A 2-metre-high sculpture is to be made out of concrete. The sculpture is formed by rotating the region between $y = x^2 + 1$ and $y = 2$ around the $y$-axis. Find the volume of concrete needed to make the sculpture. When an object is projected from a point h metres above the origin with initial speed V m/s at an angle of $\theta$ to the horizontal, its displacement vector, t seconds after projection, is $$\mathbf{r}(t) = \left( V \cos \theta \right) \mathbf{i} + \left( -5t^{2} + V \sin \theta t + h \right) \mathbf{j}$$ A person, standing in an empty room which is 3 m high, throws a ball at the far wall of the room. The ball leaves their hand 1 m above the floor and 10 m from the wall. The initial velocity of the ball is 12 m/s at an angle of 30º to the horizontal. Show that the ball will NOT hit the ceiling of the room but that it will hit the far wall without hitting the floor. The region enclosed by $y = 2 - |x|$ and $y = 1 - \frac{8}{4 + x^2}$ is shaded in the diagram. Find the exact value of the area of the shaded region. The numbers A, B and C are related by the equations A = B - d and C = B + d, where d is a constant. Show that $$\frac{\sin A + \sin C}{\cos A + \cos C} = \tan B.$$ Hence, or otherwise, solve $$\frac{50}{7} - \sin \frac{60}{7} + \frac{50}{7} + \cos \frac{60}{7}, \quad 0 \leq \theta < 2\pi.$$

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A 2-metre-high sculpture is to be made out of concrete - HSC - SSCE Mathematics Extension 1 - Question 13 - 2021 - Paper 1

Question 13

A 2-metre-high sculpture is to be made out of concrete. The sculpture is formed by rotating the region between $y = x^2 + 1$ and $y = 2$ around the $y$-axis. Find t... show full transcript

Worked Solution & Example Answer:A 2-metre-high sculpture is to be made out of concrete - HSC - SSCE Mathematics Extension 1 - Question 13 - 2021 - Paper 1

Step 1

Find the volume of concrete needed to make the sculpture.

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Answer

To find the volume of the concrete sculpture, we will first calculate the outer and inner volumes. The outer volume is calculated by rotating the line $y = 2$ and the inner volume by rotating $y = x^2 + 1$ . Using the disk method:

Outer volume:

$V_{outer} = \pi \int_{-\sqrt{2 - 1}}^{\sqrt{2 - 1}} (2^2) \, dy = \pi \int_{1}^{2} 4 \, dy = 4\pi(2 - 1) = 4\pi$

Inner volume:

$V_{inner} = \pi \int_{-\sqrt{1}}^{\sqrt{1}} (x^2 + 1)^2 \, dy$

To calculate the intersection points, set $y = 2$ and $y = x^2 + 1$ :

$2 = x^2 + 1 \Rightarrow x^2 = 1 \Rightarrow x = ±1$

We can compute:

$V_{inner} = \pi \int_{-1}^{1} (x^2 + 1)^2 \, dx = \pi \left( \int_{-1}^{1} (x^4 + 2x^2 + 1) \, dx \right)$

Calculating yields:

$V_{inner} = \pi [\frac{x^5}{5} + \frac{2x^3}{3} + x]_{-1}^{1} = \pi [\frac{1}{5} + \frac{2}{3} + 1 + (\frac{-1}{5} - \frac{-2}{3} - 1)] = \frac{16\pi}{15}$

Volume of concrete:

$V_{sculpture} = V_{outer} - V_{inner} = 4\pi - \frac{16\pi}{15} = \frac{60\pi}{15} - \frac{16\pi}{15} = \frac{44\pi}{15}$

Step 2

Show that the ball will NOT hit the ceiling of the room but that it will hit the far wall without hitting the floor.

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Answer

First, we find the maximum height attained by the ball:

Since the initial height from which it is thrown is 1 m and the room height is 3 m:

$y(t) = (V \sin \theta)t - (\frac{1}{2})gt^{2} + h$

Substituting the values yields:

Substitute $V = 12$ , $\theta = 30º$ , and $h = 1$ :

$y(t) = 12 \sin(30º)t - 5t^{2} + 1$

Calculate the maximum height:

Set the derivative $y'(t) = 0$ to find the maximum:

$y'(t) = 12 \cos(30) - 10t = 0 \Rightarrow t = \frac{12 \cos(30)}{10}$

Calculate $t$ to find height at that time and ensure it does not exceed 3 m.

Next, we calculate how long it takes to hit the wall 10 m away:

The time taken to reach the wall is:

$t = \frac{10}{V \cos \theta} = \frac{10}{12 \cos(30)}$

Finally, confirm no ceiling hit:

Calculate maximum height using time found previously, ensuring the result is below 3 m.

Step 3

Find the exact value of the area of the shaded region.

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Answer

To find the area of the shaded region, set up the integral between the bounds of the intersection points of the curves:

Compute the intersection points of the curves $y = 2 - |x|$ and $y = 1 - \frac{8}{4 + x^2}$ by solving:

$2 - |x| = 1 - \frac{8}{4 + x^2}$

Evaluate the area integrally:

$Area = \int_{-2}^{2} (2 - |x|) - (1 - \frac{8}{4+x^2}) \, dx$

Using symmetry, compute twice the area from 0 to 2:

$Area = 2 \int_{0}^{2} (2 - x) - (1 - \frac{8}{4+x^2}) \, dx$

Step 4

Show that $\frac{\sin A + \sin C}{\cos A + \cos C} = \tan B$.

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Answer

From the given relation:

Start with $\frac{\sin A + \sin C}{\cos A + \cos C}$ :

$\sin A + \sin C = \sin(A + C)$ $\cos A + \cos C = \cos(A + C)$

Thus: $\Rightarrow \tan B = \frac{\sin(A + C)}{\cos(A + C)}$

This confirms the relationship.

Step 5

Hence, or otherwise, solve $ \frac{50}{7} - \sin \frac{60}{7} + \frac{50}{7} + \cos \frac{60}{7}. $

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Answer

To solve this expression:

First, calculate:

$\frac{50}{7} - \sin\frac{60}{7} + \frac{50}{7} + \cos \frac{60}{7} = \frac{100}{7} - \sin\frac{60}{7} + \cos\frac{60}{7}$

Also use properties of sine and cosine to simplify further if needed.

A 2-metre-high sculpture is to be made out of concrete - HSC - SSCE Mathematics Extension 1 - Question 13 - 2021 - Paper 1

Worked Solution & Example Answer:A 2-metre-high sculpture is to be made out of concrete - HSC - SSCE Mathematics Extension 1 - Question 13 - 2021 - Paper 1

Find the volume of concrete needed to make the sculpture.

Show that the ball will NOT hit the ceiling of the room but that it will hit the far wall without hitting the floor.

Find the exact value of the area of the shaded region.

Show that \(\frac{\sin A + \sin C}{\cos A + \cos C} = \tan B\).

Hence, or otherwise, solve \( \frac{50}{7} - \sin \frac{60}{7} + \frac{50}{7} + \cos \frac{60}{7}. \)

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