Dr. J's Maths.com
Where the techniques of Maths
are explained in simple terms.

Trigonometric functions - differentiation - max/min questions.
Test Yourself 1 - Solutions.


 

Theoretical 1. (i)

(ii)

  2.
  3.

(i)

(ii) Length of EG is simply the x value for G - so solve the equation from (i) with y = 1 (easiest by substitution).

(iii)

For area of a trapezium (if you have forgotten) see
Geometry - Quadrilaterals.

(iv) .

θ 0 π/6 π/4
dA/dθ -0.5 0 0.414

Change of gradient direction around π/6 from -ve to +ve - therefore a minimum area at θ = π/6.

Also see the graph of this function:
  4. (i)

(ii)

(iii)

(iv)

θ π/4 π/3 π/2
dA/dθ. 1/√2 0 -1

Change of gradient from +ve to -ve
∴ max at (π/3, (3√3)/4).

θ 3π/2 5π/3 2π
dA/dθ -1 0 2

Change of gradient from -ve to +ve
∴ min at (5π/3, (-3√3)/4).

θ π/4 π 5π/4
dA/dθ -2.71 0 -1.29

No change in the sign of the gradient and both the 1st and 2nd derivatives equal zero at x = π
∴ horiz point of inflection at (π, 0).

(v)

Practical situations 5.

(i)

 

(ii)

  6. (i)

(ii)

Type 2:
2D shapes - sectors.		 7.

(i) Note that the structure of the expression to be proved has minus sign - that implies the proof involves a difference.

In the diagram, CF = OF - OC (so there is the difference!!).

In Δ OFE, with angle α,

In Δ OCD, with angle π/3, we know no side lengths.
But we do know CD = EF so:

(ii)

(iii)

  8. (i) To calculate the perimeter, we need arc lengths for both AD (xθ)and BC (4xθ).

Perimeter = BC + AD + AB + CD

= 4xθ + xθ + 6x

240 = 5xθ + 6x

(ii)

(iii)

  9. (i)

(ii)

Type 3:
3D shape.

 10.

(i)

(ii)

 

(iii)

Type 4: Rates

11.

(ii)

 

(iii)

 

(iv)