# A Level Maths Practise Paper Question

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The depth, h metres, of water in a harbour on any particular day can be modelled by the formula, h = 5+3sin (30t) where t is the time in hours after midday.

a) find the maximum depth of water in the harbour using this model (1 mark)

b) The ship can enter the harbour when the depth of water is at least 6.5 metres.

Find the times after mid-day during which the ship can enter the harbour

a) find the maximum depth of water in the harbour using this model (1 mark)

b) The ship can enter the harbour when the depth of water is at least 6.5 metres.

Find the times after mid-day during which the ship can enter the harbour

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#2

(Original post by

The depth, h metres, of water in a harbour on any particular day can be modelled by the formula, h = 5+3sin (30t) where t is the time in hours after midday.

a) find the maximum depth of water in the harbour using this model (1 mark)

b) The ship can enter the harbour when the depth of water is at least 6.5 metres.

Find the times after mid-day during which the ship can enter the harbour

**mumtaz15**)The depth, h metres, of water in a harbour on any particular day can be modelled by the formula, h = 5+3sin (30t) where t is the time in hours after midday.

a) find the maximum depth of water in the harbour using this model (1 mark)

b) The ship can enter the harbour when the depth of water is at least 6.5 metres.

Find the times after mid-day during which the ship can enter the harbour

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I haven't known what to do with part a but for part b I substituted h in as 6.5, so 6.5 = 5+3sin(30t), then took the 5 away so 1.5 = 3sin(30t) and divided by 3sin(30) and I got t=1

(Original post by

What have you tried? Please post your working.

**Notnek**)What have you tried? Please post your working.

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#4

(Original post by

I haven't known what to do with part a but for part b I substituted h in as 6.5, so 6.5 = 5+3sin(30t), then took the 5 away so 1.5 = 3sin(30t) and divided by 3sin(30) and I got t=1

**mumtaz15**)I haven't known what to do with part a but for part b I substituted h in as 6.5, so 6.5 = 5+3sin(30t), then took the 5 away so 1.5 = 3sin(30t) and divided by 3sin(30) and I got t=1

You need to find the maximum value that h can be. First what's the maximum that sin(30t) can be?

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I don't understand your question, how would I work out the maximum

(Original post by

Starting with a):

You need to find the maximum value that h can be. First what's the maximum that sin(30t) can be?

**Notnek**)Starting with a):

You need to find the maximum value that h can be. First what's the maximum that sin(30t) can be?

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#6

(Original post by

I don't understand your question, how would I work out the maximum

**mumtaz15**)I don't understand your question, how would I work out the maximum

The maximum of sin(30t) is the same as the maximum of the sin(x) graph since t can be any positive value.

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Ohh is it 1?

(Original post by

You should know from GCSE what the maximum value that sin(x) can be. Think about the graph.

The maximum of sin(30t) is the same as the maximum of the sin(x) graph since t can be any positive value.

**Notnek**)You should know from GCSE what the maximum value that sin(x) can be. Think about the graph.

The maximum of sin(30t) is the same as the maximum of the sin(x) graph since t can be any positive value.

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#8

(Original post by

Ohh is it 1?

**mumtaz15**)Ohh is it 1?

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#9

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Omg of course h=8, right?

(Original post by

Your working is confusing. You know that the maximum of sin(30t) is 1 so what's the maximum of 3sin(30t)? You shouldn't be setting t = 1.

**Notnek**)Your working is confusing. You know that the maximum of sin(30t) is 1 so what's the maximum of 3sin(30t)? You shouldn't be setting t = 1.

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#11

(Original post by

Omg of course h=8, right?

**mumtaz15**)Omg of course h=8, right?

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I'd like some help as it says 'times' not 'time' and my working out is giving me 1 answer so I don't think it's right

(Original post by

Correct Are you happy with b) or do you need any help?

**Notnek**)Correct Are you happy with b) or do you need any help?

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#13

(Original post by

I'd like some help as it says 'times' not 'time' and my working out is giving me 1 answer so I don't think it's right

**mumtaz15**)I'd like some help as it says 'times' not 'time' and my working out is giving me 1 answer so I don't think it's right

Dividing by 3 gives

When you solve this there will be multiple solutions. I'm sure you've done something like this before.

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when I inverse sin then divide by 30 my answer is 1

(Original post by

Dividing by 3 gives

When you solve this there will be multiple solutions. I'm sure you've done something like this before.

**Notnek**)Dividing by 3 gives

When you solve this there will be multiple solutions. I'm sure you've done something like this before.

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#15

(Original post by

when I inverse sin then divide by 30 my answer is 1

**mumtaz15**)when I inverse sin then divide by 30 my answer is 1

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#16

yep, to find other solutions, you would've learnt how to use the graph of sine or the CAST diagram if that rings a bell?

(Original post by

. At GCSE there would have been 1 solution to this but at A Level you will have learnt that there are many solutions. t = 1 is only one solution.

**Notnek**). At GCSE there would have been 1 solution to this but at A Level you will have learnt that there are many solutions. t = 1 is only one solution.

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#17

**mumtaz15**)

The depth, h metres, of water in a harbour on any particular day can be modelled by the formula, h = 5+3sin (30t) where t is the time in hours after midday.

a) find the maximum depth of water in the harbour using this model (1 mark)

b) The ship can enter the harbour when the depth of water is at least 6.5 metres.

Find the times after mid-day during which the ship can enter the harbour

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