# oblique impact

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Thread starter 1 month ago
#1
How to do part d ? I have created an equation for tanY using the compound angle formulae with tanY= tan(A-B) where A is alpha and B is Beta . Do I just differentiate and solve for Y ?
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Thread starter 1 month ago
#2
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1 month ago
#3
(Original post by RLangdon569)
What did you get for c?
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Thread starter 1 month ago
#4
(Original post by mqb2766)
What did you get for c?
tanY = (-2tanA)/(1+3(tanA)^2)
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Thread starter 1 month ago
#5
(Original post by mqb2766)
What did you get for c?
hopefully that is right
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1 month ago
#6
(Original post by RLangdon569)
tanY = (-2tanA)/(1+3(tanA)^2)
Not worked it through but if A is acute, tan(A) is positive so tan(Y) <= 0. The max would be at (0,0). That doesnt sound right..
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#7
(Original post by mqb2766)
Not worked it through but if A is acute, tan(A) is positive so tan(Y) <= 0. The max would be at (0,0). That doesnt sound right..
well is Y=A-B ?
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1 month ago
#8
(Original post by RLangdon569)
well is Y=A-B ?
Can you post your working and give me 5 to go over it?
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Thread starter 1 month ago
#9
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Thread starter 1 month ago
#10
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1 month ago
#11
(Original post by RLangdon569)
well is Y=A-B ?
B-A
I suspect which would make the numerator positive and make the answer more sensible..
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#12
(Original post by mqb2766)
B-A
I suspect which would make the numerator positive and make the answer more sensible..
B is less than A ?
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1 month ago
#13
(Original post by RLangdon569)
B is less than A ?
No
tan(B) = 3 tan(A)
so ...

Note also, you could write the expression as
2x / (1+3x^2)
and differentiate with respect to x. Then use atan() to get the angle. tan() is monotonic so the max values will be equivalent.
Last edited by mqb2766; 1 month ago
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Thread starter 1 month ago
#14
(Original post by mqb2766)
No
tan(B) = 3 tan(A)
so ...

Note also, you could write the expression as
2x / (1+3x^2)
and differentiate with respect to x. Then use atan() to get the angle. tan() is monotonic so the max values will be equivalent.
I am a bit confused because I get that tanY = tanA.
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Thread starter 1 month ago
#15
2-6x^2 = 0 , x^2=1/3. x= 1/rt3 . tanY = (2/rt3)/(1+3(1/3)) therefore tanY=1/rt3 . so tanY = tanA ?!?
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Thread starter 1 month ago
#16
Is the angle of deflection A+B ?
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1 month ago
#17
(Original post by RLangdon569)
2-6x^2 = 0 , x^2=1/3. x= 1/rt3 . tanY = (2/rt3)/(1+3(1/3)) therefore tanY=1/rt3 . so tanY = tanA ?!?
Assuming no errors, "yes". Yes in quotes because tan y = tan a for the specific value of a that maximizes the deflection in this question. (What the question wants is just the numerical value).
Last edited by DFranklin; 1 month ago
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1 month ago
#18
Incidentally, I'm pretty sure there's a "cute" way of getting the result by minimizing 1/tan y instead of maximizing tan y and then using the AM-GM inequality.
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1 month ago
#19
(Original post by RLangdon569)
2-6x^2 = 0 , x^2=1/3. x= 1/rt3 . tanY = (2/rt3)/(1+3(1/3)) therefore tanY=1/rt3 . so tanY = tanA ?!?
Sory been asleep. As you say, the max occurs when alpha 30 is beta is 60 so alpha+beta = 90.
Gamma=30 as well.
Last edited by mqb2766; 1 month ago
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1 month ago
#20
(Original post by RLangdon569)
Is the angle of deflection A+B ?
The diagram you should have is basically Q38 in

So B-A
Last edited by mqb2766; 1 month ago
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