10 A triangle ABC is made from two thin rods hinged together at A and a piece of elastic which joins \(B\) and \(C\). \(A B\) is a 30 cm rod and \(A C\) is a 15 cm rod. The angle \(B A C\) is \(\theta\) radians as shown in the diagram.
The angle \(\theta\) increases at a rate of 0.1 radians per second.
Show mark scheme
Show mark scheme source
Question 10:
Answer Marks
Guidance
Answer Marks
Guidance
Length BC: \(l^2 = 30^2 + 15^2 - 2 \times 30 \times 15\cos\theta\) M1
If M0 awarded here allow SC1 for \(BC = \sqrt{675} = 15\sqrt{3}\) found using \(\theta = \frac{\pi}{3}\)
\(l^2 = 1125 - 900\cos\theta\), \(l = (1125 - 900\cos\theta)^{\frac{1}{2}}\) A1
Soi. Allow equivalent in metres. If working in metres \(l^2 = 0.1125 - 0.0900\cos\theta\)
\(\frac{dl}{d\theta} = \frac{1}{2}(1125 - 900\cos\theta)^{-\frac{1}{2}} \times 900\sin\theta\) M1
Attempt to use the chain rule
(any form) A1
Any form
\(\frac{d\theta}{dt} = 0.1\) B1
Soi eg from \(\theta = 0.1t\)
\(\frac{dl}{dt} = \frac{dl}{d\theta} \times \frac{d\theta}{dt} = \frac{450\sin\theta}{(1125 - 900\cos\theta)^{\frac{1}{2}}} \times 0.1\) M1
Using the chain rule to find \(\frac{dl}{dt}\)
When \(\theta = \frac{\pi}{3}\): substitute into \(\frac{dl}{d\theta}\) M1
Substitute \(\theta = \frac{\pi}{3}\) into their \(\frac{dl}{d\theta}\)
\(\frac{dl}{dt} = \frac{45\sin\frac{\pi}{3}}{(1125 - 900\cos\frac{\pi}{3})^{\frac{1}{2}}} = \left[\frac{45\sqrt{3}}{2 \times 15\sqrt{3}} = \frac{3}{2}\right]\) M1
\(1.5\ \text{cm s}^{-1}\) A1
Must have correct unit. Allow written as cm per second. Alternative: \(0.015\ \text{m s}^{-1}\)
[8]
Alternative method:
Answer Marks
Guidance
Answer Marks
Guidance
\(l^2 = 30^2 + 15^2 - 2 \times 30 \times 15\cos\theta\) M1
\(l^2 = 1125 - 900\cos\theta\) A1
If working in metres \(= 0.1125 - 0.0900\cos\theta\)
\(2l\frac{dl}{d\theta} = 900\sin\theta\) M1
Attempt to use implicit differentiation. Any form
(any form) A1
\(\frac{d\theta}{dt} = 0.1\) B1
Soi
\(\frac{dl}{dt} = \frac{dl}{d\theta} \times \frac{d\theta}{dt} = \frac{450\sin\theta}{l} \times 0.1\) M1
Using the chain rule to find \(\frac{dl}{dt}\)
When \(\theta = \frac{\pi}{3}\): \(\frac{dl}{dt} = \frac{45\sin\frac{\pi}{3}}{15\sqrt{3}} = \frac{3}{2}\) M1
Substitute \(\theta = \frac{\pi}{3}\) into their \(\frac{dl}{d\theta}\)
\(1.5\ \text{cm s}^{-1}\) A1
Must have correct unit. Allow \(0.015\ \text{m s}^{-1}\)
[8]
Copy
## Question 10:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Length BC: $l^2 = 30^2 + 15^2 - 2 \times 30 \times 15\cos\theta$ | M1 | If M0 awarded here allow SC1 for $BC = \sqrt{675} = 15\sqrt{3}$ found using $\theta = \frac{\pi}{3}$ |
| $l^2 = 1125 - 900\cos\theta$, $l = (1125 - 900\cos\theta)^{\frac{1}{2}}$ | A1 | Soi. Allow equivalent in metres. If working in metres $l^2 = 0.1125 - 0.0900\cos\theta$ |
| $\frac{dl}{d\theta} = \frac{1}{2}(1125 - 900\cos\theta)^{-\frac{1}{2}} \times 900\sin\theta$ | M1 | Attempt to use the chain rule |
| (any form) | A1 | Any form |
| $\frac{d\theta}{dt} = 0.1$ | B1 | Soi eg from $\theta = 0.1t$ |
| $\frac{dl}{dt} = \frac{dl}{d\theta} \times \frac{d\theta}{dt} = \frac{450\sin\theta}{(1125 - 900\cos\theta)^{\frac{1}{2}}} \times 0.1$ | M1 | Using the chain rule to find $\frac{dl}{dt}$ |
| When $\theta = \frac{\pi}{3}$: substitute into $\frac{dl}{d\theta}$ | M1 | Substitute $\theta = \frac{\pi}{3}$ into their $\frac{dl}{d\theta}$ |
| $\frac{dl}{dt} = \frac{45\sin\frac{\pi}{3}}{(1125 - 900\cos\frac{\pi}{3})^{\frac{1}{2}}} = \left[\frac{45\sqrt{3}}{2 \times 15\sqrt{3}} = \frac{3}{2}\right]$ | M1 | |
| $1.5\ \text{cm s}^{-1}$ | A1 | Must have correct unit. Allow written as cm per second. Alternative: $0.015\ \text{m s}^{-1}$ |
| **[8]** | | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $l^2 = 30^2 + 15^2 - 2 \times 30 \times 15\cos\theta$ | M1 | |
| $l^2 = 1125 - 900\cos\theta$ | A1 | If working in metres $= 0.1125 - 0.0900\cos\theta$ |
| $2l\frac{dl}{d\theta} = 900\sin\theta$ | M1 | Attempt to use implicit differentiation. Any form |
| (any form) | A1 | |
| $\frac{d\theta}{dt} = 0.1$ | B1 | Soi |
| $\frac{dl}{dt} = \frac{dl}{d\theta} \times \frac{d\theta}{dt} = \frac{450\sin\theta}{l} \times 0.1$ | M1 | Using the chain rule to find $\frac{dl}{dt}$ |
| When $\theta = \frac{\pi}{3}$: $\frac{dl}{dt} = \frac{45\sin\frac{\pi}{3}}{15\sqrt{3}} = \frac{3}{2}$ | M1 | Substitute $\theta = \frac{\pi}{3}$ into their $\frac{dl}{d\theta}$ |
| $1.5\ \text{cm s}^{-1}$ | A1 | Must have correct unit. Allow $0.015\ \text{m s}^{-1}$ |
| **[8]** | | |
---
Show LaTeX source
Copy
10 A triangle ABC is made from two thin rods hinged together at A and a piece of elastic which joins $B$ and $C$. $A B$ is a 30 cm rod and $A C$ is a 15 cm rod. The angle $B A C$ is $\theta$ radians as shown in the diagram.
The angle $\theta$ increases at a rate of 0.1 radians per second.\\
Determine the rate of change of the length BC when $\theta = \frac { 1 } { 3 } \pi$.
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q10 [8]}}