Mathematics Question Thread

[fun_jirachi]

For a), you're looking at the point where x is a maximum.
The second derivative is -6, showing the that x has a maximum. The maximum is at the point where the derivative = 0, ie. where 15-6t=0
At the point t=15/6, the ball is 18.75m up the slope.
b) With the current formula of distance, you're looking at the ball going up the slope, so when you differentiate the velocity will be negative, as the ball is travelling down the slope, not up.
You need to first find the time t when it reaches the bottom of the slope. 15t-3t^2 = 3t(5-t) , so it will reach the bottom of the slope at t=5s. Subbing into the derivative 15-6t, the ball is travelling 15m/s down the slope.
c) I may be wrong for this one as the wording is a bit weird. I assume that the whole motion means the time it takes to go all the way up and all the way down. That's just 5s as I figured it part b)

Hope this helps!

[amelia20181]

if this is the graph for displacement how would you graph velocity and acceleration

[dermite]

if this is the graph for displacement how would you graph velocity and acceleration

take the gradients of each subsequent graph
taking the gradient of a displacement graph gives velocity as dx/dt = v = x(dot)
taking the gradient of a velocitry/time graph will give the acceleration as dv/dt = a = x(double_dot)

[amelia20181]

for velocity should the graph be a straight line below the x axis

[amelia20181]

for this question how would you find the total area

[fun_jirachi]

 

for velocity should the graph be a straight line below the x axis

yes for that graph the velocity should be a horizontal line below the x-axis, as the gradient of the 'curve' is constant, and negative

for this question how would you find the total area

The garden is 40x25. The path surrounding it is 2m wide, which means it adds 2m to each side of the garden's dimensions. The yard as such is 44x29m.
The percentage of the yard that is garden I'm pretty sure you can figure out for yourself.

Hope this helps

[amelia20181]

how do you describe the motion of the particle at T1 for this displacement graph

[jamonwindeyer]

how do you describe the motion of the particle at T1 for this displacement graph

Hey! So to describe the motion you need three things:

- Where is it (position)
- How fast is it moving and in what direction (velocity)
- What about acceleration? Is It speeding up or slowing down?

We can see from the graph that it is passing through \(x=0\) at that time, so we're at the origin. As for speed, well we're moving to the right, since the slope of the graph is positive (we're going from negative x to positive x). For acceleration, I throw to you, do you think it is getting faster? Slowing down? About the same?

[amelia20181]

is it getting faster

[jamonwindeyer]

is it getting faster

Yep, awesome stuff! If the line were straight it would be going a constant speed, but it is curving upwards, so it is getting faster/accelerating!

[StephTol]

Hey!
I was just wondering how to do this question, would I need to draw it out?

Thanks

[Opengangs]

Hey!
I was just wondering how to do this question, would I need to draw it out?

Thanks

Hey, StephTol.
So for these types of questions, note that solutions occur when \(\tan x = 2\) OR \(\sin x = 1/2\).
In knowing this, we note that there is exactly one solution for \(\tan x = 2\) in between 0 and \(\pi\), and that occurs at \(x = \tan^{-1}(2)\).

How many solutions are there for \(\sin x = 1/2\)?
Well, we know that \(\frac{\pi}{6}\) is another unique solution. And finally, there lies another solution: \(\frac{5\pi}{6}\). We can find this by determining the nature of the sine curve.

Notice that there is a symmetry between \(\frac{\pi}{2}\) and \(\frac{\pi}{6}\)! So, to find where it hits the sine curve, we simply note that:
\[ \begin{align*}x &= \left(\frac{\pi}{2} - \frac{\pi}{6}\right) + \frac{\pi}{2} \\ &= 1 - \frac{\pi}{6} \\ &= \frac{5\pi}{6} \\ &\leq \pi\end{align*}\]

In all, we know there are 3 unique solutions!

[amelia20181]

how do you do this question

[fun_jirachi]

how do you do this question

Integrating the second derivative gets you the derivative. As such, integrating 8x gives you the derivative of the function which is 4x^2 + C, since we aren't given any upper or lower limits. however, at (-2, 5) the derivative y = 4x^2 + C = 1 because it states that the tangent makes an angle with the x axis of 45 degrees. It's important to remember tangent means gradient ie. first derivative at a point and also that gradient = tan theta. As tan 45 = 1, we know the gradient is 1. so you sub in x = -2 and you get C = -15.

So we now know that y' = 4x^2 - 15
integrating we get y = 2x^3 - 15x + C
now because it has a point (-2, 5) on its curve, we substitute those values in
5 = -16 + 30 + C
so C = -9
so the function is y = 2x^3 - 15x -9
Hope this helps!

[isabella104]

Hi everyone!

I'm pretty new here, this is my first post! Hope I'm doing it properly haha.

I've been having trouble with these questions I have attached below from the 2017 CSSA Trial Paper. It provides solutions but I don't know how they got there!

In 13.a.i, I don't understand how they got the answer (I thought your were meant to use the product rule?)

In 14.a.ii, I don't understand where the π comes from for (5-π)/2

I'm sure there are other people out there that have been struggling with these questions because my maths class have been talking about them non-stop!