Radian Questions (1 Viewer)

[allGenreGamer]

I've gathered some challenging radian questions (well, at least I've failed to complete them ), please have a go!

1) Given that the wingspan of an aeroplane is 30m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' (minutes NOT degrees) when it is directly overhead.
(Answer: 7367m)

2) Find the volume of the solid formed when the curve y = sec (pie) x is rotated about the x-axis from x = 0 to x = 0.15
(Answer: 0.51)

3) A sector of a circle with radius 5cm and an angle of (pie)/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find its exact surface area and volume.
(Answer: SA = (25 pie) / 6
V = (125 (square root of 35 pie) / 648

4) Find lim x -> 0
(1 - cos2x) / x^2
(Answer: 2)

Thanks, I seriously need help , even if you don't do all the questions, that's fine.

 

[CM_Tutor]

Originally posted by allGenreGamer
1) Given that the wingspan of an aeroplane is 30m, find the plane's altitude to the nearest metre if the wingspan subtends an angle of 14' (minutes NOT degrees) when it is directly overhead.
(Answer: 7367m)

Draw an upside down isosceles triangle. Height, H, is height above ground, 'base' of 30 m is wingspan, 'apex' angle is 14'

Using trig, tan 7' = 15 / H
H = 15 / tan 7' = 7366.5 ... m

2) Find the volume of the solid formed when the curve y = sec (pie) x is rotated about the x-axis from x = 0 to x = 0.15
(Answer: 0.51)

V = π∫ (from 0 to 0.15) y² dx, where y = sec πx
= π∫ (from 0 to 0.15) sec²πx dx
= π[(1 / π)* tan πx] (from 0 to 0.15)
= (tan 0.15π - tan 0)
= 0.5095... cu units

3) A sector of a circle with radius 5cm and an angle of (pie)/3 subtended at the centre is cut out of cardboard. It is then curved around to form a cone. Find its exact surface area and volume.
(Answer: SA = (25 pie) / 6
V = (125 (square root of 35 pie) / 648

SA = πrs where r is the radius of the base of the cone - ie r² + h² = s², where s = 5 cm and 5π / 3 = 2πr

4) Find lim x -> 0
(1 - cos2x) / x^2
(Answer: 2)

http://www.boredofstudies.org/community/showthread.php?s=&threadid=35113

 

[allGenreGamer]

Thanks so much! Unfortunately, I've found more questions that I cannot do:

1) Simplify:

2cos((pie/4) + theta).cos((pie/4) - theta)
answer: cos^2(theta) - sin^2(theta) = cos2(theta)

2) Solve for 0 =< x =< 2(pie)

a) cosx + (square root of 3).sinx = 2
answer: x = (pie)/2

b) sin3x + sinx = 0
answer: 0, pie, 2(pie), (pie)/2, (3pie)/2

3) Find all the points of inflexion on the curve y = 3cos(2x + (pie/4)) for 0 <= x <= 2pie
answer = ((pie/8), 0), ((5pie/8), 0), ((9pie/8), 0), ((13pie/8), 0)

thanks, I owe you so much.............

 

[mojako]

Let's see if I make it before CM_Tutor, hehehe...

Q1
Method #1:
See picture attached… Typed it with MS Word but was too lazy to continue with the rest of your questions.
Second line is by compound angle, third is by difference of 2 squares.
cos^2(theta) - sin^2(theta) = cos2(theta) by double angle or compound angle

Method#2:
If you’re doing 4U, use the products to sums formula which will bring you straight to cos2(theta) after simplification.

Q2

a) cosx + (square root of 3).sinx = 2
Method #1: rewrite LHS as either sine or cosine equation.
How? Set up an identity cosx + (square root of 3).sinx = Rsin(x+alpha) –(1)
Expand the LHS of (1) using compound angle, then equate coefficients of sinx and cosx to the LHS of (1)… etc…
Now you’re solving for Rsin(x+alpha) = 2, with the values of R and alpha already known from the above steps.

Method #2:
Use the t-formulae. Also check if pi (180 degrees) is a solution (well in this question it’s not).

b) sin3x + sinx = 0
Write sin3x as sin(2x + x) then use compound angle formula. Then, use double angle to get rid of sin2x and cos2x.
You get sin3x = 3sin(x) – 4sin^3(x), so you’re now solving for
4sin(x) – 4sin^3(x) =0
sin(x) – sin^3(x) = 0
sin(x) [1-sin^2(x)] = 0
either sin(x)=0 or 1-sin^2(x)=0, i.e. sin^2(x)=1, i.e. sin(x)=plus/minus 1

Q3:

y = 3cos(2x + (pie/4))
y’ = 3 * -sin(2x + pi/4) * [derivative of 2x+pi/4]
= -6sin(2x+pi/4)
Y’’ = -6 * cos(2x+pi/4) * 2
= -12cos(2x+pi/4)
Solve for -12cos(2x+pi/4) = 0
cos(2x+pi/4) = 0

Our boundary is 0 <= x <= 2pi
So, 0 <= 2x <= 4pi
Also pi/4 <= 2x+pi/4 <= 4pi+pi/4

Now you solve for cos(u) = 0 with pi/4 <= u <= 4pi+pi/4

[it’s pi not pie]

 

[CM_Tutor]

Originally posted by allGenreGamer
b) sin3x + sinx = 0
answer: 0, pie, 2(pie), (pie)/2, (3pie)/2

There's a quicker way to do this one:

sin 3x + sin x = 0
sin 3x = -sin x = sin(x + &pi;)

Now, when sin A = sin B, A - B = 2n&pi; or A + B = (2n + 1)&pi;, where n is an integer

So, in this case 2x - &pi; = 2n&pi; or 4x - &pi; = (2n + 1)&pi;, as A = 3x and B = x + &pi;
x = (2n + 1)&pi; / 2 or x = n&pi; / 2

Thus, x = (&pi; / 2), (3&pi; / 2), ... or x = 0, (&pi; / 2), &pi;, (3&pi; / 2), ...

Noting the domain restriction of 0 &le; x &le; 2&pi;, we have: x = 0, &pi; / 2, &pi;, 3&pi; / 2, 2&pi;

Edited to correct method - answer was correct, but method was flawed.