3U Maths Question Thread

[ilikeapples]

Struggling with this projectiles question:
A projectile at the highest point of its trajectory has a velocity 8 metres per second and its position is 8 metres above the ground. Find, taking 𝑔 = 9.8ms-2. Find:
(i) The angle of projection (to the nearest degree).
(ii) The initial velocity (correct to 1 decimal place).

[ilikeapples]

And also another projectiles question.....
𝑂𝐴 is a vertical building of height 20 metres. A particle is projected horizontally from 𝐴 with speed 10m/s. At the same instant, a second particle is projected from 𝑂 with speed 10√5m/s at an angle 𝜃 above the horizontal. The two particles travel in the same plane of motion. Take 𝑔 = 10m/s2.
(i) Write down expressions for horizontal and vertical displacements relative to 𝑂 for each particle after time 𝑡 seconds.
(ii) Show that if the two particles collide, then they do so after 1 second
(iii) Show that if the two particles collide, when they do so their paths of motion are

Attached is a diagram of the question. Thankyou!

[RuiAce]

Struggling with this projectiles question:
A projectile at the highest point of its trajectory has a velocity 8 metres per second and its position is 8 metres above the ground. Find, taking 𝑔 = 9.8ms-2. Find:
(i) The angle of projection (to the nearest degree).
(ii) The initial velocity (correct to 1 decimal place).

Both of which can then be computed with the aid of the calculator

[RuiAce]

[arii]

Not sure how to do this

[RuiAce]

Not sure how to do this

[Dragomistress]

Hi,
How do I do this question for c?

[RuiAce]

Hi,
How do I do this question for c?

This question does not look doable because part b) is not doable. I had a look at the textbook answer and assuming that the maximum speed is \(7\) and consequently an amplitude of \(1\) is a bizarre and potentially false assumption. Just because the particle starts 1m to the right of the origin does not mean that it represents the maximum displacement away from the centre point \(x_0 = 0\) - take for example \( x = 2\cos \left( 7t +\frac\pi3 \right) \). (Then clearly \( \ddot{x} = -49x \) and when \(t=0\), \(x=1\), but it is obvious that the maximum speed is in fact 14.)

However, if we were to cheat and pretend that the amplitude is 1 anyway, then we can simply do this by recognising that \(x = \cos 7t\), so for part c) we'd then be solving \( \frac12 = \cos 7t\).

[Mate2425]

Hi could someone please help me with getting to the answers in the following questions below - regarding Inverse Trig, thanks!:

Q1. Find the first derivative of y = cos^-1 (Sin x) in the domain -pi < x < pi   (where for both '<' it is also representing equal to). The textbook answers have three sets of domains, but still unaware in how to obtain it?

Q2. A 6m long ladder is leaning up against a wall at a height of h and an angle of theta,.
The ladder slips down the wall at a constant rate of 0.05 ms/s. Find the rate at which the angle is changing in degrees and minutes when the height is 2.5m.

Also, with recognition of these sort of questions is there a guidance to which sort of differentiation rule you should use e.g Product rule/ Function of a Function Rule when trying to differentiate the inverse trig with variation from the general formula of e.g sin^-1(x) .

Thank you, for all your help it would be of great appreciation if someone could help me in working these questions out, thanks!
Mate2425

[jamonwindeyer]

Hi could someone please help me with getting to the answers in the following questions below - regarding Inverse Trig, thanks!:

Q1. Find the first derivative of y = cos^-1 (Sin x) in the domain -pi < x < pi   (where for both '<' it is also representing equal to). The textbook answers have three sets of domains, but still unaware in how to obtain it?

I'll take the first one! You can use the chain rule here:



The tough bit is the denominator, it could be \(\cos{x}\) or \(-\cos{x}\). We have to think about where those things happen.

If \(x\) is between \(\frac{-\pi}{2}\) and \(\frac{\pi}{2}\), then \(\cos{x}\ge0\). So, in that range, the denominator is just \(\cos{x}\), the square and square root pair doesn't change the sign. So overall, the answer is -1.

