CAIE P1 2011 June — Question 2 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2011
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeRelated rates with spheres, circles, and cubes
DifficultyModerate -0.3 This is a straightforward related rates problem requiring differentiation of the sphere volume formula with respect to time, then substituting r=10. It's slightly easier than average because it's a single-step application of the chain rule with a given formula, though it does require understanding of implicit differentiation with respect to time.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
2 The volume of a spherical balloon is increasing at a constant rate of \(50 \mathrm {~cm} ^ { 3 }\) per second. Find the rate of increase of the radius when the radius is 10 cm . [Volume of a sphere \(= \frac { 4 } { 3 } \pi r ^ { 3 }\).]
AnswerMarks Guidance
\(\left(\frac{dv}{dr}\right) 4\pi r^2\)M1
\(= 4\pi \times 10^2\)A1 SOI at any point
\(\frac{dr}{dt} = \frac{dv}{dt} \div \frac{dv}{dr}\) OE usedM1 Correct link between differentials with \(\frac{dr}{dt}\) finally as subject
\(\frac{50}{4\pi \times 10^2} = \frac{1}{8\pi}\) or 0.0398A1 Allow \(\frac{50}{400\pi}\) Non-calculus methods \(\frac{0}{4}\) 
$\left(\frac{dv}{dr}\right) 4\pi r^2$ | M1 |
$= 4\pi \times 10^2$ | A1 | SOI at any point
$\frac{dr}{dt} = \frac{dv}{dt} \div \frac{dv}{dr}$ OE used | M1 | Correct link between differentials with $\frac{dr}{dt}$ finally as subject
$\frac{50}{4\pi \times 10^2} = \frac{1}{8\pi}$ or 0.0398 | A1 | Allow $\frac{50}{400\pi}$ Non-calculus methods $\frac{0}{4}$

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2 The volume of a spherical balloon is increasing at a constant rate of $50 \mathrm {~cm} ^ { 3 }$ per second. Find the rate of increase of the radius when the radius is 10 cm . [Volume of a sphere $= \frac { 4 } { 3 } \pi r ^ { 3 }$.]

\hfill \mbox{\textit{CAIE P1 2011 Q2 [4]}}
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