IB DP Math AI -Topic SL 5.12 Area of the region enclosed by a curve IB Style Question HL Paper 2

Detail Solution: –

Part (a): Find an explicit expression for the volume of water in terms of h

The interior of the vase is formed by rotating a region around the y-axis. The region is bounded by the curve \( y = \frac{1}{2}x^2 – 1 \), the lines \( x = 0 \), \( y = 0 \), and \( y = 15 \). Since the rotation is about the y-axis, we’ll use the method of cylindrical shells or the disk method to compute the volume, but because we’re interested in the volume up to a height \( h \), the disk method fits naturally here (with \( x \) as a function of \( y \)).

First, interpret the boundaries:
 \( y = \frac{1}{2}x^2 – 1 \): This is the curve defining the vase’s shape.
 \( x = 0 \): The y-axis, which is the axis of rotation.
 \( y = 0 \): A horizontal boundary.
 \( y = 15 \): The top of the vase.

Solve for \( x \) in terms of \( y \) from the curve:
\[
y = \frac{1}{2}x^2 – 1
\]
\[
y + 1 = \frac{1}{2}x^2
\]
\[
x^2 = 2(y + 1)
\]
\[
x = \sqrt{2(y + 1)} \quad (\text{since } x \geq 0 \text{ to the right of the y-axis})
\]

Now, determine the region being rotated. Let’s find where the curve intersects the boundaries:
 At \( y = 0 \):
\[
0 = \frac{1}{2}x^2 – 1 \implies 1 = \frac{1}{2}x^2 \implies x^2 = 2 \implies x = \sqrt{2}
\]
 At \( y = 15 \):
\[
15 = \frac{1}{2}x^2 – 1 \implies 16 = \frac{1}{2}x^2 \implies x^2 = 32 \implies x = \sqrt{32} = 4\sqrt{2}
\]

The curve starts below the x-axis at \( y = -1 \) when \( x = 0 \), rises to \( ( \sqrt{2}, 0) \) at \( y = 0 \), and reaches \( (4\sqrt{2}, 15) \) at \( y = 15 \). The region “bounded by” these lines suggests we need the area between \( x = 0 \) and the curve from \( y = 0 \) to \( y = 15 \), rotated around the y-axis. However, the curve dips below \( y = 0 \) (from \( y = -1 \) to \( y = 0 \)), and we need to clarify the vase’s base. A vase typically has a solid bottom, so let’s assume the region from \( y = 0 \) to \( y = 15 \) between \( x = 0 \) and \( x = \sqrt{2(y + 1)} \) is rotated, forming a container open at \( y = 15 \).

For a solid of revolution about the y-axis, the volume from \( y = 0 \) to \( y = h \) (where \( 0 \leq h \leq 15 \)) uses the disk method:
\[
V = \pi \int_{0}^{h} [x(y)]^2 \, dy
\]
Substitute \( x = \sqrt{2(y + 1)} \):
\[
x^2 = 2(y + 1)
\]
\[
V = \pi \int_{0}^{h} 2(y + 1) \, dy
\]
\[
= 2\pi \int_{0}^{h} (y + 1) \, dy
\]
\[
= 2\pi \left[ \frac{y^2}{2} + y \right]_{0}^{h}
\]
\[
= 2\pi \left( \frac{h^2}{2} + h – 0 \right)
\]
\[
= 2\pi \left( \frac{h^2}{2} + h \right)
\]
\[
V = \pi h^2 + 2\pi h
\]

This is the volume of water when the height is \( h \) cm. Let’s verify: at \( h = 0 \), \( V = 0 \), which makes sense (no water); at \( h = 15 \), we’ll compute later to check consistency.

Part (b): Find the time taken to completely fill the vase

“Completely fill the vase” means up to \( h = 15 \) cm, the top boundary. Calculate the total volume:
\[
V = \pi (15)^2 + 2\pi (15)
\]
\[
= \pi \cdot 225 + 30\pi
\]
\[
= 255\pi \, \text{cm}^3
\]

The vase fills at a rate of \( 20 \, \text{cm}^3/\text{s} \):
\[
\text{Time} = \frac{\text{Total Volume}}{\text{Fill Rate}} = \frac{255\pi}{20}
\]
\[
= \frac{255 \cdot \pi}{20} = 12.75\pi \, \text{seconds}
\]

Numerically, \( \pi \approx 3.1416 \), so:
\[
12.75 \cdot 3.1416 \approx 40.055 \, \text{seconds}
\]
Let’s keep it exact as \( 12.75\pi \) seconds unless a decimal is required.

 Part (c): Find the rate at which the height is changing when $ h = 10 \, \text{cm}$

We need \( \frac{dh}{dt} \) when \( h = 10 \), given \( \frac{dV}{dt} = 20 \, \text{cm}^3/\text{s} \). Use the chain rule:
\[
\frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}
\]
\[
\frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dh}}
\]

Differentiate the volume:
\[
V = \pi h^2 + 2\pi h
\]
\[
\frac{dV}{dh} = 2\pi h + 2\pi = 2\pi (h + 1)
\]

At \( h = 10 \):
\[
\frac{dV}{dh} = 2\pi (10 + 1) = 2\pi \cdot 11 = 22\pi
\]

\[
\frac{dh}{dt} = \frac{20}{22\pi} = \frac{10}{11\pi} \, \text{cm/s}
\]

Numerically:
\[
\frac{10}{11 \cdot 3.1416} \approx \frac{10}{34.5576} \approx 0.289 \, \text{cm/s}
\]

————Markscheme—————–

7. (a) attempt to use $V=\pi\int x^{2}dy$
$x^{2}=2y+2$ or any reasonable attempt to find x in terms of y
\[V=\pi\int_{0}^{h}(2y+2)dy\]

$$\int2y+2~dy=y^{2}+2y$$

$$V=\pi[y^{2}+2y]_{0}^{h}$$
$$=\pi(h^{2}+2h)$$

(b) volume of vase $=\pi(15^{2}+2\times15) = 801.106…$

(time to fill vase = $\frac{801.106…}{20} = 40.1$ (40.0553…) (seconds)

(c) EITHER

$\frac{dh}{dt}=\frac{dV}{dt}\times\frac{dh}{dV}$
$\frac{dV}{dh}=\pi(2h+2)$

OR

differentiating $V=\pi(h^{2}+2h)$ implicitly

$\frac{dV}{dt}=\pi(2h+2)\frac{dh}{dt}$

THEN

$\frac{dh}{dt}=20\times\frac{1}{\pi(2h+2)}$

substituting $h=10$ seen anywhere
0.289 (0.289372…) cm $s^{-1}$