VCE Methods Question Thread!

[SE_JM]

It's alright Eulerfan, really. You can take as long as you want. But seriously, never in my life did i see anyone who was so willing to help me.

Most of my teachers at school got pretty sick of me because i kept asking too many questions, lol.

Thank you for answering. The unsolved questions torture me in my sleep (what a nerd, i know)

[psyxwar]

Hello~ I have a question i'm having trouble with. I was a bit reluctant to ask more questions (coz, no one answered me on my last question   )

BUT, the world is teeming with questions i don't know and AN has one of the most patient helpers 

So, here goes my question:
A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds is s=400t-16t². How fast is the distance changing from the rocket to an observer on the ground 1800 feet away from the launching site, when the rocket is still rising and is 2400 feet above the ground?

The correct answer is: increasing at a rate of 64 feet per second

Thank you!

Easiest way is probably just to draw a right angled triangle of hypotenuse D, base 1800 and height s, where d is distance and s is the height.




Using the chain rule:





We want dD/dt at s=2400

At s=2400, subbing into s(t) yields . Solving this yields t=10 or t=15, but we want it when it is rising. Thus we take t=10 (as t=15 is when it is falling; it is launched upwards)

We then find dD/dt at s=2400 and t=10.

Subbing into the above gives us 64.

Edit: ninja'd by my slow LaTeX skills LOL

[SE_JM]

Thanks, Psyxwar. Closely beaten by Eulerfan but your efforts are appreciated

[keltingmeith]

Great minds think alike, my friend. However, your post brought up something I might mention:

1) As said, drawing the situation out as a triangle helps - drawing out ANY situation helps so you can pick up things like that. Without drawing the situation, neither psy or myself would've known to treat the distance as a triangle.

2) Be careful when naming things "D". If you do name it D, make sure it's capital - using a little d means that when you differentiate/integrate, you'll get a "dd", and that's just weird to look at. You might end up accidentally missing a d, or the examiner not reading a d, if you do it, so you're best to use something else - either a capital D, or another letter entirely (I picked x). Of course, this is more a preference thing - it's like writing in pen for your exam instead of pencil so it doesn't rub out in transit.

[SE_JM]

Hi, Eulerfan
I was just wondering why is at t=15, the rocket is falling?

[keltingmeith]

Hi, Eulerfan
I was just wondering why is at t=15, the rocket is falling?

If you look at the graph, you can see that at t=15 the slope is negative, so it must be falling. You also could've plugged t=15 into ds/dt to see that you would've gotten a negative answer, implying the rocket is falling at that point.

[knightrider]

In this graph what would be the range?

I am confused because at the bottom one endpoint is included and one is not and they both have the same y-values.

[cosine]

In this graph what would be the range?

I am confused because at the bottom one endpoint is included and one is not and they both have the same y-values.

Hmm, well what does it mean that -3 is not included? It means that we can go up to -2.98, -2.99 and so on but never reach -3. On the other side of the graph, were -3 is included, this means we can have -2.99 AND -3.

So, if you think about it, on the left hand side of the graph, the point stops at -2.99999999999999... but on the right hand side, the graph stops on -3 exactly as it's included. SO, now, the range as we know it is the lowest y-value to the highest y-value. So which is lower, -2.9999999999999.... or -3?

Obviously, -3. So the range would be [-3, 0]

Hope it helped!

[knightrider]

Hmm, well what does it mean that -3 is not included? It means that we can go up to -2.98, -2.99 and so on but never reach -3. On the other side of the graph, were -3 is included, this means we can have -2.99 AND -3.

So, if you think about it, on the left hand side of the graph, the point stops at -2.99999999999999... but on the right hand side, the graph stops on -3 exactly as it's included. SO, now, the range as we know it is the lowest y-value to the highest y-value. So which is lower, -2.9999999999999.... or -3?

Obviously, -3. So the range would be [-3, 0]

Hope it helped!

Thankyou cosine 

[Zues]

for the graph

y=2sin2(x+pi/2), im a bit confused by why the graph begans at the origin and goes like a graph that has been reflected in the x axis.

the points of interest ive got
- intercept is pi - pi2 = pi/2
- starts at x=0
y=2sin(pi) = 0, i can only see that it starts at original if it was a -pi translation, but pi/2 im a bit confused. can someone show me how they would go about this? i didnt do translations in year 11 so im just getting important points and adding/subtracting the angle. 

[AirLandBus]

Can anyone supply the answer to part b) and each sub section for this question. The answers in the book seem a bit suss and what to confirm whether or not there are right.
Taaaa

[Zues]

how do you know how the graph starts?

[brightsky]

for the graph

y=2sin2(x+pi/2), im a bit confused by why the graph begans at the origin and goes like a graph that has been reflected in the x axis.

the points of interest ive got
- intercept is pi - pi2 = pi/2
- starts at x=0
y=2sin(pi) = 0, i can only see that it starts at original if it was a -pi translation, but pi/2 im a bit confused. can someone show me how they would go about this? i didnt do translations in year 11 so im just getting important points and adding/subtracting the angle. 

The most efficient way to obtain the graph of y = 2sin[2(x+pi/2)] is not to find 'points of interest', but rather to sketch the graph of y = sin(x) and then apply an appropriate sequence of transformations. In order to get from y = sin(x) to y = 2sin[2(x+pi/2)], we know that the following transformations need to be performed:

1. Dilation by a factor of 2 from the x-axis.
2. Dilation by a factor of 1/2 from the y-axis.
3. Translation of pi/2 units in the negative direction of the x-axis.

We know already what y = sin(x) looks like. It has an amplitude of 1 and period of 2pi. Now, sketch the graph of the resultant function after each transformation has been applied.

1. The amplitude becomes 2. In other words, the graph becomes 2 times taller.
2. The period shrinks to pi. In other words, the graph becomes 1/2 as wide.
3. The entire graph is shifted to the left by pi/2 units.

Hence, the graph of y = 2sin[2(x+pi/2)] should look like this: http://www.wolframalpha.com/input/?i=+y+%3D+2sin[2%28x%2Bpi%2F2%29].

[Zues]

The most efficient way to obtain the graph of y = 2sin[2(x+pi/2)] is not to find 'points of interest', but rather to sketch the graph of y = sin(x) and then apply an appropriate sequence of transformations. In order to get from y = sin(x) to y = 2sin[2(x+pi/2)], we know that the following transformations need to be performed:

1. Dilation by a factor of 2 from the x-axis.
2. Dilation by a factor of 1/2 from the y-axis.
3. Translation of pi/2 units in the negative direction of the x-axis.

We know already what y = sin(x) looks like. It has an amplitude of 1 and period of 2pi. Now, sketch the graph of the resultant function after each transformation has been applied.

1. The amplitude becomes 2. In other words, the graph becomes 2 times taller.
2. The period shrinks to pi. In other words, the graph becomes 1/2 as wide.
3. The entire graph is shifted to the left by pi/2 units.

Hence, the graph of y = 2sin[2(x+pi/2)] should look like this: http://www.wolframalpha.com/input/?i=+y+%3D+2sin[2%28x%2Bpi%2F2%29].

then how would i go about finding x intercepts?

would you be able to show me why y=−1/2 cos3(x+π)+1 also starts like that depicted in the attachment?

[Zues]

is this how its done?

sorry for messy its hard on a comp haha.

what happens if i had a translation of like -pi/6 it would be hard to see if the graph goes down then up or how it begins.