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Lecture 15: Related Rates (Trigonometric Relations)

Assume all variables are functions of time.

1 Pythagorean Relations

As we saw in last lecture, if quantities are related, we can see how their derivatives are related.

We already saw some examples of how variables can be related. For examples, the perimeter, area, surface area, or volume of a shape. But, these are far from the only examples.

For another example, suppose \( x, y, \) and \( z \) are the side lengths of a right triangle.

Right triangle with horizontal leg x, vertical leg y, and hypotenuse z.
Figure 1
A hand-drawn right triangle. The horizontal bottom side is labeled \( x \). The vertical right side is labeled \( y \), with a small square at the bottom-right corner marking the right angle between the horizontal and vertical sides. The slanted side from the bottom-left vertex up to the top-right vertex is the hypotenuse, labeled \( z \).

Problem 1.

Suppose \( x = x(t) \), \( y = y(t) \), and \( z = z(t) \) are related by the Pythagorean Theorem. So,

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

How are \( \frac{dx}{dt} \), \( \frac{dy}{dt} \), and \( \frac{dz}{dt} \) related? In particular, what happens when \( \frac{dz}{dt} = 0 \)?

Like last time, we see how the derivatives of \( x, y, \) and \( z \) are related by taking the derivative of both sides of \( x^2 + y^2 = z^2 \).

\[ \begin{gathered} x^2 + y^2 = z^2 \\[0.5em] 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \\[0.5em] x \frac{dx}{dt} + y \frac{dy}{dt} = z \frac{dz}{dt} \end{gathered} \]

In particular, if \( \frac{dz}{dt} = 0 \).

\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

Problem 2.

A ladder 10m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6m from the wall?

Right triangle: vertical leg y against wall, horizontal leg x on floor, ladder of length 10 as hypotenuse.
Figure 2
A hand-drawn right triangle representing a ladder leaning against a vertical wall. The vertical side on the left, labeled \( y \), is the wall from the floor to the top of the ladder. The horizontal side along the bottom, labeled \( x \), is the distance along the floor from the wall to the bottom of the ladder. The slanted hypotenuse, labeled 10, is the ladder. A small square at the lower-left corner marks the right angle between the wall and the floor.

As with any Related Rates problem, we run through the checklist (see Lecture 14).

  1. Let \( x \) be the displacement from the wall to the bottom of the ladder. Let \( y \) be the displacement from the floor to the top of the ladder.
  2. Yes, \( x^2 + y^2 = 10^2 \)
  3. K: \( \frac{dx}{dt} = 1 \)
  4. NTK: \( \frac{dy}{dt} \) when \( x = 6 \). As we will see later, we also NTK \( y \) when \( x = 6 \).
\[ \begin{gathered} x^2 + y^2 = 10^2 \\[0.5em] 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \\[0.5em] x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \\[0.5em] y \frac{dy}{dt} = -x \frac{dx}{dt} \\[0.5em] \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \end{gathered} \]

Find \( y \) when \( x = 6 \): As \( y > 0 \),

\[ \begin{gathered} y = \sqrt{10^2 - x^2} \\[0.5em] y\big|_{x=6} = \sqrt{10^2 - 6^2} \\[0.5em] = \sqrt{100 - 36} \\[0.5em] = \sqrt{64} \\[0.5em] = 8 \end{gathered} \]

So,

\[ \begin{gathered} \frac{dy}{dt} = -\frac{x}{y} \frac{dy}{dt} \\[0.5em] \frac{dy}{dt}\bigg|_{\substack{x=6 \\ x'=1}} = -\frac{6}{8}(1) \\[0.5em] = -\frac{6}{8} \\[0.5em] = -\frac{3}{4} \end{gathered} \]

Therefore, the top of the ladder is falling at a rate of \( \frac{3}{4} \) m/s.

Problem 3.

Two cars leave from the same point at the same time. Car A is traveling east at 50km/h and car B is traveling north at 60km/h. At what rate is the distance between the two cars changing when car A is 0.3km and car B is 0.4km from their starting point?

