Ho Ho... No

[blacksanta62]

Decided to start a topic where I would place some questions I need help with (mostly application stuff but sometimes casual misunderstandings).

This thread is sorta like (a rip off) of Pi's thread: pi's Specialist Maths Questions

I know others have probably done it too, this was just the first example I could remember 

First question: In the Delfino Square race course in Mario Kart, the players race along narrow city streets, several of which have low archways through which the karts must drive. Suppose each of these arches has been encoded by the programmers according to the equation: \frac{1}{a^2x^2-4 } + 6,a>0
If the largest kart in the game, Browsers Koopa kart, has a cross sectional rectangular are with height 5 units and width 2 units, then: a) Establish the value of a for the smallest possible arch on the track.
b) What is the highest point on this arch?

My first guess was "it must be a negative number so that it becomes a reflected parabola", an arch looking thing. Wrong! Read straight after that it said a>0 and now I'm confused. Hence the genesis ( ) of this thread.

Edit: The above equation came out nothing like I wanted it too: 1/(a^2 x^2-4 )+6,a>0

[silverpixeli]

First question: In the Delfino Square race course in Mario Kart, the players race along narrow city streets, several of which have low archways through which the karts must drive. Suppose each of these arches has been encoded by the programmers according to the equation: \frac{1}{a^2x^2-4 } + 6,a>0
If the largest kart in the game, Browsers Koopa kart, has a cross sectional rectangular are with height 5 units and width 2 units, then: a) Establish the value of a for the smallest possible arch on the track.
b) What is the highest point on this arch?

My first guess was "it must be a negative number so that it becomes a reflected parabola", an arch looking thing. Wrong! Read straight after that it said a>0 and now I'm confused. Hence the genesis ( ) of this thread.

Edit: The above equation came out nothing like I wanted it too: 1/(a^2 x^2-4 )+6,a>0

try [ tex] \frac{1}{a^2x^2-4 } + 6,a>0 [/tex] (without the space in the first tag)



as for your thoughts on it being a reflected parabola... this is actually a reciprocal graph! if you sketch it for a few values of a you will see what the question is getting at.



edit: here's a version you can play with on desmos: just slide for different values of a

https://www.desmos.com/calculator/fk9yqdpczc

[blacksanta62]

try [ tex] \frac{1}{a^2x^2-4 } + 6,a>0 [/tex] (without the space in the first tag)



as for your thoughts on it being a reflected parabola... this is actually a reciprocal graph! if you sketch it for a few values of a you will see what the question is getting at.

(Image removed from quote.)

edit: here's a version you can play with on desmos: just slide for different values of a

https://www.desmos.com/calculator/fk9yqdpczc

Cool, I had been sketching them separately before but there wasn't another variable (a^2) and it never asked me to find a value which would give a small arch so it kinda blew up in my face

"Stupid" question: Can it be even smaller than 3? Or is 3 the final answer? Why? I haven't done it yet because I have an early day tomorrow at UoM but I'll give it a try then. 

[silverpixeli]

Cool, I had been sketching them separately before but there wasn't another variable (a^2) and it never asked me to find a value which would give a small arch so it kinda blew up in my face

"Stupid" question: Can it be even smaller than 3? Or is 3 the final answer? Why? I haven't done it yet because I have an early day tomorrow at UoM but I'll give it a try then.

Sorry! you can make it as small as you like I just set it to stop at 3 because it seemed skinny enough on the graph. I have adjusted the link to now have bowser's kart on it so you can see how skinny to make it

[pi]

This thread is sorta like (a rip off) of Pi's thread: pi's Specialist Maths Questions

Oh gawd, the memories hahahahaha

[blacksanta62]

It's Pi, I've actually read through a decent chunk of that thread (way back in term 2 2015) but had idea what was going on! Make's a bit more sense now and should make even more sense in the future. "as for your thoughts on it being a reflected parabola... this is actually a reciprocal graph! if you sketch it for a few values of a you will see what the question is getting at.", I'm adding this to my derp files   . It's right up there with my friends attempt answer of "negative zero"   

[blacksanta62]

More of a methods question: Will the graphing of logarithms and exponential graphs like this one: 3 -3 Loge (2x) or 2e^-2x - 1 be in the methods 3/4 or spesh 3/4 course? I just brushed up of my graphing skills today and I'm starting to enjoy it more 

[blacksanta62]

And what should be my first step when doing this question from the 2010 VCAA spesh exam 1:
It's question 9a

[blacksanta62]

Not going to give up on that tex frac thing @silverpixeli. Thats meant to be x^2

[pi]

And what should be my first step when doing this question from the 2010 VCAA spesh exam 1:
It's question 9a

The very first thing to do would be to recognise what type of graph it is

Do you know what equations describing circles, ellipses, and hyperbolas look like?

[blacksanta62]

I've only just started the hyperbola section questions so I know a bit... is it: x^2/a^2-y^2/b^2?
The ellipses one is x^2+y^2=1^2. Don't think it's a ellipse though so it must be the hyperbola

[pi]

Yup that's right

The general formula for a hyperbola that looks like >< is:

Where:
- Centre:
- Domain:
- Range:
- Asymptotes:
- Vertices:

Plug in a, b, h, k; plot the vertices; draw the asymptotes; find the axial intercepts, and the sketch away

[blacksanta62]

Cool. I'll give it a whirl and post what I get as a final answer. I wasn't given any solutions so I plug it into desmos and link it on here

[blacksanta62]

Today was packed with maths! Here s one of the questions I need help with:

Differentiate

1) g(x) = tan(4x^2 - 3) - e^x/2x + 1

I can do this question but would this (- e^x/2x + 1) be -(e^x - 2x - 1)^-1
The e^x has just confused me 

2) 'party poppers' are plastic devices which use gun powder to launch streamers into the air. As the streamers unfurl, they get more wind resistance. The height of the centre of the streamers (y meters, measured from the solid ground) is modelled t seconds after pulling the cord, and has the following equation:

                                    y(t) = 1/6 (12t - t^2)e^2/5(2-t)

a) State a sensible domain for this equation (0<x<2 ??)
b) Find the highest point the streamers reach
c) The velocity at time t is given by y'(t). What is the maximum velocity, and when is it reached? (Would I differentiate to find the gradient function and the set it equal to 0?)

Thanks. If some working out or guidance is provided I thank you

[blacksanta62]

Integral calculus (pretty sure it's definite but correct me if wrong...)
Anyway: Calculate the area under the following functions in the indicated region:
f(x) = 3sin(2x) dx from x=0 to x=3pi

I have two answers but I'm confused so I would like some help (the answer isn't needed but some guidance in the right direction)

1) I antidiff. the function to get: -1/6 cos (6x) + c (am I even allowed to change the thing inside the bracket to begin with, left me thinking this solution isn't correct)
2) [-1/6 cos(6x)]0-3 - [-1/6 cos(6x)]0-3 (I don't know how to use the equation writer but this is from 0 to 3pi)
3) I then got confused, do I sub in 3pi? and 0 for the other one which will result in 0? or do I just sub in 3 alone?

The second solution I got was:
1) Antidiff. to get -3/2 cos(2x) + c (according to the formula sheet this is what I do)
2) [-3/2 cos(2x)]0-3 - [-3/2 cos(2x)]0-3
3) I'm still not sure whether to sub in 3pi or something else

Help will be +1 and appreciated. And if someone can help with the question above thank you

What's the difference between definite and indefinite (is it that one is specific i.e. it's defined to a specific area?)