Standard +0.3 This is a straightforward related rates problem requiring differentiation of the surface area formula with respect to time and substitution of given values. While it involves the chain rule (dA/dt = dA/dr × dr/dt), the formula is provided, the setup is standard, and the calculation is direct with no conceptual obstacles—slightly easier than average for C3.
3 A giant spherical balloon is being inflated in a theme park. The radius of the balloon is increasing at a rate of 12 cm per hour. Find the rate at which the surface area of the balloon is increasing at the instant when the radius is 150 cm . Give your answer in \(\mathrm { cm } ^ { 2 }\) per hour correct to 2 significant figures. [0pt]
[Surface area of sphere \(= 4 \pi r ^ { 2 }\).]
numerical or algebraic; using multiplication or division
Obtain \(8r \times 150 \times 12\) and hence 45000 or 14400\(\pi\) or 14000\(\pi\)
A1
3 or equiv; or greater accuracy (45239); condone absence of units or use of wrong units
Or: Use \(r = 12t\) to show \(S = 576\pi t^2\)
B1
Attempt \(\frac{dS}{dt}\) and substitute for \(t\)
M1
Obtain \(1152\pi \times \frac{150}{12}\) and hence 45000 or 14400\(\pi\) or 14000\(\pi\)
A1
(3) or equiv; or greater accuracy (45239); condone absence of units or use of wrong units
Either: State or imply $8\pi r$ as derivative | B1 | or equiv
Attempt to connect 12 and their derivative | M1 | numerical or algebraic; using multiplication or division
Obtain $8r \times 150 \times 12$ and hence 45000 or 14400$\pi$ or 14000$\pi$ | A1 | 3 or equiv; or greater accuracy (45239); condone absence of units or use of wrong units
Or: Use $r = 12t$ to show $S = 576\pi t^2$ | B1 |
Attempt $\frac{dS}{dt}$ and substitute for $t$ | M1 |
Obtain $1152\pi \times \frac{150}{12}$ and hence 45000 or 14400$\pi$ or 14000$\pi$ | A1 | (3) or equiv; or greater accuracy (45239); condone absence of units or use of wrong units
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3 A giant spherical balloon is being inflated in a theme park. The radius of the balloon is increasing at a rate of 12 cm per hour. Find the rate at which the surface area of the balloon is increasing at the instant when the radius is 150 cm . Give your answer in $\mathrm { cm } ^ { 2 }$ per hour correct to 2 significant figures.\\[0pt]
[Surface area of sphere $= 4 \pi r ^ { 2 }$.]
\hfill \mbox{\textit{OCR C3 2011 Q3 [3]}}