dextrous63
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#1
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Came across this question. Confused about why part a is worth 3 marks for a start. More to the point, what does it mean by "Average gradient"?
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RDKGames
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(Original post by dextrous63)
Came across this question. Confused about why part a is worth 3 marks for a start. More to the point, what does it mean by "Average gradient"?
Sounds to me like finding the average value of a function f(x) over the domain [a,b] as \displaystyle \dfrac{1}{b-a} \int_a^b f(x) dx

Just in this case, it's f' which we are considering and only for the portion of motion concerning descent, i.e. from t=\dfrac{\pi}{2} to \pi
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old_engineer
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(Original post by RDKGames)
Sounds to me like finding the average value of a function f(x) over the domain [a,b] as \displaystyle \dfrac{1}{b-a} \int_a^b f(x) dx

Just in this case, it's f' which we are considering and only for the portion of motion concerning descent, i.e. from \dfrac{\pi}{2} to \pi
...or it could just be the straight line gradient from the maximum point to the axis crossing to the right of it...
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dextrous63
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Thanks. I'll have a play with it later and get back to you.
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ghostwalker
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(Original post by dextrous63)
Thanks. I'll have a play with it later and get back to you.
And the difference between the two suggestions is....
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dextrous63
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(Original post by ghostwalker)
And the difference between the two suggestions is....
None whatsoever. They yield the same correct answer. Thanks.
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ghostwalker
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(Original post by dextrous63)
None whatsoever. They yield the same correct answer. Thanks.
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dextrous63
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#8
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(Original post by ghostwalker)
Still a mystery why part (a) is worth 3 marks.
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ghostwalker
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#9
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(Original post by dextrous63)
Still a mystery why part (a) is worth 3 marks.
Aye - money for old rope, as they say.
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