Hifumi's Study Notes

Steps

Draw the situation.
Establish the equation that relates the quantities of interest.
Implicitly differentiate the equation with respect to $t$ .
Substitute the given information and solve for the rate.

Examples

Two people on bikes are separated by 350 meters horizontally. Person A starts riding north at a rate of 5 m/sec and 7 minutes later Person B starts riding south at 3 m/sec. At what rate is the distance separating the two people changing 25 minutes after Person A starts riding?

300

After 25 mins:

\begin{aligned} y & = 3 \cdot 18 \cdot 60 & = 3240 m \\ z & = 5 \cdot 25 \cdot 60 & = 7500 m \\ x & = \sqrt{350^{2} + (3240 + 7500)^{2}} & = 10745.7015 \dots m \end{aligned}

We have

\begin{aligned} x^{2} & = 350^{2} + (y + z)^{2} \\ ⟹ \frac{d}{d t} x^{2} & = \frac{d}{d t} (350^{2} + (y + z)^{2}) \\ 2 x x^{'} & = 2 (y + z) (y^{'} + z^{'}) \\ x^{'} & = \frac{(y + z) (y^{'} + z^{'})}{x} \\ = \frac{(3240 + 7500) (3 + 5)}{10745.7015 \dots} \\ \approx 7.9958 \frac{m}{s} \end{aligned}

A plane flying with a constant speed of 29 km/min passes over a ground radar station at an altitude of 11 km and climbs at an angle of 35 degrees. At what rate is the distance from the plane to the radar station increasing 3 minutes later?

\begin{aligned} d & = \sqrt{(\cos 35 ° x)^{2} + (11 + \sin 35 ° x)^{2}} \\ ⟹ \frac{d d}{d t} & = \frac{d}{d t} \sqrt{(\cos 35 ° x)^{2} + (11 + \sin 35 ° x)^{2}} \\ = \frac{1}{2} ((\cos 35 ° x)^{2} + (11 + \sin 35 ° x)^{2})^{- \frac{1}{2}} (2 (\cos 35 ° x) \cos 35 ° x^{'} + 2 (11 + \sin 35 ° x) \sin 35 ° x^{'}) \\ = \frac{1}{2} ((\cos 35 ° \cdot 3 \cdot 29)^{2} + (11 + \sin 35 ° \cdot 3 \cdot 29)^{2})^{- \frac{1}{2}} (2 (\cos 35 ° \cdot 3 \cdot 29) \cos 35 ° \cdot 29 + 2 (11 + \sin 35 ° \cdot 3 \cdot 29) \sin 35 ° \cdot 29) \\ \approx 28.8657 \frac{km}{\min} \end{aligned}

Two parallel sides of a rectangle are being increased at a rate of 2 cm per second while the other 2 sides are being shortened so that the area remains constant at $50 {cm}^{2}$ .
(a) What is the rate of change of the perimeter when the length of an increasing side is 5 cm?
(b) What are the dimensions at the instant when the perimeter stops decreasing?

We know that

\begin{aligned} x y & = 50 & (1) \\ P & = 2 x + 2 y & (2) \end{aligned}

Using (1),

y = \frac{50}{x}

Using (2),

\begin{aligned} P & = 2 x + 2 (\frac{50}{x}) \\ = 2 x + \frac{100}{x} \end{aligned}

Then,

\begin{aligned} \frac{d}{d t} P & = \frac{d}{d t} (2 x + \frac{100}{x}) \\ P^{'} & = 2 x^{'} - \frac{100}{x^{2}} x^{'} \end{aligned}

When $x = 5$ and $x^{'} = 2$ ,

\begin{aligned} P^{'} & = 2 \cdot 2 - \frac{100}{5^{2}} \cdot 2 \\ = - 4 \frac{cm}{s} \end{aligned}

(b) we want $x$ and $y$ when $P^{'} = 0$ .

\begin{aligned} P^{'} & = 0 \\ ⟹ 2 x^{'} - \frac{100}{x^{2}} x^{'} & = 0 \\ 2 - \frac{100}{x^{2}} & = 0 \\ 2 x^{2} & = 100 \\ x & = 5 \sqrt{2} \end{aligned}

Note that $x > 0$ .

0	$5 \sqrt{2}$	$\infty$
P'	-	+
P	dec	inc

So, $P$ stops decreasing when $x = 5 \sqrt{2}$ .
By (1),

\begin{aligned} 5 \sqrt{2} y & = 50 \\ y & = 5 \sqrt{2} \end{aligned}

A rain gutter is to be constructed from a sheet of metal of width 30 cm bending up one third of the sheet on each side through an angle of 𝜃 . Find 𝜃 so the gutter will carry the maximum amount of water.

Note that

{\begin{cases} A & = \frac{(B + b) h}{2} \\ h & = \sin θ \cdot 10 \\ b & = 10 \\ B & = 10 + \cos θ \cdot 10 \end{cases}

Using the above equations,

\begin{aligned} A (θ) & = \frac{(10 + \cos θ \cdot 10 + 10) \sin θ \cdot 10}{2} \\ = 100 (1 + \cos θ) \sin θ \end{aligned}

As the sides close down, an equilateral triangle is formed.
So,

\begin{aligned} θ_{max} & = \frac{π}{2} + \frac{π}{6} = \frac{2 π}{3} \\ ∴ θ \in [0, \frac{2 π}{3}] \end{aligned}

Then,

\begin{aligned} A^{'} (θ) & = 100 (- \sin θ \sin θ + (1 + \cos θ) \cos θ) \\ = 100 (\cos^{2} θ - 1 + \cos θ + \cos^{2} θ) \\ = 100 (2 \cos^{2} θ + \cos θ - 1) \end{aligned}

$A^{'}$ is never undefined. $A^{'} = 0$ if

\begin{aligned} 2 \cos^{2} θ + \cos θ - 1 & = 0 \\ 2 \cos^{2} θ + 2 \cos θ - \cos θ - 1 & = 0 \\ 2 \cos θ (\cos θ + 1) - (\cos θ + 1) & = 0 \\ (2 \cos θ - 1) (\cos θ + 1) & = 0 \\ ⟹ 2 \cos θ - 1 & = 0 or & \cos θ + 1 & = 0 \\ 2 \cos θ & = 1 & \cos θ & = - 1 \\ \cos θ & = \frac{1}{2} \\ θ & = π & θ & = \frac{π}{3} or \frac{5 π}{3} \end{aligned}

Only $\frac{π}{3} \in [0, \frac{2 π}{3}]$ . Then,

0	$\frac{π}{3}$	$\frac{2 π}{3}$
A'	+	-
A	inc	dec

By the first derivative test, $A (θ)$ is maximized when $θ = \frac{π}{3}$ .

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Ajitani Hifumi

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