Answer :
Let's solve this step by step.
1. Define the Geometry and Given Values:
- [tex]\( PQ = 20 \)[/tex] cm (fixed length)
- [tex]\( QR = x \)[/tex] cm (variable length)
- The rate at which [tex]\( QR \)[/tex] is increasing: [tex]\( \frac{dx}{dt} = 2 \)[/tex] cm/s
- At the given instant, [tex]\( QR = 15 \)[/tex] cm
2. Formulate the Relationship Using Pythagoras' Theorem:
- In a rectangle, the diagonals are equal in length.
- Using Pythagoras' theorem, find the length of diagonal [tex]\( PR \)[/tex]:
[tex]\[ PR = \sqrt{PQ^2 + QR^2} = \sqrt{20^2 + x^2} = \sqrt{400 + x^2} \][/tex]
3. Apply the Law of Cosines to Find the Acute Angle Between the Diagonals:
- Let's denote the acute angle between the diagonals [tex]\( PR \)[/tex] and [tex]\( QS \)[/tex] as [tex]\( \theta \)[/tex].
- Using the law of cosines for the diagonals, which split the rectangle into two congruent right triangles:
[tex]\[ PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\theta) \][/tex]
Since [tex]\( PR \)[/tex] is a diagonal and [tex]\( QS \)[/tex] is another diagonal, both are equal:
[tex]\[ PR = QS \][/tex]
- Therefore, we use:
[tex]\[ \cos(\theta) = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} \][/tex]
Substitute [tex]\( PR \)[/tex]:
[tex]\[ PR = \sqrt{400 + x^2} \][/tex]
[tex]\[ \cos(\theta) = \frac{400 + x^2 - (400 + x^2)}{2 \cdot 20 \cdot x} = \frac{0}{40x} = 0 \][/tex]
4. Find the Rate of Change of the Angle [tex]\( \theta \)[/tex]:
- Differentiate implicitly with respect to time [tex]\( t \)[/tex].
[tex]\[ \frac{d}{dt} \left( \cos(\theta) \right) = \frac{d}{dt} \left( \frac{400 + x^2 - (400 + x^2)}{2 \cdot 20 \cdot x} \right) \][/tex]
[tex]\[ -\sin(\theta) \cdot \frac{d\theta}{dt} = 0 \][/tex]
[tex]\[ -\sin(\theta) \cdot \frac{d\theta}{dt} = 0 \][/tex]
we get:
[tex]\[ -\sin(\theta) \cdot \frac{d\theta}{dt} = \frac{-x}{40000} \binom{}{}2 \text cm/s \][/tex]
At this point , [tex]\(\cos \theta = 0.6 , \binom{}{} \sin \theta = -0.8 So at this point \frac{d \theta}}{dt} = - 8^(-1)/2^ (-0.8 \ ) . 2 = 0. 1 /4 = 2 \left(\frac{d \theta)}{dt} = \)[/tex]
\ fim^ly therefore :
So now substitute the value PR = sqrt{400 + 15^2 } = 25
Using the concept
The \theta was changing at a rate of
The \(\frac{d}{dt} 20)
]= \binom{}{1} cm/sec was second)
1. Define the Geometry and Given Values:
- [tex]\( PQ = 20 \)[/tex] cm (fixed length)
- [tex]\( QR = x \)[/tex] cm (variable length)
- The rate at which [tex]\( QR \)[/tex] is increasing: [tex]\( \frac{dx}{dt} = 2 \)[/tex] cm/s
- At the given instant, [tex]\( QR = 15 \)[/tex] cm
2. Formulate the Relationship Using Pythagoras' Theorem:
- In a rectangle, the diagonals are equal in length.
- Using Pythagoras' theorem, find the length of diagonal [tex]\( PR \)[/tex]:
[tex]\[ PR = \sqrt{PQ^2 + QR^2} = \sqrt{20^2 + x^2} = \sqrt{400 + x^2} \][/tex]
3. Apply the Law of Cosines to Find the Acute Angle Between the Diagonals:
- Let's denote the acute angle between the diagonals [tex]\( PR \)[/tex] and [tex]\( QS \)[/tex] as [tex]\( \theta \)[/tex].
- Using the law of cosines for the diagonals, which split the rectangle into two congruent right triangles:
[tex]\[ PR^2 = PQ^2 + QR^2 - 2 \cdot PQ \cdot QR \cdot \cos(\theta) \][/tex]
Since [tex]\( PR \)[/tex] is a diagonal and [tex]\( QS \)[/tex] is another diagonal, both are equal:
[tex]\[ PR = QS \][/tex]
- Therefore, we use:
[tex]\[ \cos(\theta) = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} \][/tex]
Substitute [tex]\( PR \)[/tex]:
[tex]\[ PR = \sqrt{400 + x^2} \][/tex]
[tex]\[ \cos(\theta) = \frac{400 + x^2 - (400 + x^2)}{2 \cdot 20 \cdot x} = \frac{0}{40x} = 0 \][/tex]
4. Find the Rate of Change of the Angle [tex]\( \theta \)[/tex]:
- Differentiate implicitly with respect to time [tex]\( t \)[/tex].
[tex]\[ \frac{d}{dt} \left( \cos(\theta) \right) = \frac{d}{dt} \left( \frac{400 + x^2 - (400 + x^2)}{2 \cdot 20 \cdot x} \right) \][/tex]
[tex]\[ -\sin(\theta) \cdot \frac{d\theta}{dt} = 0 \][/tex]
[tex]\[ -\sin(\theta) \cdot \frac{d\theta}{dt} = 0 \][/tex]
we get:
[tex]\[ -\sin(\theta) \cdot \frac{d\theta}{dt} = \frac{-x}{40000} \binom{}{}2 \text cm/s \][/tex]
At this point , [tex]\(\cos \theta = 0.6 , \binom{}{} \sin \theta = -0.8 So at this point \frac{d \theta}}{dt} = - 8^(-1)/2^ (-0.8 \ ) . 2 = 0. 1 /4 = 2 \left(\frac{d \theta)}{dt} = \)[/tex]
\ fim^ly therefore :
So now substitute the value PR = sqrt{400 + 15^2 } = 25
Using the concept
The \theta was changing at a rate of
The \(\frac{d}{dt} 20)
]= \binom{}{1} cm/sec was second)
