VCE Methods Question Thread!

[Mr. Study]

Thanks for that kamil.

What a unique short cut!

[1ne]

Sketch the graph of f:R->R, f(x)=4 cos (3x) for [0, 2pi]
The reason I have trouble sketching is because I don't know how to solve for x-ints.. I mean I know the period is 2pi/3 but it is saying that the domain is [0, 2pi] so how would you find the x-ints..



Sketch the graph of function f:R->R where y=2 sin (2x) for [-pi, pi]
Same problem with the question before.. does the -pi mean that all the x-ints will be the same as pi but just negative?

Thanks

[Aurelian]

Sketch the graph of f:R->R, f(x)=4 cos (3x) for [0, 2pi]
The reason I have trouble sketching is because I don't know how to solve for x-ints.. I mean I know the period is 2pi/3 but it is saying that the domain is [0, 2pi] so how would you find the x-ints..

Sketch the graph of function f:R->R where y=2 sin (2x) for [-pi, pi]
Same problem with the question before.. does the -pi mean that all the x-ints will be the same as pi but just negative?

Thanks

Solve for x-intercepts algebraically, by setting the equation equal to zero. Thus, for the first one;

4cos(3x) = 0
cos(3x) = 0
3x = ...-3pi/2, -pi/2, pi/2, 3pi/2, 5pi/2, 7pi/2, 9pi/2, 11pi/2, 13pi/2 etc...
x = ...-3pi/6, -pi/6, pi/6, 3pi/6, 5pi/6, 7pi/6, 9pi/6, 11pi/6, 13pi/6 etc...

However, x must be within the interval [0, 2pi]. Thus we must exclude anything outside this.

This leaves us with x = pi/6, 3pi/6, 5pi/6, 7pi/6, 9pi/6, 11pi/6.

Hence your x-intercepts will be (pi/6, 0), (pi/2, 0), (5pi/6, 0), (7pi/6, 0), (9pi/6, 0) and (11pi/6, 0).

Does that help?

[dinosaur93]

Sketch the graph of f:R->R, f(x)=4 cos (3x) for [0, 2pi]
The reason I have trouble sketching is because I don't know how to solve for x-ints.. I mean I know the period is 2pi/3 but it is saying that the domain is [0, 2pi] so how would you find the x-ints..

Basic angle =



Quardant 1 and 4 cosine are positive (based on CAST rule)











You can check your answer on the calculator, by restricting the domain, you could confirm the number of domain...but be aware that the graph will not give you exact answers...

[1ne]

how come not all the x-ints that you guys solved for are in the graph??

[monkeywantsabanana]

Hi, I'm having trouble with linear approximations... Can someone help me with this question?

The area of a circular disc increases from 100pi cm^2 to 101pi cm^2. Find the corresponding increase in the radius.

Thanks in advance.

[brightsky]

we know that A(r) = pi*r^2, A'(r) = 2 pi r
when A = 100pi, r = 10
A(10+h) = 101pi = hA'(10)+A(10) = h(20pi) + 100 pi
h=(101 pi - 100 pi)/20pi=1/20
technically i think the textbook's asking for an approx increase.

[1ne]

How do I graph y=1/2 sin (2x) + 3 for one cycle

I understand what all the transformations but I don't no how to find the x-intercepts?

Can anyone tell me if I'm doing it right..

y=1/2sin (2x) + 3 = 0
y=1/2sin (2x) = -3
y=sin (2x) = -6

Is this it? I'm really confused, please help

[TrueTears]

There are no x intercepts for this, think about it, the amplitude is 1/2 and the median line is at y = 3. so the lowest value the sin graph has is 3-1/2 and biggest value it obtains is 3+1/2.

another way to think about it is look at the equation you derived: sin (2x) = -6

remember that -1<=sin(x)<=1 so -1<=sin(2x)<=1

[b^3]

Firstly there is something different about this graph compared to a lot of trig graphs

Notice that the amplitude of the graph is .
For y=sin(x) the function's minimum and maximums are -1 and 1.
So for the minimum and maximums will be and .
The function has been shifted upwards by three so now our minimum and maximums are and .

So that means there are no x-intercepts, and it makes sense because when you get to the last step in your calculations, you cannot do it since you'll get , which you can't do since whats in the brackets has to be between -1 and 1.


EDIT: beaten by TrueTears by 27 seconds, now I'll just fix that image up

[1ne]

I need to, how you guys know what values the sin graph will go through.. for example in the pic you can see that the graph goes through pi.. but how do I do that for 1/2 sin (2x) + 3.

Sorry if i'm confusing you guys, is suck at trig.

[TrueTears]

note that the 1/2 and 3 does not affect x values

let X = 2x

note that sin(X) passes through the values of 0, pi/2, pi, 3pi/2 and 2pi

now x = X/2 -> x passes through the values 0/2, (pi/2)/2, (pi)/2, (3pi/2)/2 and (2pi)/2

[1ne]

omg! thanks.. now I get it!

[dinosaur93]

A Silly Question to ask but...

QUESTION 1

By writing as , show that

STEP 1:

STEP 2:

STEP 3:

STEP 4:

could someone please help me figure out how to get from step 1 down to step 4?




QUESTION 2
At the mouth of the sea there is a tidal river, where the height, h, in metres of the river varies according to the rule , where t is the time in hours after 2.00am on a sunday.

When the water level first drops to o.5 metres in the daylight hours, Charlie climbs a tree that is in the river. Charlie cannot swim very well and knows that she must get down from the tree when the water level rises to 0.5 metres again or she will be trapped in the tree.

Fisherman Fred has a large boat that requires a depth of atleast 'd' metres for him to get out of the river to go fishing in the sea.

a. What is the largest value of 'd' if Fred requires a continuous period of 1 hour t get in or out of the river ( to two decimal places.)

b. What time of the day does Charlie get rescued by Fred if he picks her up in his boat as soon as he has the maximum depth required for his boat?

[TrueTears]

step 1 is just the compound angle formula: sin(x+y) = sin(x)cos(y)+cos(x)sin(y) where x = 2theta and y = theta

step 2 first involves the double angle formula for sin, ie sin(2x) = 2sin(x)cos(x) where x = theta

step 2 also involves the double angle formula for cos, ie, cos(2x) = 1-2sin^2(x) where x = theta [note there are 2 other formulas, cos(2x) = 2cos^2(x)-1 and cos(2x) = cos^2(x)-sin^2(x)]

step 3 and 4 are just expansion and simplification