June 2005 - C4Question 1: Re-write as: (4 - 9x)½ We can't expand this, because the term on the LEFT is not "1", so first we have to re-write it: (4 - 9x)½ => [4(1 - 9x/4)]½ => 4½(1 - 9x/4)½ => 2(1 - 9x/4)½ So, we expand (1 - 9x/4)½ first, then we multiply through by "2": Starting with the Pascal Coefficients: 1 ½ ½(-½) ½(-½)(-3/2)
2! 3!
Then adding in the powers of (-9x/4): 1(-9x/4)0 ½(-9x/4)1 ½(-½)(-9x/4)2 ½(-½)(-3/2)(-9x/4)3
2! 3!
Then simplifying [taking lots of care with things like: And remembering we need: 2(1 - 9x/4)½ so we have to multiply our answer by "2"... The validity of our expansion is: -1 < 9x/4 < 1 => ... < x < ...
Question 2: Differentiating: x² + 2xy - 3y² + 16 = 0 => 2x + [(2)(y) + (dy/dx)(2x)] - 6y dy/dx = 0 We want the points where dy/dx = 0: => 2x + [(2)(y) + (0) (2x)] - 6y (0) = 0 => 2x + 2y = 0 => x = -y Now we substitute this back into the original equation: x² + 2x(-x) - 3(-x)² + 16 = 0 => x = ... or x = ... And, remembering that x=-y, the points are: (..., -...) and (-..., ...)
Question 4: This is easy if you follow the steps carefully... Step 1: Change the limits: x = sin θ
x=½ => θ=π/6
x=0 => θ=0
Step 2: Replace the "dx": x = sin θ
dx = cos θ dθ
⌠π/6
│ 1 cos θ dθ
│ (1-x²)3/2
⌡0
Step 3: Now replace the "x": x = sin θ
⌠π/6
│ 1 cos θ dθ
│ (1-sin²θ)3/2
⌡0
Step 4: Simplify and integrate. You wont have to substitute back
because you changed the limits...
[Don't forget that ∫ sec²x = tan x]
Question 5, part (a): We have to use parts: [Leave 1st][Integrate 2nd] - ∫ [Differentiate 1st][Integrate 2nd] Which gives: ∫ xe2x = [x][½e2x] - ∫ [1][½e2x]
= ½xe2x - ½∫ e2x
= ½xe2x - ¼e2x
= ¼(2x-1)e2x
Putting back the limits: ┌ ┐1
│ ¼(2x-1)e2x │
└ ┘0
And working out the answer...
Question 6, part (a): Easy peasy! dy/dt = 4 cos t sin t
dx/dt = -2 cosec²t
=> dy/dx = 4 cos t sin t = 4 cos t sin t = 4 cos t sin³t
-2 cosec²t -2/sin²t -2
Which cancels down... Question 6, part (b): So, to find the equation of the tangent when x = 2 cot (¼π) = ...
y = 2 sin²(¼π) = ....
dy/dx = -2 cos (¼π) sin³(¼π) = .....
So, the tangent has gradient ..... through (..., ....) Question 6, part (c): Now, when we want to convert a (trig) parametric equation into a Cartesian equation, we: Step 1: Make sure the angles are the same
Which they already are...
Step 2: Make the trig function the subject of each equation:
cot t = x/2 sin²t = y/2
Step 3: Now find a trig ID that connects these trig fns
1 + cot²t = cosec²t
Step 4: Substitute into the trig ID:
1 + cot²t = cosec²t
1 + (x/2)² = (2/y)
Then simplify...
Question 7, part (a): To find where they meet, we make the equations equal: ┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐
│ 3 │ │ 1 │ │ 0 │ │ 1 │
│ 1 │ + λ│-1 │ = │ 4 │ + μ│-1 │
│ 2 │ │ 4 │ │-2 │ │ 0 │
└ ┘ └ ┘ └ ┘ └ ┘
Then we normally read Row1 and Row2: ┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐
│ 3 │ │ 1 │ │ 0 │ │ 1 │ <- 3 + λ = 0 + μ
│ 1 │ + λ│-1 │ = │ 4 │ + μ│-1 │ <- 1 - λ = 4 - μ
│ 2 │ │ 4 │ │-2 │ │ 0 │
└ ┘ └ ┘ └ ┘ └ ┘
The problem in this particular question is that you can't solve these two equations because when you add the equations, all the λ's and the μ's disappear... Never mind, let's use Row2 and Row3 instead: ┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐
│ 3 │ │ 1 │ │ 0 │ │ 1 │
│ 1 │ + λ│-1 │ = │ 4 │ + μ│-1 │ <- 1 - λ = 4 - μ
│ 2 │ │ 4 │ │-2 │ │ 0 │ <- 4 + 4λ = -2
└ ┘ └ ┘ └ ┘ └ ┘
R3 solves to give λ = ... Check in R1 [don't forget the check!!!] Finally, putting λ=... back into equation (1) to find B... Question 7, part (b): To find the angle between two lines, we use the dot product of the direction vectors: ┌ ┐ ┌ ┐
│ 1 │ │ 1 │
│-1 │ ● │-1 │ = √(1)²+(-1)²+(4)² √(1)²+(-1)² cos θ
│ 4 │ │ 0 │
└ ┘ └ ┘
Which gives cos θ = ... Question 7, part (d): Draw two lines (called l1 and l2) and mark the point where they intersect as "B". Mark a point "A" which lies on l1 and a point "C" which lies on l2 Now, by adding in two parallel lines (dotted), locate "D" so that ABCD is a parallelogram... The first thing you will realise is that AD is the same as BC We can easily find the vector BC: BC = C - B ┌ ┐ ┌ ┐ ┌ ┐ │ 5 │ │ 2 │ │ 3 │ BC = │-1 │ - │ 2 │ = │-3 │ │-2 │ │-2 │ │ 0 │ └ ┘ └ ┘ └ ┘ Which means that AD is the same... AD = D - A
┌ ┐ ┌ ┐ ┌ ┐
│ 3 │ │ x │ │ 3 │
│-3 │ = │ y │ - │ 1 │
│ 0 │ │ z │ │ 2 │
└ ┘ └ ┘ └ ┘
So D = (..., ..., ...)
Question 8, part (a): The rate at which liquid is going "in" =20 and the rate at which liquid is going "out" = kV So the overall change in volume (dV/dt) is the "in" minus the "out" Question 8, part (b): Separating the variables: dV = 20 - kV . . . . . (1)
dt
⌠ 1 dV = ⌠ 1 dt
⌡ 20-kV ⌡
Our guess for the left integral is: ln | 20 - kV |: differentiate
ln | 20 - kV | -------------------> -k
<------------------- 20 - kV
integrate
So, we need to divide by "-k": -1/k ln A | 20-kV | = t
▲
│
constant placed inside the ln
Now, putting in t=0, V=0 (initially empty), to find A... Then re-arranging to make V the subject... -1/k ln 1/20 |20 - kV| = t => ln 1/20 |20 - kV| = -kt => 1/20 (20 - kV) = e-kt => 20 - kV = ... => kV = 20 - ... => V = 20/k - ... . . . . (2) Question 8, part (c): We are told that, when t=5, dV/dt = 10 But, we can't put that into equation (1), because equation (1) gives dV/dt in terms of "V" and not in terms of "t" So, instead, we differentiate equation (2): dV/dt = 20 e-kt And then substitute t=5, dV/dt = 10 into this to find k... Then we can re-write equation (2) without any unknowns... |