9.1.1

Maximum capacity has to do with maximum volume

That means we have to find the function describing maximum volume and solve for h

If you check on the answer sheet below you will notice that I had first two incorrect attempts,

The idea here is that I know my theory, I know what I have to do, but must figure out how

The first attempt would not work because it has two variables

The second one also made my life difficult and I just had to find another way, which was to express r in terms of h and be left with only h as a variable

Since h is exactly what I want

Then, the first derivative of my volume is equal to zero at maximum, and then I solve for h as in the answer sheet below

Exam tip:

It is usual and common that you don’t know how you are going to find the answer, but if you know your principles, sometimes you need to keep looking for that answer and you will get it.

Hence it is necessary to beat time on easy questions so that you can tackle difficult questions with more time.

9.1.2

On this one we simply have to substitute h using the given 6 root of 3, and then we find our volume in cm cubed

9.2.1

H(t) is displacement, and the first derivative is velocity

What is velocity?

Speed with direction. Velocity will have a positive or negative sign indicating direction

In this case, positive will be speed on increasing gradient, moving upwards. Negative will be speed on declining gradient.

At maximum, the turning point of this negative parabola, velocity (the first derivative) is equal to zero

And then we solve for t as in the answer sheet below

9.2.2

This one is a bit tricky for me, and I don’t have the memo as well.

But here is my attempt:

First, I skipped this question, calculated the maximum height and the use distance over time to find speed

But that is average speed throughout the whole distance from when it is thrown upto maximum

Then, I calculated using velocity, speed after 1 second and speed at zero seconds.

I believe that speed at zero seconds is the correct speed exactly when the stone is thrown

9.2.3

To find maximum height, just substitute the time it takes to get to maximum, substitute it on the original height function

9.2.4

Once again, this one is tricky also.

But we have two times when the stone is on the ground, just before it is thrown and when it hits the ground again

At both those times, height is equal to zero

So, equate the height function and solve for t

As in the answer sheet below, the height is zero when time is zero seconds and when it is 9 seconds

That means the stone hit the ground after 9 seconds

Substitute 9 seconds on the velocity, the first derivative function, to find the speed exactly at 9 seconds on the clock.

The stone seems to hit the ground at the same speed as it was thrown.

I am not sure about 9.2.2 and 9.2.4 but I think my logic is brilliant and we are very close to the memo.

Happy studying 😊