Chapter 3.1, Problem 68E

(a.)

To determine

The approximate day of the year on which, the daily temperature shows the greatest increase from the previous day.

Answer to Problem 68E

It has been determined that the daily temperature shows the greatest increase from the previous day on approximately the ${101}^{\text{st}}$ day of the year.

Explanation of Solution

Given:

The equation that approximates the Fahrenheit temperature on day $x$ is:

  $y=19.3\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]+70$

Concept used:

The approximate day of the year on which, the daily temperature shows the greatest increase from the previous day is the value of $x$ for which $\frac{dy}{dx}$ is maximum.

Calculation:

It is given that,

  $y=19.3\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]+70$

Differentiating,

  $\frac{dy}{dx}=19.3\left(\frac{2\pi }{365}\right)\cos \left[\frac{2\pi }{365}\left(x-101\right)\right]$

Differentiating again,

  $\frac{d^{2}y}{d{x}^2}=-19.3{\left(\frac{2\pi }{365}\right)}^2\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]$

Differentiating again,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\frac{2\pi }{365}\left(x-101\right)\right]$

Now, $\frac{dy}{dx}$ is maximum when $\frac{d^{2}y}{d{x}^2}=0$ and $\frac{d^{3}y}{d{x}^3}<0$ .

Put $\frac{d^{2}y}{d{x}^2}=0$ in $\frac{d^{2}y}{d{x}^2}=-19.3{\left(\frac{2\pi }{365}\right)}^2\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]$ to get,

  $-19.3{\left(\frac{2\pi }{365}\right)}^2\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]=0$

Simplifying,

  $\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]=0$

On further simplification,

  $\frac{2\pi }{365}\left(x-101\right)=0$ or $\frac{2\pi }{365}\left(x-101\right)=\pi$

Solving,

  $x=101$ or $x=101+\frac{365}{2}$

Put $x=101$ in $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\frac{2\pi }{365}\left(x-101\right)\right]$ to get,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\frac{2\pi }{365}\left(101-101\right)\right]$

Simplifying,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[0\right]$

On further simplification,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3<0$

Put $x=101+\frac{365}{2}$ in $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\frac{2\pi }{365}\left(x-101\right)\right]$ to get,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\frac{2\pi }{365}\left(101+\frac{365}{2}-101\right)\right]$

Simplifying,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\frac{2\pi }{365}\left(\frac{365}{2}\right)\right]$

On further simplification,

  $\frac{d^{3}y}{d{x}^3}=-19.3{\left(\frac{2\pi }{365}\right)}^3\cos \left[\pi \right]$

Finally,

  $\frac{d^{3}y}{d{x}^3}=19.3{\left(\frac{2\pi }{365}\right)}^3>0$

This implies that $\frac{dy}{dx}$ is maximum for $x=101$ .

So, the daily temperature shows the greatest increase from the previous day on approximately the ${101}^{\text{st}}$ day of the year.

Conclusion:

It has been determined that the daily temperature shows the greatest increase from the previous day on approximately the ${101}^{\text{st}}$ day of the year.

(b.)

To determine

The approximate number of degrees per day that the temperature is increasing by at that time of the year.

Answer to Problem 68E

It has been determined that the temperature is increasing by approximately $0.33°F$ per day, at that time of the year.

Explanation of Solution

Given:

The equation that approximates the Fahrenheit temperature on day $x$ is:

  $y=19.3\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]+70$

Concept used:

The approximate number of degrees per day that the temperature is increasing by at that time of the year is the maximum value of $\frac{dy}{dx}$ .

Calculation:

It is given that,

  $y=19.3\sin \left[\frac{2\pi }{365}\left(x-101\right)\right]+70$

Differentiating,

  $\frac{dy}{dx}=19.3\left(\frac{2\pi }{365}\right)\cos \left[\frac{2\pi }{365}\left(x-101\right)\right]$

As determined previously, $\frac{dy}{dx}$ is maximum for $x=101$ .

Put $x=101$ in $\frac{dy}{dx}=19.3\left(\frac{2\pi }{365}\right)\cos \left[\frac{2\pi }{365}\left(x-101\right)\right]$ to get,

  $\frac{dy}{dx}=19.3\left(\frac{2\pi }{365}\right)\cos \left[\frac{2\pi }{365}\left(101-101\right)\right]$

Simplifying,

  $\frac{dy}{dx}=19.3\left(\frac{2\pi }{365}\right)\cos \left[0\right]$

On further simplification,

  $\frac{dy}{dx}=19.3\left(\frac{2\pi }{365}\right)$

Solving,

  $\frac{dy}{dx}\approx 0.33$

So, the temperature is increasing by approximately $0.33°F$ per day, at that time of the year.

Conclusion:

It has been determined that the temperature is increasing by approximately $0.33°F$ per day, at that time of the year.