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March 29, 2024, 01:01:08 am

Author Topic: 3U Maths Question Thread  (Read 1230263 times)  Share 

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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #255 on: June 29, 2016, 10:38:01 am »
+1


Hmm yep you're right, the subtraction being there implies the sign we should take in this case. No idea what I was thinking above, had a bad streak of mistakes  :P

140498

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Re: 3U Maths Question Thread
« Reply #256 on: June 29, 2016, 11:11:16 am »
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If dy/dx = 5x and dx/dt = -2
Find:    dy/dt and d^2y/dt^2

jakesilove

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Re: 3U Maths Question Thread
« Reply #257 on: June 29, 2016, 11:22:39 am »
+1
If dy/dx = 5x and dx/dt = -2
Find:    dy/dt and d^2y/dt^2

We know that




To find dy/dt, we can use the relationship that we learn in 3U:



Since we have all the information required, we can just sub everything in!



Similarly,






Therefore, subbing all this in, you get



Hope that made sense!
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RuiAce

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Re: 3U Maths Question Thread
« Reply #258 on: June 29, 2016, 11:41:40 am »
+1
To find dy/dt, we can use the relationship that we learn in 3U:


Lol that's just the chain rule.

Also I'm not too sure if it works like that for the second derivative... https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives
Which makes me also question the validity of this problem at the Extension 1 level.


jakesilove

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Re: 3U Maths Question Thread
« Reply #259 on: June 29, 2016, 11:54:13 am »
+1
Lol that's just the chain rule.

Also I'm not too sure if it works like that for the second derivative... https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives
Which makes me also question the validity of this problem at the Extension 1 level.



To be honest, I couldn't remember whether they call it the Chain rule in High School or they just told you it was a thing that was true. I also assumed they couldn't do the method you wrote out above, so assumed mine had to be true. I was probably/definitely wrong with that second half, sorry!
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Re: 3U Maths Question Thread
« Reply #260 on: June 29, 2016, 11:59:11 am »
+1
To be honest, I couldn't remember whether they call it the Chain rule in High School or they just told you it was a thing that was true. I also assumed they couldn't do the method you wrote out above, so assumed mine had to be true. I was probably/definitely wrong with that second half, sorry!
Haha they explicitly teach the chain, product and quotient rules in 2U for a reason.

Yeah like I said, the validity of the problem at 3U seems dodgy. At the 4U level this is a trivial implicit differentiation question.

Whilst the method I used still sticks within the boundaries of 3U, the ability to see the change of position of 'dx' can be hard.

140498

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Re: 3U Maths Question Thread
« Reply #261 on: June 29, 2016, 02:03:10 pm »
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Prove:  cos^-1(-x) = pi - cos^-1(x)

RuiAce

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Re: 3U Maths Question Thread
« Reply #262 on: June 29, 2016, 02:04:20 pm »
+1
Prove:  cos^-1(-x) = pi - cos^-1(x)


You will need to provide context if you want a specific proof.

140498

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Re: 3U Maths Question Thread
« Reply #263 on: June 29, 2016, 02:28:36 pm »
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Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right.

Would just like to check

RuiAce

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Re: 3U Maths Question Thread
« Reply #264 on: June 29, 2016, 02:36:47 pm »
+1
Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right.

Would just like to check
Sure thing. (Also, lol I made a mistake you don't really need compound angles; you just need ASTC for this particular one which eases out the proof)

Note that arccos is just the international notation for cos-1


Note that your working does not have to be set out exactly identically. If you want me to provide feedback on your working out then provide it.

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Re: 3U Maths Question Thread
« Reply #265 on: June 29, 2016, 02:58:30 pm »
+1
Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right.

Would just like to check

If it's within inverse functions, I think the easiest (and neatest) way to prove this relationship is to create a function



Differentiate the thing, using the formulas you've learnt in the inverse function section of the course. You'll see it is equal to zero, suggesting that the gradient of this function is zero (first derivative = gradient). If the gradient is zero, then the original function can only be a horizontal line (ie. for any value of x, there is only one value for y). From there, proving it equals Pi (as the question asks) is simple: sub in any value for x. Whether you put x=1, x=3.14 or x=1.6893*10^8, you'll get Pi out as the y value. As the function has a gradient of 0, this solution holds true for any value of x, therefore the original relationship must be true!

Jake
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Re: 3U Maths Question Thread
« Reply #266 on: June 29, 2016, 03:29:58 pm »
+1
If it's within inverse functions, I think the easiest (and neatest) way to prove this relationship is to create a function



Differentiate the thing, using the formulas you've learnt in the inverse function section of the course. You'll see it is equal to zero, suggesting that the gradient of this function is zero (first derivative = gradient). If the gradient is zero, then the original function can only be a horizontal line (ie. for any value of x, there is only one value for y). From there, proving it equals Pi (as the question asks) is simple: sub in any value for x. Whether you put x=1, x=3.14 or x=1.6893*10^8, you'll get Pi out as the y value. As the function has a gradient of 0, this solution holds true for any value of x, therefore the original relationship must be true!

Jake
Yeah I agree that calculus is the faster method here. He just wanted a compound angles proof.

WLalex

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Re: 3U Maths Question Thread
« Reply #267 on: June 29, 2016, 03:38:13 pm »
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Pulling strings here having me do a whole page...!







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Thank you so so much! Very quick as well, impressive :P
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Re: 3U Maths Question Thread
« Reply #268 on: June 29, 2016, 04:15:28 pm »
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Thanks so much for the previous answer
 
Thanks for the help, this should be my last one in a while:

an inverted cone of radius 20cm and depth is filled with water. The water flows out through the apex of the cone at a constant rate of 10pi cm^3s^-1. Find the rate at which the water is falling when the depth of the water is 20cm.

Just like my other questions i have an answer, but it is a take home task and i would just like to check my answer.

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Re: 3U Maths Question Thread
« Reply #269 on: June 29, 2016, 04:25:39 pm »
+1
Thanks so much for the previous answer
 
Thanks for the help, this should be my last one in a while:

an inverted cone of radius 20cm and depth is filled with water. The water flows out through the apex of the cone at a constant rate of 10pi cm^3s^-1. Find the rate at which the water is falling when the depth of the water is 20cm.

Just like my other questions i have an answer, but it is a take home task and i would just like to check my answer.
Some clarity needed.

Are we saying that the depth of the full cone is also 20cm? (And by consequence we're basically saying the initial rate that water is falling out?)