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Parents Pricing Home Leaving Cert Applied Maths Differential Equations If \( \frac{dy}{dx} = 3 \sin 3x + \cos 5x \) and \( y = 1 \) when \( x = \frac{\pi}{2} \), find the value of \( y \) when \( x = \frac{\pi}{4} \)
If \( \frac{dy}{dx} = 3 \sin 3x + \cos 5x \) and \( y = 1 \) when \( x = \frac{\pi}{2} \), find the value of \( y \) when \( x = \frac{\pi}{4} \) - Leaving Cert Applied Maths - Question 10 - 2018 Question 10
View full question If \( \frac{dy}{dx} = 3 \sin 3x + \cos 5x \) and \( y = 1 \) when \( x = \frac{\pi}{2} \), find the value of \( y \) when \( x = \frac{\pi}{4} \).
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View marking scheme Worked Solution & Example Answer:If \( \frac{dy}{dx} = 3 \sin 3x + \cos 5x \) and \( y = 1 \) when \( x = \frac{\pi}{2} \), find the value of \( y \) when \( x = \frac{\pi}{4} \) - Leaving Cert Applied Maths - Question 10 - 2018
Find the value of \( y \) when \( x = \frac{\pi}{4} \) Only available for registered users.
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To find the value of ( y ) given ( \frac{dy}{dx} = 3 \sin 3x + \cos 5x ), we start by integrating this expression:
∫ d y = ∫ ( 3 sin 3 x + cos 5 x ) d x \int dy = \int (3 \sin 3x + \cos 5x) \, dx ∫ d y = ∫ ( 3 sin 3 x + cos 5 x ) d x This results in:
y = − cos 3 x + sin 5 x 5 + C y = -\cos 3x + \frac{\sin 5x}{5} + C y = − cos 3 x + 5 sin 5 x + C Next, we need to find the constant ( C ) using the initial condition ( y = 1 ) when ( x = \frac{\pi}{2} ):
1 = − cos ( 3 ⋅ π 2 ) + sin ( 5 ⋅ π 2 ) 5 + C 1 = -\cos(3 \cdot \frac{\pi}{2}) + \frac{\sin(5 \cdot \frac{\pi}{2})}{5} + C 1 = − cos ( 3 ⋅ 2 π ) + 5 sin ( 5 ⋅ 2 π ) + C Calculating this gives:
1 = 0 + 1 5 + C ⇒ C = 1 − 1 5 = 4 5 1 = 0 + \frac{1}{5} + C \Rightarrow C = 1 - \frac{1}{5} = \frac{4}{5} 1 = 0 + 5 1 + C ⇒ C = 1 − 5 1 = 5 4 Now substituting ( C ) back into the equation:
y = − cos 3 x + sin 5 x 5 + 4 5 y = -\cos 3x + \frac{\sin 5x}{5} + \frac{4}{5} y = − cos 3 x + 5 sin 5 x + 5 4 Finally, we find ( y ) at ( x = \frac{\pi}{4} ):
Calculate ( -\cos(\frac{3\pi}{4}) = -\left(-\frac{\sqrt{2}}{2}\right) = \frac{\sqrt{2}}{2} )
Calculate ( \sin(\frac{5\pi}{4}) = -\frac{\sqrt{2}}{2} )
Thus,
y = 2 2 − 2 10 + 4 5 = 0.63 (to 2 decimal places) y = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{10} + \frac{4}{5} = 0.63 \text{ (to 2 decimal places)} y = 2 2 − 10 2 + 5 4 = 0.63 (to 2 decimal places)
If initially the population is \( P \) people, find in terms of \( n, P \) and \( t \), the population after \( t \) years. Only available for registered users.
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From the given rate of change of population:
d x d t = − x 40 − n \frac{dx}{dt} = -\frac{x}{40} - n d t d x = − 40 x − n This can be rearranged to:
40 d x d t + x = − 40 n 40 \frac{dx}{dt} + x = -40n 40 d t d x + x = − 40 n Now integratin gives:
∫ d x x + 40 n = − ∫ d t 40 \int \frac{dx}{x + 40n} = -\int \frac{dt}{40} ∫ x + 40 n d x = − ∫ 40 d t Therefore:
ln ( x − 40 n ) = − t 40 + C \ln(x - 40n) = -\frac{t}{40} + C ln ( x − 40 n ) = − 40 t + C Exponentiating both sides leads to:
x − 40 n = C e − t / 40 x - 40n = Ce^{-t/40} x − 40 n = C e − t /40 Substituting for the initial condition when ( t = 0, x = P):
P − 40 n = C i g h t a r r o w C = P − 40 n P - 40n = C
ightarrow C = P - 40n P − 40 n = C i g h t a rro wC = P − 40 n Thus the formula becomes:
x = ( P − 40 n ) e − t / 40 + 40 n x = (P - 40n)e^{-t/40} + 40n x = ( P − 40 n ) e − t /40 + 40 n
Given that \( n = 800 \) and \( P = 30000 \), find the value of \( t \) when the population is 29734. Only available for registered users.
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Substituting ( n = 800 ) and ( P = 30000 ) into the derived formula results in:
x = ( 30000 − 32000 ) e − t / 40 + 32000 x = (30000 - 32000)e^{-t/40} + 32000 x = ( 30000 − 32000 ) e − t /40 + 32000 That simplifies to:
x = − 2000 e − t / 40 + 32000 x = -2000 e^{-t/40} + 32000 x = − 2000 e − t /40 + 32000 We are given ( x = 29734 ):
29734 = − 2000 e − t / 40 + 32000 29734 = -2000 e^{-t/40} + 32000 29734 = − 2000 e − t /40 + 32000 Rearranging gives:
− 2000 e − t / 40 = 29734 − 32000 -2000 e^{-t/40} = 29734 - 32000 − 2000 e − t /40 = 29734 − 32000 − 2000 e − t / 40 = − 2266 i g h t a r r o w e − t / 40 = 2266 2000 -2000 e^{-t/40} = -2266
ightarrow e^{-t/40} = \frac{2266}{2000} − 2000 e − t /40 = − 2266 i g h t a rro w e − t /40 = 2000 2266 Taking the natural logarithm:
− t 40 = ln ( 1.133 ) -\frac{t}{40} = \ln(1.133) − 40 t = ln ( 1.133 ) Therefore:
t = − 40 ln ( 1.133 ) e x t w h i c h e v a l u a t e s t o 4.99 e x t y e a r s . t = -40 \ln(1.133) ext{ which evaluates to } 4.99 ext{ years.} t = − 40 ln ( 1.133 ) e x t w hi c h e v a l u a t es t o 4.99 e x t ye a rs . Join the Leaving Cert students using SimpleStudy...97% of StudentsReport Improved Results
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