Right triangle with vertical leg y to Car B, horizontal leg x to Car A, and hypotenuse z.
Figure 3
A hand-drawn right triangle representing the positions of two cars. The right angle is at the bottom-left corner, which is the common starting point. The vertical leg, labeled \( y \), rises from the starting point to a dot at the top labeled Car B. The horizontal leg, labeled \( x \), runs right from the starting point to a dot labeled Car A. The hypotenuse joins Car B to Car A and is labeled \( z \), the distance between the two cars.

Run through the checklist:

  1. Let \( x \) and \( y \) be Car A's and B's distance from their starting point. Let \( z \) be the distance between Car A and B.
  2. \( x^2 + y^2 = z^2 \)
  3. K: \( \frac{dx}{dt} = 50 \) and \( \frac{dy}{dt} = 60 \)
  4. NTK: \( \frac{dz}{dt} \) when \( x = 0.3 \) and \( y = 0.4 \). Also, the value of \( z \) when \( x = 0.3 \) and \( y = 0.4 \).
\[ \begin{gathered} x^2 + y^2 = z^2 \\[0.5em] 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \\[0.5em] \frac{dz}{dt} = \frac{1}{z}\left[ x \frac{dx}{dt} + y \frac{dy}{dt} \right] \end{gathered} \]

Find \( z \):

Since \( z > 0 \), \( z = \sqrt{x^2 + y^2} \).

\[ \begin{gathered} z = \sqrt{x^2 + y^2} \\[0.5em] z\bigg|_{\substack{x=0.3 \\ y=0.4}} = \sqrt{(0.3)^2 + (0.4)^2} \\[0.5em] = \sqrt{0.25} \\[0.5em] = 0.5 \end{gathered} \]

Now,

\[ \begin{gathered} \frac{dz}{dt} = \frac{1}{z}\left[ x \frac{dx}{dt} + y \frac{dy}{dt} \right] \\[0.5em] = \frac{1}{0.5}\left[ 0.3(50) + 0.4(60) \right] \\[0.5em] = 78 \end{gathered} \]

Therefore, the distance between cars A and B is increasing at a rate of 78 \( \frac{km}{h} \).

Problem 4.

A plane flying horizontally at an altitude of 1 km and a speed of 500km/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 2km away from the station.

Right triangle: vertical leg 1, horizontal leg x to Plane, hypotenuse z from Radar Station.
Figure 4
A hand-drawn right triangle. A dotted vertical segment labeled 1 rises from a dot at the bottom labeled Radar Station to a right-angle corner at the top left. From that corner, a dotted horizontal segment labeled x runs to the right, ending at a dot labeled Plane. A solid line labeled z runs diagonally from the Radar Station dot up to the Plane dot, forming the hypotenuse. The right angle is marked at the top-left corner where the altitude of 1 km meets the horizontal distance x.

Run through the check list:

  1. Let \( z \) denote the distance from the plane and the radar station and \( x \) the horizontal distance.
  2. \( x^2 + 1^2 = z^2 \)
  3. K: \( \frac{dx}{dt} = 500 \)
  4. NTK: \( \frac{dz}{dt} \) (and \( x \)) when \( z = 2 \).
\[ \begin{gathered} x^2 + 1^2 = z^2 \\[0.5em] 2x \frac{dx}{dt} = 2z \frac{dz}{dt} \\[0.5em] \frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt} \end{gathered} \]

Find \( x \): When \( x > 0 \), \( x = \sqrt{z^2 - 1} \). So, when \( z = 2 \).

\[ \begin{gathered} x\big|_{z=2} = \sqrt{2^2 - 1} \\[0.5em] = \sqrt{3} \end{gathered} \]

So,

\[ \begin{gathered} \frac{dz}{dt} = \frac{x}{z} \frac{dx}{dt} \\[0.5em] \frac{dz}{dt}\bigg|_{\substack{z=2 \\ z'=500}} = \frac{\sqrt{3}}{2}(500) \\[0.5em] = 250\sqrt{3} \end{gathered} \]

Therefore, the distance between the plane and the radar station is increasing at a rate of \( 250\sqrt{3} \) \( \frac{km}{h} \).

