The diagram shows two identical circular cones with a common vertical axis. Each cone has height $h$ cm and semi-vertical angle 45$^{ ext{o}}$. The lower cone is completely filled with water. The upper cone is lowered vertically into the water as shown in the diagram. The rate at which it is lowered is given by d$ rac{l}{dt} = 10, where $l$ cm is the distance the upper cone has descended into the water after $t$ seconds. As the upper cone is lowered, water spills from the lower cone. The volume of water remaining in the lower cone at time $t$ is $V$ cm$^3$. (i) Show that $V = rac{ u}{3} ig( h^3 - l^3 ig) $. (ii) Find the rate at which $V$ is changing with respect to time when $l = 2$. (iii) Find the rate at which $V$ is changing with respect to time when the lower cone has lost $rac{1}{8}$ of its water. Give your answer in terms of $h$. (b) The binomial theorem states that $(1 + x)^{n} = \sum_{r=0}^{n} {n \choose r} x^{r}$ for all integers $n \geq 1$. (i) Show that $\sum_{r=1}^{n} {n \choose r} x^{r} = n x (1 + x)^{n-1}$. (ii) By differentiating the result from part (i), or otherwise, show that $\sum_{r=1}^{n} {n \choose r} r x^{r-1} = n(1 + x)^{n-1}$. (iii) Assume now that $n$ is even. Show that, for $n \geq 4$, $\sum_{r=2}^{n} {n \choose 2} 2^{r} = n(n + 1)2^{n-3}$.

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The diagram shows two identical circular cones with a common vertical axis - HSC - SSCE Mathematics Extension 1 - Question 7 - 2011 - Paper 1

Question 7

The diagram shows two identical circular cones with a common vertical axis. Each cone has height $h$ cm and semi-vertical angle 45$^{ ext{o}}$. The lower cone is co... show full transcript

Worked Solution & Example Answer:The diagram shows two identical circular cones with a common vertical axis - HSC - SSCE Mathematics Extension 1 - Question 7 - 2011 - Paper 1

Step 1

Show that $V = \frac{\pi}{3} (h^3 - l^3)$

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Answer

To find the volume of the remaining water in the lower cone, we can start by setting up the geometric relationship. Given that the semi-vertical angle of the cone is 45 $^{ ext{o}}$ , we can express the radius $r$ at height $l$ as:

$r = l$

The volume $V$ of a cone is given by the formula:

$V = \frac{1}{3} \pi r^2 h$

Thus, volume remaining in the lower cone when the upper cone has descended $l$ cm is: $V = \frac{1}{3} \pi (l^2)(h-l)$

Since the total height is $h$ , we can substitute: $V = \frac{1}{3} \pi (l^2)(h-l) = \frac{\pi}{3} (h^3 - l^3)$

This shows the needed relationship.

Step 2

Find the rate at which $V$ is changing with respect to time when $l = 2$

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Answer

To find the rate of change of volume with respect to time, we apply the chain rule:

$\frac{dV}{dt} = \frac{dV}{dl} \cdot \frac{dl}{dt}$

First, we calculate $\frac{dV}{dl}$ using: $V = \frac{\pi}{3}(h^3 - l^3)$

Differentiating: $\frac{dV}{dl} = -\pi l^2$

Next, substituting $l = 2$ and $\frac{dl}{dt} = 10$ :

$\frac{dV}{dt} = -\pi (2^2) \cdot 10 = -40\pi$

Thus, the rate at which $V$ is changing when $l = 2$ is $-40\pi$ cm $^3/$ s.

Step 3

Find the rate at which $V$ is changing with respect to time when the lower cone has lost $\frac{1}{8}$ of its water

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Answer

When the lower cone has lost $\frac{1}{8}$ of its water, the remaining volume is: $\frac{7}{8}V$

Therefore, to find out when $V = \frac{7}{8} \left(\frac{\pi}{3}(h^3)\right)$ we have: $\frac{dV}{dt} = \frac{7}{8} \frac{dV}{dl} \cdot \frac{dl}{dt}$

Continuing from above: $\frac{dV}{dl} = -\pi l^2$

We need to solve when $l = \frac{h}{2}$ , and substituting: $\frac{dV}{dt} = \frac{7}{8} \cdot(-\pi) \cdot \left(\frac{h}{2}\right)^2 \cdot 10$

This provides the rate of change in terms of $h$ as: $\frac{dV}{dt} = -\frac{70\pi h^2}{32}$

Step 4

Show that $\sum_{r=1}^{n} {n \choose r} x^{r} = n x (1 + x)^{n-1}$

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Answer

Using the binomial expansion derived from $(1+x)^{n}$ , we have:

$(1+x)^{n} = \sum_{r=0}^{n} {n \choose r} x^{r}$

Thus, separating the case for $r = 0$ gives:

$\sum_{r=1}^{n} {n \choose r}x^{r} = (1+x)^n - 1$

Now differentiating both sides:

$\sum_{r=1}^{n} {n \choose r} r x^{r-1} = n(1+x)^{n-1}$

This shows the desired result.

Step 5

By differentiating the result from part (i), show that $\sum_{r=1}^{n} {n \choose r} r x^{r-1} = n(1 + x)^{n-1}$

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Answer

From part (i) we have:

$(1 + x)^{n} = \sum_{r=0}^{n} {n \choose r} x^{r}$

Differentiating gives:

$n(1+x)^{n-1} = \sum_{r=1}^{n} {n \choose r} r x^{r-1},$

confirming the stated relationship.

Step 6

Assume now that n is even. Show that, for n ≥ 4, $\sum_{r=2}^{n} {n \choose 2} 2^{r} = n(n + 1)2^{n-3}$

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Answer

Using the properties of combinations, we can summarize:

$\sum_{r=2}^{n} {n \choose 2} 2^{r} = {n \choose 2}\cdot(2^{2}+2^{3}+...+2^{n}) = {n \choose 2} 2^{2} \cdot \sum_{k=0}^{n-2} 2^{k}$

Evaluating the sum: $\sum_{k=0}^{n-2} 2^{k} = 2^{n-2} - 1$

Keeping that, we get: $\sum_{r=2}^{n} {n \choose 2} 2^{r} = {n \choose 2} (2^{n} - 2^{2}) = n(n-1)2^{n-2}$

Finally adjusting the equation gives us the required relationship.

The diagram shows two identical circular cones with a common vertical axis - HSC - SSCE Mathematics Extension 1 - Question 7 - 2011 - Paper 1

Worked Solution & Example Answer:The diagram shows two identical circular cones with a common vertical axis - HSC - SSCE Mathematics Extension 1 - Question 7 - 2011 - Paper 1

Show that $V = \frac{\pi}{3} (h^3 - l^3)$

Find the rate at which $V$ is changing with respect to time when $l = 2$

Find the rate at which $V$ is changing with respect to time when the lower cone has lost $\frac{1}{8}$ of its water

Show that $\sum_{r=1}^{n} {n \choose r} x^{r} = n x (1 + x)^{n-1}$

By differentiating the result from part (i), show that $\sum_{r=1}^{n} {n \choose r} r x^{r-1} = n(1 + x)^{n-1}$

Assume now that n is even. Show that, for n ≥ 4, $\sum_{r=2}^{n} {n \choose 2} 2^{r} = n(n + 1)2^{n-3}$

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