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9709 P12 - Jun 2014 - Q9
1291
The diagram shows part of the curve \(y = 8 - \sqrt{4 - x}\) and the tangent to the curve at \(P(3, 7)\).
(i) Find expressions for \(\frac{dy}{dx}\) and \(\int y \, dx\).
(ii) Find the equation of the tangent to the curve at \(P\) in the form \(y = mx + c\).
(iii) Find, showing all necessary working, the area of the shaded region.
Solution
(i) To find \(\frac{dy}{dx}\), differentiate \(y = 8 - \sqrt{4-x}\). Using the chain rule, \(\frac{dy}{dx} = -\frac{1}{2}(4-x)^{-\frac{1}{2}} \times (-1)\).
For \(\int y \, dx\), integrate \(y = 8 - \sqrt{4-x}\). The integral is \(8x - \frac{(4-x)^{\frac{3}{2}}}{\frac{3}{2}} \times (-1)\).
(ii) The equation of the tangent at \(P(3, 7)\) is found using the point-slope form. The slope \(m = \frac{1}{2}\), so the equation is \(y - 7 = \frac{1}{2}(x - 3)\), which simplifies to \(y = \frac{1}{2}x + \frac{5}{2}\).
(iii) The area under the curve from 0 to 3 is \(\int_{0}^{3} (8 - \sqrt{4-x}) \, dx = \frac{58}{3}\). The area under the line is \(\frac{1}{2}(\frac{5}{2} + 7) \times 3 = \frac{75}{4}\). The area of the shaded region is \(\frac{58}{3} - \frac{75}{4} = \frac{7}{12}\).
