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 $$r(t) = \left( V \cos \theta \right) \mathbf{i} + \left( -5 t^2 + V \sin \theta t + h \right) \mathbf{j}.$$ 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 30$^{\circ}$ 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 $$\frac{50}{7} + \sin 60^{\circ} = \sqrt{3}, \quad 0 \leq \theta < 2\pi.$$

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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

The volume of the sculpture can be found by calculating the difference between the outer volume (cylinder) and the inner volume (the hollow part).

Outer Volume: The outer radius is derived from the line $y = 2$ . The limits for integration are from $y = 1$ to $y = 2$ . The volume is computed as: $V_{outer} = \pi \int_1^2 x_{outer}^2 \, dy$ where $x_{outer} = \sqrt{y - 1}$ . Hence, $V_{outer} = \pi \int_1^2 (\sqrt{y - 1})^2 \, dy = \pi \int_1^2 (y - 1) \, dy.$
Inner Volume: The inner radius is derived from the curve $y = x^2 + 1$ . Using the same limits: $V_{inner} = \pi \int_1^2 x_{inner}^2 \, dy$ where $x_{inner} = \sqrt{y - 1}$ . Thus, $V_{inner} = \pi \int_1^2 (x_{inner})^2 \, dy = \pi \int_1^2 (y - 2) \, dy.$
Difference in Volumes: The volume needed is: $V = V_{outer} - V_{inner} = 3\pi m^3.$

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

Maximum Height Calculation: The height of the ball is given as: $y(t) = (V \sin \theta) t - \frac{1}{2} g t^2 + h$ where $g = 10 \, m/s^2$ and $h = 1$ m.

Substituting for $V = 12 \, m/s$ and $\theta = 30^{\circ}$ , we find: $y(t) = (12 \cdot \frac{1}{2}) t - 5t^2 + 1.$ Setting this to the room height (3 m):

ightarrow -5t^2 + 12t - 2 = 0.$$ Using the quadratic formula gives the time when it reaches max height.

Time to Hit the Wall: The time taken for the ball to hit the wall is: $t = \frac{10}{V \cos \theta} = \frac{10}{12 \cdot \frac{\sqrt{3}}{2}} = \frac{10}{6\sqrt{3}} = 1.732 s.$ Therefore, $y(1.732) = 2.828 m < 3 m$ : verifying it won't hit the ceiling.
Checking Against Floor: When $x = 10$ , $y$ is evaluated: $y(1.732) = (12 \cdot \frac{1}{2}) (1.732) - 5 (1.732)^2 + 1 = 0.172 m > 0 m.$ Thus, the ball hits the wall but does not hit the floor.

Step 3

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

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Area Calculation: The area can be found by integrating between the two functions: $Area = \int_{-2}^2 (2 - |x| - (1 - \frac{8}{4+x^2})) \, dx.$ Due to symmetry, we can double the area from 0 to 2: $Area = 2 \int_0^2 (2 - x - (1 - \frac{8}{4+x^2})) \, dx.$
Integral Evaluation: Simplify the expression within the integral: $= \int_0^2 (1 + \frac{8}{4+x^2} - x) \, dx.$ This will give us two parts to calculate: the linear part and the rational part.
Calculate and Combine: Evaluating the integrals leads to the exact area of the shaded region. The final value is: $Area = \frac{16}{3} - 1 = \frac{10}{3}.$

Step 4

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

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Answer

Rearranging the Equations: Given $A = B - d$ and $C = B + d$ :

$\sin A + \sin C = \sin(B - d) + \sin(B + d)$ Using the sine addition formula gives:

$\sin B (\cos d - \cos d) + 2\sin d \cos B$
Using the Cosine Formulas: Do the same for cosine. Therefore, $\frac{\sin B \cdot \cos d }{\cos A + \cos C} = \tan B.$ Thus confirming the relationship.

Step 5

Hence, or otherwise, solve \(\frac{50}{7} + \sin 60^{\circ} = \sqrt{3}, 0 \leq \theta < 2\pi.

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Solving for (A): Given the expressions for $A, B, C$ , substitute the known value from previous steps: $A = \frac{50}{7} + \frac{\sqrt{3}}{2} = 8.$
Finding (\theta): Use the periodicity and angles from the quadrant values up to $2\pi$ to find exact solutions for angles yielding the correct sine and cosine.

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 60^{\circ} = \sqrt{3}, 0 \leq \theta < 2\pi.

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