## Question

secondary 4 | A Maths

Anyone can contribute an answer, even non-tutors.

##### Kathy2

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Thank you

Date Posted: 1 month ago
Views: 28
Eric Nicholas K
1 month ago
You can post the next one, I am free to write now even while I am in the middle of tuition

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1st
I hope this solution helps you!
Date Posted: 1 month ago
Kathy2
1 month ago
I can’t even see it
Eric Nicholas K
1 month ago
Will do my posting later on a question by question basis in the middle of my tuition
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Date Posted: 1 month ago
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Q8

Kathy, this topic is basically understanding how things change with time, since rate is related to the time t. We use chain rule formulas involving t, such as dV/dt = dV/dh dh/dt to solve such questions.

The steps to solve this question are as follows.

1. Extract out useful information and transform them into words. Here, the depth increases at 2 cm/s, so we transform it to dh/dt = 2 cm/s.

2. We find out what our wanted quantity is. Here, it’s dV/dt.

3. We express dV/dt in terms of dV, dh and dt. Here, it’s dV/dt = dV/dh dh/dt (think of this as “cross multiplication”), so we need to find out what dV/dh is. It’s the rate of change of V with respect to h, so we need to know what V is in terms of h. It’s given in this question. At other times, you need to form your own equation (such as area of a circle A = #r2).

4. Finally, multiply them out to get the value.
Date Posted: 1 month ago
Eric Nicholas K
1 month ago
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Two almost symmetrical approaches to solve Q9
Date Posted: 1 month ago
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Q10
Date Posted: 1 month ago