If \(x\) is outside that range, \(\cos{x}\) is negative. So, \(\sqrt{\cos^2{x}}=-\cos{x}\), it's like changing the sign. So the overall answer is 1 on the outer domains

^^ Reading back, that explanation of why the sign flips in certain ranges wasn't the best, does it help at all or should I go in a little more depth?

Edit: Anything like these questions is definitely going to use the chain rule!!

[arii]

Not sure about this one

[RuiAce]

Q2. A 6m long ladder is leaning up against a wall at a height of h and an angle of theta,.
The ladder slips down the wall at a constant rate of 0.05 ms/s. Find the rate at which the angle is changing in degrees and minutes when the height is 2.5m.

Also, with recognition of these sort of questions is there a guidance to which sort of differentiation rule you should use e.g Product rule/ Function of a Function Rule when trying to differentiate the inverse trig with variation from the general formula of e.g sin^-1(x) .

Thank you, for all your help it would be of great appreciation if someone could help me in working these questions out, thanks!
Mate2425

Not quite sure where the concern is with your follow-up question. Q2 was more of a related rates question than an inverse trig question so for that one the chain rule just made sense. But if you're talking about ordinary differentiation, you're really supposed to do the same thing as with all other differentiation questions, e.g. product rule for something like \( \frac{d}{dx} e^x \sin^{-1}x\).

^^ Reading back, that explanation of why the sign flips in certain ranges wasn't the best, does it help at all or should I go in a little more depth?

I guess when that sort of thing happens I just rely on firstly recalling that \( \sqrt{x^2} = |x|\) and then just quoting that \( |x| = \begin{cases}x&\text{ if }x\geq 0\\ -x &\text{if }x < 0 \end{cases} \)

[RuiAce]

Not sure about this one

Parts i) and ii) are just typical binomial probability questions with corresponding solutions \(\binom{5}{2} \left( \frac{1}{40} \right)^2 \left( \frac{39}{40} \right)^3\) and \(\binom{5}{2} \left( \frac{1}{40} \right)^2 \left( \frac{39}{40} \right)^3+\binom{5}{3} \left( \frac{1}{40} \right)^3 \left( \frac{39}{40} \right)^2+\binom{5}{4} \left( \frac{1}{40} \right)^4 \left( \frac{39}{40} \right)+ \left( \frac{1}{40} \right)^5\). This should be relatively straightforward to see.

I kept staring at iii) and it seems like I can't understand the wording at all. What's the final answer they provide so that I can backtrack with it?

[Mate2425]

I'll take the first one! You can use the chain rule here:



The tough bit is the denominator, it could be \(\cos{x}\) or \(-\cos{x}\). We have to think about where those things happen.

If \(x\) is between \(\frac{-\pi}{2}\) and \(\frac{\pi}{2}\), then \(\cos{x}\ge0\). So, in that range, the denominator is just \(\cos{x}\), the square and square root pair doesn't change the sign. So overall, the answer is -1.

If \(x\) is outside that range, \(\cos{x}\) is negative. So, \(\sqrt{\cos^2{x}}=-\cos{x}\), it's like changing the sign. So the overall answer is 1 on the outer domains

^^ Reading back, that explanation of why the sign flips in certain ranges wasn't the best, does it help at all or should I go in a little more depth?

Edit: Anything like these questions is definitely going to use the chain rule!!

Hi Jamon,
Thanks for your help so far!
Could you please go into a bit of further depth into why the sign flips and also how you arrive at the correct final answers.
Thank you so much 

[Opengangs]

Hi Jamon,
Thanks for your help so far!
Could you please go into a bit of further depth into why the sign flips and also how you arrive at the correct final answers.
Thank you so much 

Note that:
\[ \sqrt{\cos^2(x)} = \left|\cos(x)\right|\]

So when \(x\) lies outside of the domain where \(\cos(x)\) is normally positive, then it becomes \(-\cos(x)\).
It's the same idea with \(\left|x\right|\). If \(x\) is negative, then \(\left|x\right|\) becomes \(-x\) and if \(x\) is positive, then we just keep the same sign