Problem 5.

A Little League baseball diamond is a square with 60ft on each side. A player runs from 1st base to 2nd base at 12ft/s. At what rate is the player's distance from home plate increasing if he is halfway between 1st and 2nd base?

Baseball diamond square with bases labeled; red triangle from home to player between 1st and 2nd base.
Figure 5
A diamond-shaped square drawn with vertices labeled 2nd Base at the top, 3rd Base at the left, 1st Base at the right, and Home at the bottom. The side from Home to 1st Base is labeled 60 ft. Drawn in red is a point on the segment between 1st Base and 2nd Base representing the player, with a short red segment labeled y running from 1st Base up to the player, and a longer red segment labeled z running from Home diagonally up to the player, forming a right triangle with the 60 ft side from Home to 1st Base.

Run through the checklist:

  1. Let \( y \) be the distance from 1st Base to the player, and \( z \) the distance from home to the player.
  2. \( 60^2 + y^2 = z^2 \)
  3. K: \( \frac{dy}{dt} = 12 \)
  4. NTK: \( \frac{dz}{dt} \) (and \( z \)) when \( y = \frac{1}{2}(60) = 30 \).
\[ \begin{gathered} 60^2 + y^2 = z^2 \\[0.5em] 2y \frac{dy}{dt} = 2z \frac{dz}{dt} \\[0.5em] \frac{dz}{dt} = \frac{y}{z} \frac{dz}{dt} \end{gathered} \]

Find \( z \): Similar to other problems, when \( y = 30 \)

\[ z = \sqrt{60^2 + 30^2} = 30\sqrt{5} \]

Thus,

\[ \begin{gathered} \frac{dz}{dt} = \frac{y}{z} \frac{dy}{dt} \\[0.5em] \frac{dz}{dt}\bigg|_{\substack{y=30 \\ y'=12}} = \frac{30}{30\sqrt{5}}(12) \\[0.5em] = \frac{12}{\sqrt{5}} \end{gathered} \]

Therefore, the distance between the player and home base is increasing at a rate of \( \frac{12}{\sqrt{5}} \) \( \frac{ft}{s} \).

Problem 6.

A boat is pulled into a dock by a rope attached to the bow of the boat and passing through a pulley on the dock that is 1m higher than the bow of the boat. If the rope is pulled in at a rate of 1m/s, how fast is the boat approaching the dock when it is 8m from the dock?

Sketch of a boat with rope of length z to a pulley 1 m above the dock, horizontal distance x.
Figure 6
A hand-drawn diagram of a boat at the left, drawn as a half-circle hull with a small flag-like mast. From a dot at the bow of the boat, a straight line labeled z slants upward to the right to a dot labeled Pulley. From the pulley a short vertical dotted segment labeled 1 drops down to a horizontal line labeled Dock, with a small right-angle mark at the corner. A horizontal dashed line labeled x runs from the bow of the boat to the base of the vertical segment, forming a right triangle with legs x and 1 and hypotenuse z.

Run through the checklist:

  1. Let \( z \) be the length of the rope and \( x \) the boat's distance from the dock.
  2. \( x^2 + 1^2 = z^2 \)
  3. K: \( \frac{dz}{dt} = -1 \)
  4. NTK: \( \frac{dx}{dt} \) (and \( z \)) when \( x = 8 \)
\[ \begin{gathered} x^2 + 1^2 = z^2 \\[0.5em] x \frac{dx}{dt} = z \frac{dz}{dt} \\[0.5em] \frac{dx}{dt} = \frac{z}{x} \frac{dz}{dt} \end{gathered} \]

Find \( z \):

\[ \begin{gathered} z = \sqrt{x^2 + 1} \\[0.5em] z\big|_{x=8} = \sqrt{64 + 1} \\[0.5em] = \sqrt{65} \end{gathered} \]

So,

\[ \begin{gathered} \frac{dx}{dt} = \frac{z}{x} \frac{dz}{dt} \\[0.5em] \frac{dx}{dt}\bigg|_{\substack{z'=-1 \\ x=8}} = \frac{\sqrt{65}}{8}(-1) \\[0.5em] = -\frac{\sqrt{65}}{8} \end{gathered} \]

Therefore, the boat is approaching the dock at a rate of \( \sqrt{65}/8 \) \( \frac{m}{s} \).

