Question 9
In this question, all lengths are in metres.
(a) A particle $P$ has position vector $ \binom{2+12t}{5-5t} $ at time $t$ seconds, $t\geq0$.
(i) Write down the initial position vector of $P$. [1]
(ii) Find the speed of $P$. [2]
(iii) Determine whether $P$ passes through the point with position vector $ \binom{158}{-48}. $ [2]
(b)The diagram shows the triangle $OAC$. The point $B$ lies on $AC$ such that $ AB:AC=1:4. $ Given that $ \overrightarrow{OA}=\mathbf{a}, \quad \overrightarrow{OB}=\mathbf{b}, \quad \overrightarrow{OC}=\mathbf{c}, $ find $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. [3]
Solution
Step 1: Initial position vector of $P$
The position vector of $P$ is:
$$ \binom{2+12t}{5-5t} $$At $t=0$:
$$ \binom{2+12(0)}{5-5(0)} = \binom{2}{5} $$Thus, the initial position vector is:
$$ \boxed{\binom{2}{5}} $$Step 2: Find the speed of $P$
Differentiate the position vector with respect to $t$:
$$ \mathbf{v}(t) = \binom{\frac{d}{dt}(2+12t)}{\frac{d}{dt}(5-5t)} = \binom{12}{-5} $$The speed is the magnitude of the velocity vector:
$$ \text{Speed} = \sqrt{12^2+(-5)^2} $$ $$ = \sqrt{144+25} $$ $$ = \sqrt{169} $$ $$ =13 $$Thus, the speed of $P$ is:
$$ \boxed{13} $$metres per second.
Step 3: Determine whether $P$ passes through $\binom{158}{-48}$
Set the position vector equal to $ \binom{158}{-48}. $
$$ \binom{2+12t}{5-5t} = \binom{158}{-48} $$Equating components:
$$ 2+12t=158 $$ $$ 5-5t=-48 $$From the first equation:
$$ 12t=156 $$ $$ t=13 $$From the second equation:
$$ -5t=-53 $$ $$ t=\frac{53}{5}=10.6 $$The values of $t$ are different, so the particle does not pass through the point.
The particle $P$ does not pass through $\boxed{\binom{158}{-48}}$Step 4: Find $\mathbf{c}$ in terms of $\mathbf{a}$ and $\mathbf{b}$
Given:
$$ \overrightarrow{OA}=\mathbf{a}, \quad \overrightarrow{OB}=\mathbf{b}, \quad \overrightarrow{OC}=\mathbf{c} $$Since $ AB:AC=1:4, $ point $B$ divides $AC$ in the ratio $1:3$.
Now:
$$ \overrightarrow{AC} = \mathbf{c}-\mathbf{a} $$Therefore:
$$ \overrightarrow{OB} = \overrightarrow{OA} + \frac14\overrightarrow{AC} $$ $$ \mathbf{b} = \mathbf{a} + \frac14(\mathbf{c}-\mathbf{a}) $$Expand:
$$ \mathbf{b} = \mathbf{a} + \frac14\mathbf{c} - \frac14\mathbf{a} $$ $$ \mathbf{b} = \frac34\mathbf{a} + \frac14\mathbf{c} $$Multiply by $4$:
$$ 4\mathbf{b} = 3\mathbf{a} + \mathbf{c} $$Hence:
$$ \boxed{ \mathbf{c} = 4\mathbf{b} - 3\mathbf{a} } $$Conclusion
(i) Initial position vector: $ \binom{2}{5} $
(ii) Speed: $ 13 $ metres per second
(iii) The particle does not pass through $ \binom{158}{-48} $
(iv) $ \mathbf{c}=4\mathbf{b}-3\mathb
Question 10
(a) It is given that
2 + cos θ = x for 1 < x < 3 and 2 cosec θ = y for y > 2.
Find y in terms of x. [4]
(b) Solve the equation
3 cos(φ/2) = √3 sin(φ/2)
for -4π < φ < 4π. [5]
Solution
Step (a): Expressing y in terms of x
We are given:
2 + cos θ = x, 1 < x < 3
2 cosec θ = y, y > 2
Step 1: Solve for cos θ
cos θ = x - 2
Step 2: Use identity
cosec θ = 1 / sin θ
sin² θ + cos² θ = 1
sin² θ = 1 - (x - 2)²
sin θ = ±√(1 - (x - 2)²)
Step 3: Choose correct sign
Since y > 2, we take sin θ > 0:
sin θ = √(1 - (x - 2)²)
Step 4: Find y
y = 2 cosec θ = 2 / sin θ
y = 2 / √(1 - (x - 2)²)
Thus,
y = 2 / √(1 - (x - 2)²)
Step (b): Solve 3 cos(φ/2) = √3 sin(φ/2)
Step 1: Divide by cos(φ/2)
3 = √3 tan(φ/2)
tan(φ/2) = √3
Step 2: General solution
φ/2 = nπ + π/3
Step 3: Multiply by 2
φ = 2nπ + 2π/3
Step 4: Apply interval -4π < φ < 4π
-4π < 2nπ + 2π/3 < 4π
Divide by 2π:
-2 < n + 1/3 < 2
-7/3 < n < 5/3
n = -2, -1, 0, 1
Step 5: Find values of φ
n = -2: φ = -10π/3
n = -1: φ = -4π/3
n = 0: φ = 2π/3
n = 1: φ = 8π/3
Final answer:
φ = -10π/3, -4π/3, 2π/3, 8π/3
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