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1.7 Differentiation

1.7 Differentiation

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Question 26

A curved profile for a glass sculpture C C\,C is modeled by the parametric equations

x=14t2+1,y=2t−8t,t>0 x = \frac{1}{4}t^2 + 1, \quad y = 2t - \frac{8}{t}, \quad t > 0 x=41​t2+1,y=2t−t8​,t>0

The curve C C\,C intersects the xxx-axis at the point QQQ.

a.

Find the xxx-coordinate of QQQ.

[1]
b.

The line l l\,l is the normal to C C\,C at the point PPP. Given that t=4t = 4t=4 at PPP:

Write down the coordinates of PPP.

[2]
c.

Using calculus, show that an equation of l l\,l is

4x+5y=50 4x + 5y = 50 4x+5y=50
[4]
d.

The region R R\,R is bounded by the curve CCC, the line lll, and the xxx-axis.

Using algebraic integration, find the exact volume of the solid of revolution formed when the region R R\,R is rotated through 2π 2\pi\,2π radians about the xxx-axis.

[7]
Markscheme

1.7 Differentiation Questions

  1. A Level
  2. /Maths
  3. /1.7 Differentiation

425 exam-style questions on OCR A Level Maths 1.7 Differentiation, covering 1.7.1 Derivative as gradient of the tangent, 1.7.2 Gradient of the tangent at a point, 1.7.3 Sketching the gradient function, 1.7.4 Second derivatives, 1.7.5 Second derivative as rate of change of gradient, 1.7.6 Convex, concave and points of inflection (A-level only), 1.7.7 Differentiation from first principles for powers of x, 1.7.8 Differentiation from first principles for sin x and cos x (A-level only), 1.7.9 Differentiating x^n, 1.7.10 Differentiating e^(kx) and a^(kx) (A-level only), 1.7.11 Differentiating trigonometric functions (A-level only), 1.7.12 Derivative of ln x (A-level only), 1.7.13 Tangents and normals, 1.7.14 Stationary points, 1.7.15 Increasing and decreasing functions, 1.7.16 Points of inflection (A-level only), 1.7.17 Product and quotient rules (A-level only), 1.7.18 Chain rule (A-level only), 1.7.19 Parametric and implicit differentiation (A-level only), 1.7.20 Constructing differential equations (A-level only), and 1.7 Differentiation. Each one has a worked solution and a mark scheme showing where the marks go.

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