PAGE 1

3.11 Related Rates (part 1)

relating the rate of change (derivative) of one thing to that of another

skills needed: Chain rule, implicit differentiation

example

If the base is increasing at \( 1 \text{ cm/s} \) when the base is \( 3 \text{ cm} \) and height is \( 4 \text{ cm} \). How should the height change for the area to remain unchanged?

Figure: A right-angled triangle with vertical side labeled h and horizontal base labeled b.

here, both \( b \) and \( h \) are changing with time

\( \rightarrow \) functions of time: \( h = h(t) \), \( b = b(t) \)

at a particular instant \( b = 3 \) and \( h = 4 \) and \( \frac{db}{dt} = 1 \)

find \( \frac{dh}{dt} \) such that the area \( A \) is unchanged \( (\frac{dA}{dt} = 0) \)

PAGE 2

first, relate base \( b \), height \( h \), and the area \( A \)

for a triangle, \( \text{area} = \frac{1}{2} \cdot \text{base} \cdot \text{height} \)

\[ A(t) = \frac{1}{2} \cdot b(t) \cdot h(t) \]

now differentiate both sides with respect to \( t \)

\[ \frac{d}{dt} A = \frac{d}{dt} \left( \frac{1}{2} \cdot b \cdot h \right) \]

both are functions of \( t \) so need product rule

\[ \frac{dA}{dt} = \frac{1}{2} \cdot \left( b \cdot \frac{dh}{dt} + h \cdot \frac{db}{dt} \right) \]

at the moment when \( b = 3, h = 4, \frac{db}{dt} = 1 \), we want \( \frac{dA}{dt} = 0 \)

plug in these numbers AFTER we related the rates

\[ 0 = \frac{1}{2} \left( 3 \cdot \frac{dh}{dt} + 4 \cdot 1 \right) = \frac{3}{2} \frac{dh}{dt} + 2 = 0 \]

solve for \( \frac{dh}{dt} \):

\[ \frac{dh}{dt} = \frac{-2}{3/2} = \boxed{-\frac{4}{3}} \]

negative rate so it means \( h \) is decreasing at \( 4/3 \text{ cm/s} \)

PAGE 3

Example: Related Rates of a Circle

How fast is the area of a circle changing if the circumference of the circle is decreasing at the rate of \(2 \text{ cm/s}\) when the circumference is \(3 \text{ cm}\)?

Figure: A simple hand-drawn circle with a radius line labeled r.

\[ A = \pi r^2 \quad \text{relates area to radius} \]

Want to relate area to circumference:

\[ \text{circumference: } C = 2\pi r \implies r = \frac{C}{2\pi} \]

\[ A = \pi \left( \frac{C}{2\pi} \right)^2 = \pi \cdot \frac{C^2}{4\pi^2} = \frac{1}{4\pi} C^2 \]

\[ A = \frac{1}{4\pi} C^2 \]

Both \(A\) and \(C\) are functions of \(t\).

Differentiate with respect to \(t\)

\[ \frac{d}{dt} A = \frac{d}{dt} \left( \frac{1}{4\pi} C^2 \right) \]

\[ \frac{dA}{dt} = \frac{1}{4\pi} \cdot 2C \cdot \frac{dC}{dt} \]

Now we related rates we can plug in numbers: \(\frac{dC}{dt} = -2\), \(C = 3\)

\[ \frac{dA}{dt} = \frac{1}{2\pi} \cdot 3 \cdot -2 = -3/\pi \]

Area decreasing

PAGE 4

Example: Boat and Dock Related Rates

A boat is being pulled toward a dock at a rate of \(4 \text{ ft/s}\), by a rope through a point \(6 \text{ ft}\) above water on the dock. How fast is the boat traveling when the rope is \(14 \text{ ft}\)?

Figure: Sketch of a boat on water connected by a rope to a dock 6 feet high.

Figure: Right triangle model with vertical side 6, horizontal side x, and hypotenuse y.

\(x\): distance between boat and dock
\(y\): rope length

We know: \(\frac{dy}{dt} = -4\) (negative rate because we pull in rope)

We want: \(\frac{dx}{dt}\) when \(y = 14\)

PAGE 5

We need to relate \(y\) to \(x\).

Pythagorean Theorem is the easiest:

\[y^2 = x^2 + 6^2\]

Both \(x\) and \(y\) depend on \(t\).

Now Differentiate

\[\frac{d}{dt}(y^2) = \frac{d}{dt}(x^2 + 6^2)\]\[2y \frac{dy}{dt} = 2x \frac{dx}{dt}\]

Solve for what we want: \(\frac{dx}{dt}\)

\[\frac{dx}{dt} = \frac{y}{x} \frac{dy}{dt}\]

Now rates are related, put in numbers:

\(y = 14\) (rope is 14 ft)
\(\frac{dy}{dt} = -4\)
\(x = ?\)

\[x^2 = y^2 - 36\]\[x = \sqrt{y^2 - 36} = \sqrt{14^2 - 36} = \sqrt{160}\]

\[\frac{dx}{dt} = \frac{14}{\sqrt{160}}(-4) = -4.43 \text{ ft/s}\]

Boat is approaching dock at 4.43 ft/s