The point $$\((4,7)\)$$ lies on the curve $$\(y=\mathrm{f}(x)\)$$ and it is given that $$\(\mathrm{f}^{\prime}(x)=6 x^{-\frac{1}{2}}-4 x^{-\frac{3}{2}}\)$$. A point moves along the curve in such a way that the $$\(x\)$$-coordinate is increasing at a constant rate of $$\(0.12\)$$ units per second. Find the rate of increase of the $$\(y\)$$-coordinate when $$\(x=4\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_w20_qp_12 Year：2020 Question No：7(a)
Answer:
\(f^{\prime}(4)\left(=\frac{5}{2}\right)\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{\mathrm{d} y}{\mathrm{~d} x} \times \frac{\mathrm{d} x}{\mathrm{~d} t}\right) \rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} t}\right)=\frac{5}{2} \times 0.12\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} t}=\right) 0.3\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{\mathrm{d} y}{\mathrm{~d} x} \times \frac{\mathrm{d} x}{\mathrm{~d} t}\right) \rightarrow\left(\frac{\mathrm{d} y}{\mathrm{~d} t}\right)=\frac{5}{2} \times 0.12\)
\(\left(\frac{\mathrm{d} y}{\mathrm{~d} t}=\right) 0.3\)
Knowledge points:
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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