## FPM 20130301 

This is specifically for 20130321, but is likely to grow, simply because five questions does not a webpage make.                                                                                                               footnote 1

DJS 20130320

FPM Examples 6

1.  Differentiate y = e5x + 8e-4x wrt x;
Find and classify the stationary value (x and y).

2.   Show that      e3ln2 = 2³.
Integrate y = e5x + 8e-4x wrt x between ln 2 and ln 4, giving an exact rational answer.

3. Solve 2 cos (5θ - π/6) = 1 for 0<θ<π

4.  A circle has area 16 sq units. a) What is the length of the circumference of the circle? b) Develop the formula for the perimeter of a sector.
c) What angle is subtended by a sector of perimeter of 15 units? Exact answers will do nicely. Show that this is near to three quadrants.

5. a) Show that 4633 is a product of two prime factors.
b) Find exactly tan 2φ if cos φ = 36/85 and
c)  express 4633
tan 2φ  as a product and quotient of prime factors,   2a3b5c7d etc., where a,b,c,d are integer.

6. Write down the 3 expansions of cos 2θ; you have been told repeatedly to learn these!

7.  Find an expression for tan 3θ entirely in terms of tan θ.

8.  A portion of a large hill can be modelled as a rectangular-based pyramid ABCDE. Base plane ABCD has E vertically above A. AB=18, BC=39, AE=2. Units are kilometres.
a)  Find length EC
b)  Find angle CEBto the nearest degree
c)  A climb of tan-10.04 is the upper limit for some traffic. If a straight route at this limit starts at C, going up face CBE, and reaches CE first at point P [vertically above point Q on AB], what distance is BP?                                                                                              footnote 2

Extension: add a turning and now head for EC at the limitng design angle. where is this?

1 “One swallow doth not a summer make” from Aristotle, presumably in translation: "One swallow does not a summer make, nor one fine day; similarly one day or brief time of happiness does not make a person entirely happy."

2   Call BQ x and PQ h.    CQ=25h.   Draw ΔAPQ, ΔCQB and ΔBPQ. ΔBPQ is similar to ΔBEA.   Find h, find BP.

2.   127051/640
3.   2/5 nπ +π/30 ± π/15
4.   8√π,   2r + r𝜃,      2 + 𝜃 = 15/r = 15√π /4 =>
𝜃=1.479π => close to 1.5 π, 3 quadrants.
5.  4633=41x113
tan φ = 77/36 so tan 2φ = -77x72 / 41x113
7.  (3t - t ³) /  (1 - 3t ²)
8.  EC=43 exactly;  sin ACE = 2/43 => cos ECB = 3042/3354 => BEC =65.1º sin rule
h = 1.67211, BP=15.05. Call this 15km for an answer.
Hard. R lies on EC, T below it on AC and S lies on RT  so that ST = h. Let RS = y.
from
ΔRTC,  RC = 43(y+h)/2.  I eventually got (1845-625²)/18²= (1-(y+h)/2)² (1+h/2) and we know h. LHS of this is very close to -1200. Let k =y+h and this reduces to
y(k-y)²+9600=0, which has solutions of k±√(9600/k), entirely too big. Wrong !!

DJS 20130402

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