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How do you differentiate (x/v) + (√(a-x)^2+(a/2)^2)/(v/3) WRT x where v and a are constants?
Edit: Don't worry I've solved it.
Edit: Don't worry I've solved it.
Last edited by domm1; 4 weeks ago
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#2
(Original post by domm1)
How do you differentiate (x/v) + (√(a-x)^2+(a/2)^2)/(v/3) WRT x where v and a are constants?
How do you differentiate (x/v) + (√(a-x)^2+(a/2)^2)/(v/3) WRT x where v and a are constants?
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(Original post by mqb2766)
First term is easy, chain rule for the second. Have a go?
First term is easy, chain rule for the second. Have a go?
P.S. I have not learnt the Chain Rule yet.
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#4
(Original post by domm1)
Yeah I was being stupid as I was trying to get an expression for x in terms of a and v when the derivative was equal to 0, but ended up with an equation that was of the form of a quadratic but in terms of x, a and v, but already have the values for a and v, oops.
P.S. I have not learnt the Chain Rule yet.
Yeah I was being stupid as I was trying to get an expression for x in terms of a and v when the derivative was equal to 0, but ended up with an equation that was of the form of a quadratic but in terms of x, a and v, but already have the values for a and v, oops.
P.S. I have not learnt the Chain Rule yet.
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(Original post by mqb2766)
Do you want to describe what the problem is / upload your working?
Do you want to describe what the problem is / upload your working?
Part B. I am just worried I have differentiated the second term incorrectly.
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#6
(Original post by domm1)
https://isaacphysics.org/questions/l...a-1ea18244028f
Part B. I am just worried I have differentiated the second term incorrectly.
https://isaacphysics.org/questions/l...a-1ea18244028f
Part B. I am just worried I have differentiated the second term incorrectly.
You can also solve with some trig/geometry.
Last edited by mqb2766; 4 weeks ago
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(Original post by mqb2766)
Possibly. It's a reasonably common problem, but not worked that particular one through. Upload what you tried if you want it checked. One of the hints is relevant for the chainrule/differentiation part.
Possibly. It's a reasonably common problem, but not worked that particular one through. Upload what you tried if you want it checked. One of the hints is relevant for the chainrule/differentiation part.
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#8
(Original post by domm1)
Yeah, I used that hint to the best of my ability and thought that it would mean my second term would differentiate to (xv)/(3√(a-x)^2+(a/2)^2), but then when I took the square root part out of the fraction and had it as (3/v)((a-x)^2+(a/2)^2))^1/2, I got a very different term.
Yeah, I used that hint to the best of my ability and thought that it would mean my second term would differentiate to (xv)/(3√(a-x)^2+(a/2)^2), but then when I took the square root part out of the fraction and had it as (3/v)((a-x)^2+(a/2)^2))^1/2, I got a very different term.
time on beach + time in water
...
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(Original post by mqb2766)
Can you upload a pic of what you started with etc? It's got to be
time on beach + time in water
...
Can you upload a pic of what you started with etc? It's got to be
time on beach + time in water
...
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#11
(Original post by domm1)
But on #1 I mentioned what I started with. For part B, t=(x/v) + (√(a-x)^2+(a/2)^2)/(v/3)
But on #1 I mentioned what I started with. For part B, t=(x/v) + (√(a-x)^2+(a/2)^2)/(v/3)
Sometimes it's easier to see someone's work than to ask about all formatting problems.
Last edited by mqb2766; 4 weeks ago
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(Original post by mqb2766)
I could say the brackets are wrong in the root ...
Sometimes it's easier to see someone's work than to ask about all formatting problems.
I could say the brackets are wrong in the root ...
Sometimes it's easier to see someone's work than to ask about all formatting problems.
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#13
Then you have
x + 3sqrt((x-a)^2 + (a/2))^2)
The chain rule has the numerator for the second terms derivative
3/2 * d(x-a)^2 /dx
can you do that? I've flipped the sign inside the brackets as it's unchanged.
Last edited by mqb2766; 4 weeks ago
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(Original post by mqb2766)
Ok. Multiply through by v. It's a constant so won't affect the minimum time.
Then you have
x + 3sqrt((x-a)^2 + (a/2))^2)
The chain rule has the numerator for the second terms derivative
3/2 * d(x-a)^2 /dx
can you do that? I've flipped the sign inside the brackets as it's unchanged.
Ok. Multiply through by v. It's a constant so won't affect the minimum time.
Then you have
x + 3sqrt((x-a)^2 + (a/2))^2)
The chain rule has the numerator for the second terms derivative
3/2 * d(x-a)^2 /dx
can you do that? I've flipped the sign inside the brackets as it's unchanged.
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#15
(Original post by domm1)
I'm sorry I don't understand the parts I have boldened.
I'm sorry I don't understand the parts I have boldened.
It's slightly easier to differentiate the left expression. Can you do that? It is the ~numerator you get using the chain rule.
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(Original post by mqb2766)
(x-a)^2 = (a-x)^2
It's slightly easier to differentiate the left expression. Can you do that? It is the ~numerator you get using the chain rule.
(x-a)^2 = (a-x)^2
It's slightly easier to differentiate the left expression. Can you do that? It is the ~numerator you get using the chain rule.
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#17
(Original post by domm1)
d(x-a)^2 / dx = 2x-2a. Sorry that shouldn't of needed repeating, it was just the way it was written that threw me off.
d(x-a)^2 / dx = 2x-2a. Sorry that shouldn't of needed repeating, it was just the way it was written that threw me off.
You divide by 2 because of the sqrt(...) and multiply by 3 because of the velocity.
So the derivative of 3sqrt(..) Is
3(x-a)/sqrt(...)
So you have
0 = 1 + 3(x-a)/sqrt(...)
Are you ok and can you proceed?
Last edited by mqb2766; 4 weeks ago
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(Original post by mqb2766)
Which is 2(x-a).
You divide by 2 because of the sqrt(...) and multiply by 3 because of the velocity.
So the derivative of 3sqrt(..) Is
3(x-a)/sqrt(...)
So you have
0 = x + 3(x-a)/sqrt(...)
Are you ok and can you proceed?
Which is 2(x-a).
You divide by 2 because of the sqrt(...) and multiply by 3 because of the velocity.
So the derivative of 3sqrt(..) Is
3(x-a)/sqrt(...)
So you have
0 = x + 3(x-a)/sqrt(...)
Are you ok and can you proceed?
I've gotten a quadratic, x^2 -40x +410=0 as my equation, however it gives no real roots, I am confused how this is still not correct.
Last edited by domm1; 4 weeks ago
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#19
(Original post by domm1)
Edit: Wait, I thought the derivative should be 1 + (Boldened part) as you have the single x term at the start?
I've gotten a quadratic, x^2 -40x +410=0 as my equation, however it gives no real roots, I am confused how this is still not correct.
Edit: Wait, I thought the derivative should be 1 + (Boldened part) as you have the single x term at the start?
I've gotten a quadratic, x^2 -40x +410=0 as my equation, however it gives no real roots, I am confused how this is still not correct.
How did you get that?
Last edited by mqb2766; 4 weeks ago
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then squared everything to get (...) + 9(x-a)^2 (Think this may of been where I went wrong)
Expanded everything out, added like terms and multiplied by 4 just to get rid of the (a^2)/4 fraction to get 40x^2 - 80ax +41a^2=0
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