# Maths

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#1
http://prntscr.com/16cd1n6 I need help with part B only for this question (My answer 270 seconds) Not sure if right.
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#2
http://prntscr.com/16cd1n6 I need help with part B only for this question (My answer 270 seconds) Not sure if right.
help
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1 month ago
#3
http://prntscr.com/16cd1n6 I need help with part B only for this question (My answer 270 seconds) Not sure if right.
The first question is just an exercise in maximization, which requires calculus. Have you covered differentiation yet?
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#4
(Original post by davros)
The first question is just an exercise in maximization, which requires calculus. Have you covered differentiation yet?
We have done differentiation in pure but not mechanics. (Edexcel)
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1 month ago
#5
We have done differentiation in pure but not mechanics. (Edexcel)
well it's the same principle - you can actually forget the "mechanics" here You have a function of t and you want to find its maximum value, so what do you do?
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#6
(Original post by davros)
well it's the same principle - you can actually forget the "mechanics" here You have a function of t and you want to find its maximum value, so what do you do?
Differentiation (1-t^2) which becomes -2t. After this idk what to do
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1 month ago
#7
Differentiation (1-t^2) which becomes -2t. After this idk what to do
That's not quite right is it? The function is t(1-t^2), not just 1-t^2
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#8
(Original post by davros)
That's not quite right is it? The function is t(1-t^2), not just 1-t^2
1-2t. After this what do you do ?
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1 month ago
#9
1-2t. After this what do you do ?
Where does the "-2t" come from? Your original function is t(1-t^2) = t - t^3, so what do you get when you differentiate?
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#10
(Original post by davros)
Where does the "-2t" come from? Your original function is t(1-t^2) = t - t^3, so what do you get when you differentiate?
Oh just realised i never times the t. differentiation = 1-3t^2. Then I guess you find the value of t which is sqrt 3/3. After this what do you do?
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1 month ago
#11
Oh just realised i never times the t. differentiation = 1-3t^2. Then I guess you find the value of t which is sqrt 3/3. After this what do you do?
Well technically you should check that your value is going to give a maximum, not a minimum - do you know what you have to do to check this?

Then the question wants the actual maximum value of the original function, so if you're happy you've got the correct value of t, just substitute this into the original function for the velocity and calculate the result - it's just manipulating some surds which should be straightforward
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#12
(Original post by davros)
Well technically you should check that your value is going to give a maximum, not a minimum - do you know what you have to do to check this?

Then the question wants the actual maximum value of the original function, so if you're happy you've got the correct value of t, just substitute this into the original function for the velocity and calculate the result - it's just manipulating some surds which should be straightforward
Thank you I got to the answer by putting the sqrt 3/3 into the original question
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