The volume $$\(V \mathrm{~m}^{3}\)$$ of a large circular mound of iron ore of radius $$\(r \mathrm{~m}\)$$ is modelled by the equation $$\(V=\frac{3}{2}\left(r-\frac{1}{2}\right)^{3}-1\)$$ for $$\(r \geqslant 2\)$$. Iron ore is added to the mound at a constant rate of $$\(1.5 \mathrm{~m}^{3}\)$$ per second. Find the volume of the mound at the instant when the radius is increasing at $$\(0.1 \mathrm{~m}\)$$ per second. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w21_qp_12 Year：2021 Question No：9(b)
Answer:
\(\frac{\mathrm{d} V}{\mathrm{~d} r}\) or their \(\frac{\mathrm{d} V}{\mathrm{~d} r}=\frac{1.5}{0.1}\) or 15 OR \(0.1=\frac{1.5}{\text { their } \frac{\mathrm{d} V}{\mathrm{~d} r}}\left[=\frac{2 \times 1.5}{9\left(r-\frac{1}{2}\right)^{2}}\right.\) OE \(]\)
\(\left[\frac{9}{2}\left(r-\frac{1}{2}\right)^{2}=15 \Rightarrow\right] r=\frac{1}{2}+\sqrt{\frac{10}{3}}\)
[Volume =] 8.13 AWRT
\(\left[\frac{9}{2}\left(r-\frac{1}{2}\right)^{2}=15 \Rightarrow\right] r=\frac{1}{2}+\sqrt{\frac{10}{3}}\)
[Volume =] 8.13 AWRT
Knowledge points:
1.1.3 solve quadratic equations, and quadratic inequalities, in one unknown (By factorising, completing the square and using the formula.)
1.7.3 apply differentiation to gradients, tangents and normals, increasing and decreasing functions and rates of change (Including connected rates of change, e.g. given the rate of increase of the radius of a circle, find the rate of increase of the area for a specific value of one of the variables.)
Solution:
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