16N.2.SL.TZ0.S_6

pestleMathematicsAASLPaper 216N· sl-5-5-integration-introduction-areas-between-curve-and-x-axis, ahl-5-17-areas-under-curve-onto-y-axis-volume-of-revolution-(about-x-and-y-axes)source ↗

All lengths in this question are in metres.

Let $f(x) = - 0.8{x^2} + 0.5$ , for $- 0.5 \leqslant x \leqslant 0.5$ . Mark uses $f(x)$ as a model to create a barrel. The region enclosed by the graph of $f$ , the $x$ -axis, the line $x = - 0.5$ and the line $x = 0.5$ is rotated 360° about the $x$ -axis. This is shown in the following diagram.

N16/5/MATME/SP2/ENG/TZ0/06

Use the model to find the volume of the barrel.

[3]

The empty barrel is being filled with water. The volume $V{\text{ }}{{\text{m}}^3}$ of water in the barrel after $t$ minutes is given by $V = 0.8(1 - {{\text{e}}^{ - 0.1t}})$ . How long will it take for the barrel to be half-full?

[3]

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

attempt to substitute correct limits or the function into the formula involving

${y^2}$

eg $\,\,\,\,\,$ $\pi \int_{ - 0.5}^{0.5} {{y^2}{\text{d}}x,{\text{ }}\pi \int {{{( - 0.8{x^2} + 0.5)}^2}{\text{d}}x} }$

0.601091

volume $= 0.601{\text{ }}({{\text{m}}^3})$ A2 N3

[3 marks]

attempt to equate half their volume to $V$ (M1)

eg $\,\,\,\,\,$ $0.30055 = 0.8(1 - {{\text{e}}^{ - 0.1t}})$ , graph

4.71104

4.71 (minutes) A2 N3

[3 marks]

Examiners’ report

[N/A]