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The gradient of the the perpendicular line to a curve at any point `P(x,y)` is twice the gradient of the line joining `P` and the point `Q(1,1)`.
The coordinate of points on the curve satisfy the differential equation
`A`
`m_T = dy/dx`♦♦♦ Mean mark 25%.
COMMENT: Easily the lowest mean mark in the 2011 MC section.
`m_⊥ = −1/(m_T) = – (dx)/(dy)`
`m_(PQ) = (y-1)/(x-1)`
| `- (dx)/(dy)` | `= 2m_(PQ)` |
| `-(dx)/(dy)` | `= (2(y-1))/(x-1)` |
| `- (dy)/(dx)` | `= (x – 1)/(2(y-1))` |
| `0` | `=dy/dx + (x – 1)/(2(y-1))` |
`=> A`
Find the equation of the line perpendicular to the graph of `cos (y) + y sin(x) = x^2` at `(0, -pi/2)`. (4 marks)
`y_N = (-2)/pi x – pi/2`
`d/(dx) (cos(y)) + d/(dx) (y sin(x)) = d/(dx) (x^2)`
`-sin(y) *(dy)/(dx) + sin(x) *(dy)/(dx) + y cos (x) = 2x`
| `dy/dx(sin(x)-sin(y))` | `=2x-ycos(x)` | |
| `dy/dx` | `=(2x-ycos(x))/(sin(x)-sin(y))` |
`dy/dx |_(x=0, y=pi/2) =(0-(pi/2)(1))/(0-1)=pi/2`
`m_T = pi/2\ \ =>\ \ m_⊥ = (-2)/pi`
`:.\ text(Equation of ⊥ line:)`
| `y – ((-pi)/2)` | `=(-2)/pi (x-0)` | |
| `y` | `=((-2)/pi) x – pi/2` |
Find the gradient of the line perpendicular to the tangent to the curve defined by `y = -3e^(3x) e^y` at the point `(1, -3)`. (3 marks)
`4/9`
| `(dy)/(dx)` | `= -3 d/(dx) (e^(3x) e^y)` |
| `(dy)/(dx)` | `= -9e^(3x) e^y + -3 e^(3x) e^y (dy)/(dx)` |
| ` -9e^(3x) e^y` | `=(dy)/(dx) (1 + 3e^(3x) e^y)` |
| `(dy)/(dx)` | `= (-9e^(3x) e^y)/(1 + 3e^(3x) e^(-3))` |
| `text{At (1, –3):}` | |
| `:. m_text(norm)` | `= (1 + 3e^(3 xx 1) e^(-3))/(9e^(3 xx 1) e^(-3))` |
| `= (1 + 3e^3 e^(-3))/(9e^3 e^(-3))` | |
| `= 4/9` |
The gradient of the line that is perpendicular to the graph of the relation `3y^2 - 5xy - x^2 = 1` at the point `(1, 2)` is
A. `-1/12`
B. `12/7`
C. `21`
D. `-7/12`
E. `-7/13`
`D`
| `6y *(dy)/(dx) – (5y + 5x* (dy)/(dx)) – 2x` | `= 0` |
| `6y *(dy)/(dx) – 5x *(dy)/(dx)` | `= 2x + 5y` |
| `6 xx 2 xx m_T – 5 xx 1 xx m_T` | `= 2 xx 1 + 5 xx 2` |
| `(12 – 5)m_T` | `= 12` |
| `7 m_T` | `= 12` |
| `m_T` | `= 12/7` |
`:.m_N = -7/12`
`=> D`