a) dV/dr = dV/dh * dh/dr
V = pi*r^2 h
But h = 2r (which means dh/dr = 2), r = h/2
So V = pi*(h/2)^2 h = pi/4*h^3 (which means dV/dh = 3pi/4*h^2)
So dV/dr = 3pi/4 * h^2*2 = 3pi/2*h^2

b) dV/dh = dV/dr * dr/dh
V = pi*r^2 h = pi*r^2(2r) = 2pi*r^3, dV/dr = 6pi*r^2
r = h/2, dr/dh = 1/2
So dV/dh = 1/2*6pi*r^2 = 3pi*r^2

[onur369]

Thanks Derrick Ha jnr

[kefoo]

having some trouble here..

tan (3x - pi/6) = -1
solve for x..
i get the wrong answer :S