IB Mathematics SL 3.1 The distance between two points AA SL Paper 2- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)(i)
The total volume is the sum of the volume of the hemisphere and the volume of the cone.
\(V=\dfrac{2}{3}\pi r^3+\dfrac{1}{3}\pi r^2h\)
Substitute \(r=3\) and \(h=8\).
\(V=\dfrac{2}{3}\pi(3)^3+\dfrac{1}{3}\pi(3)^2(8)\)
\(V=18\pi+24\pi\)
\(V=42\pi=131.946\ldots\)
Therefore, correct to 3 significant figures:
✅ \(V=132\text{ cm}^3\)
(a)(ii)
The wafer coating covers the curved surface of the cone.
First, use Pythagoras’ theorem to find the slant height \(l\).
\(l^2=r^2+h^2\)
\(l=\sqrt{3^2+8^2}\)
\(l=\sqrt{73}\)
The curved surface area of a cone is \(S=\pi rl\).
\(S=\pi(3)\sqrt{73}\)
\(S=3\pi\sqrt{73}=80.523\ldots\)
✅ Answer: \(S=80.5\text{ cm}^2\), correct to 3 significant figures
(b)
The total volume of the hemisphere and cone is \(120\text{ cm}^3\).
\(\dfrac{2}{3}\pi r^3+\dfrac{1}{3}\pi r^2h=120\)
Multiply throughout by \(3\).
\(2\pi r^3+\pi r^2h=360\)
\(\pi r^2h=360-2\pi r^3\)
Divide by \(\pi r^2\).
✅ Hence, \(h=\dfrac{360-2\pi r^3}{\pi r^2}\).
(c)
The slant height of the cone is:
\(l=\sqrt{r^2+h^2}\)
Substitute the expression for \(h\) from part (b).
\(l=\sqrt{r^2+\left(\dfrac{360-2\pi r^3}{\pi r^2}\right)^2}\)
The curved surface area of the cone is \(S=\pi rl\).
Therefore:
✅ \(S=\pi r\sqrt{r^2+\left(\dfrac{360-2\pi r^3}{\pi r^2}\right)^2}\)
(d)
Since \(r>0\) and \(h>0\), the possible values of \(r\) satisfy:
\(360-2\pi r^3>0\)
\(0<r<\sqrt[3]{\dfrac{180}{\pi}}\)
Using graphing technology to find the minimum of
\(S(r)=\pi r\sqrt{r^2+\left(\dfrac{360-2\pi r^3}{\pi r^2}\right)^2}\)
on this domain gives:
\(r=3.67084\ldots\)
\(S=44.4045\ldots\)
Therefore, to 3 significant figures:
✅ Answer: The minimum surface area is \(44.4\text{ cm}^2\), corresponding to \(r=3.67\text{ cm}\).