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3 years ago

15

Michelle and her father want to make a triangular sail for their boat. The height of the sail will be 24 ft., and the base will

be 18 ft. What will be the area of the sail in square feet?

1 answer:

Lemur [1.5K]3 years ago

5 0

Answer:

The area of the triangular sail is 216 ft²

Step-by-step explanation:

Since the sail is triangular, we solve the area using the area of a triangle

Area of triangle = 1/2 of base × height

base is 18 and height is 24

Area = 1/2 × 18 × 24

= 9×24

= 216 ft²

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First look for the fundamental solutions by solving the homogeneous version of the ODE:

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${y_p}=at^5+bt^4+ct^3+dt^2+et$

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(You could include a constant term <em>f</em> here, but it would get absorbed by the first solution $C_1$ anyway.)

Substitute these into the ODE:

$(20at^3+12bt^2+6ct+2d)+3(5at^4+4bt^3+3ct^2+2dt+e)=2t^4$

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$\implies\begin{cases}15a=2\\20a+12b=0\\12b+9c=0\\6c+6d=0\\2d+e=0\end{cases}\implies a=\dfrac2{15},b=-\dfrac29,c=\dfrac8{27},d=-\dfrac8{27},e=\dfrac{16}{81}$

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$e^{-3t}$ is already accounted for, so assume an ansatz of the form

$y_p=(at^3+bt^2+ct)e^{-3t}$

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$\implies {y_p}''=(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}$

Substitute into the ODE:

$(9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c)e^{-3t}+3(-3at^3+(3a-3b)t^2+(2b-3c)t+c)e^{-3t}=t^2e^{-3t}$

$9at^3+(9b-18a)t^2+(9c-12b+6a)t+2b-6c-9at^3+(9a-9b)t^2+(6b-9c)t+3c=t^2$

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Assume an ansatz solution

$y_p=a\sin(3t)+b\cos(3t)$

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$\implies {y_p}''=-9a\sin(3t)-9b\cos(3t)$

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