Please Help, I do not have much time! A series of locks manages the water height along a water source used to produce energy. As the locks are opened and closed, the water height between two consecutive locks fluctuates.

The height of the water at point B located between two locks is observed. Water height measurements are made every 10 minutes beginning at 8:00 a.m.


It is determined that the height of the water at B can be modeled by the function f(x)=−11cos(πx/48 − 5π/12)+28 , where the height of water is measured in feet and x is measured in minutes.

What is the maximum and minimum water height at B, and when do these heights first occur?

Please Help I do not have much time A series of locks manages the water height along a water source used to produce energy As the locks are opened and closed th class=

Respuesta :

Answer:

See the attached figure.

Step-by-step explanation:

The given function is [tex]f(x) = -11 \ cos(\frac{\pi x}{48} -\frac{5 \pi}{12} )+28[/tex]

We should know that:e

y = cos x

So, the maximum is y = 1 at x = 0 and the minimum is y = -1 at x = π

So, for the given function

The maximum of f(x) will be at [tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = -1[/tex]

And f(x) = -11 * -1 + 28 = 11 + 28 = 39

[tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = -1[/tex]

∴ [tex](\frac{\pi x}{48} -\frac{5 \pi}{12}) = \pi[/tex]

[tex]\frac{\pi x}{48} =\pi + \frac{5 \pi}{12} = \frac{17}{12} \pi[/tex]

x = 48 * 17/12 = 68 minutes = 1 hour and 8 minutes

The results beginning at 8:00 a.m

So, the maximum will occurs at 9:08 a.m

The minimum of f(x) will be at [tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = 1[/tex]

And f(x) = -11 * 1 + 28 = -11 + 28 = 17

[tex]cos(\frac{\pi x}{48} -\frac{5 \pi}{12}) = 1[/tex]

[tex]\frac{\pi x}{48} -\frac{5 \pi}{12}=0[/tex]

[tex]\frac{\pi x}{48} =\frac{5 \pi}{12}[/tex]

x = 48*5/12 = 20 minutes

The results beginning at 8:00 a.m

So, the minimum will occurs at 8:20 a.m

Ver imagen Matheng

The variable of the given function that determines the height at point B

is the angle which is the argument of cosine function.

Response:

  • The first minimum water height of 17 feet occurs at 8:20 a.m. The first maximum water height of 39 feet occurs at 9:08 a.m.

How can the maximum height be found from the given function?

Given:

[tex]The \ function \ for \ the \ water \ height \ is; f(x) = -11 \cdot cos \left(\dfrac{\pi \cdot x}{48} - \dfrac{5 \cdot \pi }{12} \right) + 28[/tex]

The above function is a sinusoidal function.

The maximum and minimum water height are given by the maximum

and minimum value of cos(θ) as follows;

Maximum value of cos(θ) = 1

Minimum value of cos(θ) = -1

Which gives;

[tex]Maximum \ height \ f(x)_{max} = -11 \times (-1) + 28 = \mathbf{39}[/tex]

  • The maximum height at B is 39 feet

Similarly;

[tex]Minimum \ height \ f(x)_{min} = -11 \times (1) + 28 = \mathbf{17}[/tex]

  • The minimum height at B is 17 feet

[tex]\theta = \mathbf{ \left(\dfrac{\pi \cdot x}{48} - \dfrac{5 \cdot \pi }{12} \right)}[/tex]

The value of x at the maximum and minimum points are;

At the maximum point, θ = arccos(-1) = π

Which gives;

[tex]\pi = \mathbf{ \left(\dfrac{\pi \cdot x}{48} - \dfrac{5 \cdot \pi }{12} \right)}[/tex]

Which gives;

[tex]x = \dfrac{48}{\pi} \times \left(\pi}+ \dfrac{5 \cdot \pi }{12} \right)= \mathbf{\dfrac{48}{\pi} \times \pi \left( \dfrac{12 + 5 }{12} \right)} = 68[/tex]

  • The time at which the maximum height occurs is 68 minutes after 8:00 a.m., which is at 9:08 a.m.

At the minimum height, we have; cos(θ) = -1

θ = arccos(1) = 0

Which gives;

[tex]0 = \mathbf{\left(\dfrac{\pi \cdot x}{48} - \dfrac{5 \cdot \pi }{12} \right)}[/tex]

[tex]\dfrac{\pi \cdot x}{48} = \dfrac{5 \cdot \pi }{12}[/tex]

12 × π·x = 48 × 5·π

[tex]x = \dfrac{48 \times 5\cdot \pi}{12 \times \pi} = \mathbf{20}[/tex]

  • The time at which the maximum height occurs is 20 minutes after 8:00 a.m., which is at 8:20 a.m.

Which gives;

  • The first minimum water height of 17 feet occurs at 8:20 a.m. The first maximum water height of 39 feet occurs at 9:08 a.m.

Learn more about sinusoidal functions here:

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