The cross-section of a precision-engineered lens is modeled by a curve y=f(x)y = f(x)y=f(x), for x>0x > 0x>0. The rate of change of the gradient of the profile is given by
f′′(x)=154x7−6x f''(x) = \frac{15}{4\sqrt{x^7}} - 6x f′′(x)=4x715−6xA point P(1,1.5)P(1, 1.5)P(1,1.5) lies on the boundary of the lens profile.
Given that the gradient of the curve f′(x)=0.75f'(x) = 0.75f′(x)=0.75 at point PPP,
find the equation of the normal at PPP, writing your answer in the form y=mx+cy = mx + cy=mx+c, where mmm and ccc are constants,
find f(x)f(x)f(x).
864 exam-style questions on Edexcel A Level Maths Integration, covering 11.1 Integrating Standard Functions, 11.2 Integrating f(ax + b), 11.3 Using Trigonometric Identities, 11.4 Reverse Chain Rule, 11.5 Integration by Substitution, 11.6 Integration by Parts, 11.7 Partial Fractions, 11.8 Finding Areas, 11.9 The Trapezium Rule, 11.10 Solving Differential Equations, and 11.11 Modelling with Differential Equations. Each one has a worked solution and a mark scheme showing where the marks go.