The following function gives the temperature (in degrees Celsius) at the beach in Miami, Florida, t hours after midnight on a certain day: M(t)=-6*sin((pi)/(12)t)+18 What is the instantaneous rate of change of the temperature at 9 a.m.? Choose 1 answer: (A) 1.11 degrees Celsius per hour (B) 1.11 degrees Celsius (C) 13.76 degrees Celsius per hour (D) 13.76 degrees Celsius

Calculate Derivative of M(t): To find the instantaneous rate of change of the temperature at 9 a.m., we need to calculate the derivative of M(t) with respect to t and then evaluate it at t = 9 hours.Find Derivative Using Chain Rule: First, let's find the derivative of M(t): M'(t) = d/dt[-6*sin((pi/12)t) + 18] Using the chain rule, the derivative of -6*sin((pi/12)t) is -6*cos((pi/12)t) * (pi/12).Evaluate Derivative at t = 9: So, M'(t) = -6 * (pi/12) * cos((pi/12)t) Simplify the constant: -6 * (pi/12) = -pi/2 M'(t) = -pi/2 * cos((pi/12)t)Calculate Cosine Term: Now, let's evaluate M'(t) at t = 9 hours to find the rate of change at 9 a.m.: M'(9) = -pi/2 * cos((pi/12)*9)Simplify Expression: Calculate the cosine term: cos((pi/12)*9) = cos((3/4)pi) = cos(135 degrees)Multiply Constants: Cosine of 135 degrees is -sqrt(2)/2. M'(9) = -pi/2 * (-sqrt(2)/2)Simplify the expression: M'(9) = (pi/2) * (sqrt(2)/2)Multiply the constants: M'(9) = (pi * sqrt(2))/4

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The following function gives the temperature (in degrees Celsius) at the beach in Miami, Florida,
t hours after midnight on a certain day:

M(t)=-6*sin((pi)/(12)t)+18
What is the instantaneous rate of change of the temperature at 9 a.m.?
Choose 1 answer:
(A) 1.11 degrees Celsius per hour
(B)
1.11 degrees Celsius
(C)
13.76 degrees Celsius per hour
(D)
13.76 degrees Celsius

The following function gives the temperature (in degrees Celsius) at the beach in Miami, Florida, $t$ hours after midnight on a certain day: $\newline$ $M(t)=-6 \cdot \sin \left(\frac{\pi}{12} t\right)+18$ $\newline$ What is the instantaneous rate of change of the temperature at $9$ a.m.? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ . $11$ degrees Celsius per hour $\newline$ (B) $\mathbf{1 . 1 1}$ degrees Celsius $\newline$ (C) $\mathbf{1 3 . 7 6}$ degrees Celsius per hour $\newline$ (D) $\mathbf{1 3 . 7 6}$ degrees Celsius

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Calculus

Find derivatives of sine and cosine functions

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Q. The following function gives the temperature (in degrees Celsius) at the beach in Miami, Florida,

t

hours after midnight on a certain day:

\newline

M(t)=-6 \cdot \sin \left(\frac{\pi}{12} t\right)+18

\newline

What is the instantaneous rate of change of the temperature at

9

a.m.?

\newline

Choose

1

answer:

\newline

(A)

1

11

degrees Celsius per hour

\newline

(B)

\mathbf{1 . 1 1}

degrees Celsius

\newline

(C)

\mathbf{1 3 . 7 6}

degrees Celsius per hour

\newline

(D)

\mathbf{1 3 . 7 6}

degrees Celsius

Calculate Derivative of M(t): To find the instantaneous rate of change of the temperature at $9$ a.m., we need to calculate the derivative of $M(t)$ with respect to $t$ and then evaluate it at $t = 9$ hours.
Find Derivative Using Chain Rule: First, let's find the derivative of $M(t)$ : $M'(t) = \frac{d}{dt}[-6\sin\left(\frac{\pi}{12}t\right) + 18]$ Using the chain rule, the derivative of $-6\sin\left(\frac{\pi}{12}t\right)$ is $-6\cos\left(\frac{\pi}{12}t\right) \cdot \left(\frac{\pi}{12}\right)$ .
Evaluate Derivative at $t = 9$ : So, $M'(t) = -6 \times (\pi/12) \times \cos((\pi/12)t)$ $\newline$ Simplify the constant: $-6 \times (\pi/12) = -\pi/2$ $\newline$ $M'(t) = -\pi/2 \times \cos((\pi/12)t)$
Calculate Cosine Term: Now, let's evaluate $M'(t)$ at $t = 9$ hours to find the rate of change at $9$ a.m.: $M'(9) = -\frac{\pi}{2} \cdot \cos\left(\frac{\pi}{12}\cdot9\right)$
Simplify Expression: Calculate the cosine term: $\cos\left(\frac{\pi}{12}\cdot 9\right) = \cos\left(\frac{3}{4}\pi\right) = \cos(135^\circ)$
Multiply Constants: Cosine of $135$ degrees is $-\sqrt{2}/2$ . $\newline$ $M'(9) = -\pi/2 \times (-\sqrt{2}/2)$
Multiply Constants: Cosine of $135$ degrees is $-\sqrt{2}/2$ . $M'(9) = -\pi/2 \times (-\sqrt{2}/2)$ Simplify the expression: $M'(9) = (\pi/2) \times (\sqrt{2}/2)$
Multiply Constants: Cosine of $135$ degrees is $-\sqrt{2}/2$ . $M'(9) = -\pi/2 \times (-\sqrt{2}/2)$ Simplify the expression: $M'(9) = (\pi/2) \times (\sqrt{2}/2)$ Multiply the constants: $M'(9) = (\pi \times \sqrt{2})/4$