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 e³ˡˡⁿ² = 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., [that's 2^a 3^b 5^c 7^d 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