2 Trigonometric Relations

Problem 7.

A man walks along a straight path at a speed of 4 ft/s. A searchlight is located on the ground 20 ft from the path and is kept focused on the man. At what rate is the searchlight rotating when the man is 15ft from the point on the path closest to the searchlight?

Searchlight on ground 20 ft from a straight path, beam at angle theta to a walking man at distance x.
Figure 7
A shaded gray illustration showing a man walking to the right along a horizontal straight path. A searchlight sits on the ground at the lower left, below the path. A vertical segment labeled 20 connects the searchlight up to the path, meeting it at a right angle marked by a small square at the point on the path closest to the searchlight. From that closest point, the horizontal distance to the man along the path is labeled \( x \). A beam of light fans out from the searchlight toward the man, and the angle at the searchlight between the 20 ft segment and the line of sight to the man is labeled \( \theta \).

Run through the checklist:

  1. Let \( x \) be the horizontal distance from the man and the search light and \( \theta \) the angle between the man, the search light, and the closest point on the man's path closest to the search light (see Figure 7).
  2. \( \frac{x}{20} = \tan\theta \), or \( x = 20\tan\theta \).
  3. K: \( \frac{dx}{dt} = 4 \)
  4. NTK: \( \frac{d\theta}{dt} \) (and \( \cos\theta \)) when \( x = 15 \).
\[ \begin{gathered} x = 20\tan\theta \\[0.5em] \frac{dx}{dt} = 20\sec^2\theta\, \frac{d\theta}{dt} \\[0.5em] \frac{d\theta}{dt} = \frac{1}{20} \cos^2\theta\, \frac{dx}{dt} \end{gathered} \]

Find \( \cos\theta \) when \( x = 15 \):

\[ \begin{gathered} x = 20\tan\theta \\[0.5em] 15 = 20\tan\theta \\[0.5em] \tan\theta = \frac{15}{20} \\[0.5em] \tan\theta = \frac{3}{4} \end{gathered} \]

From the right triangle,

Right triangle with legs 4 and 3, hypotenuse 5, and angle theta at the vertex between side 4 and the hypotenuse.
Figure 8
A hand-drawn right triangle sketched in red. The horizontal bottom side is labeled 4, the vertical right side is labeled 3, and the slanted hypotenuse rising from the lower-left vertex to the upper-right vertex is labeled 5. The angle at the lower-left vertex, between the horizontal side of length 4 and the hypotenuse, is marked \( \theta \). The right angle is at the lower-right vertex. This 3-4-5 triangle gives \( \tan\theta = \frac{3}{4} \) and \( \cos\theta = \frac{4}{5} \).

We get \( \cos\theta = \frac{4}{5} \)

So,

\[ \begin{gathered} \frac{d\theta}{dt} = \frac{1}{20} \cos^2\theta\, \frac{dx}{dt} \\[0.5em] \frac{d\theta}{dt}\bigg|_{\substack{x=15 \\ x'=4}} = \frac{1}{20}\left(\frac{4}{5}\right)^2 (4) \\[0.5em] = \frac{16}{125} \end{gathered} \]

Therefore, the searchlight is rotating clockwise at a rate of \( \frac{16}{125} \) \( \frac{\mathrm{rad}}{\mathrm{s}} \).

Problem 8.

Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0.06rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is \( \frac{\pi}{3} \).

Triangle with sides 5 and 4 meeting at angle theta, with dashed height h drawn to the base.
Figure 9
A hand-drawn triangle. The side labeled 5 slants upward from the left vertex to the top right vertex; the base labeled 4 runs horizontally from the left vertex to the right. The angle between these two sides, at the left vertex, is marked \( \theta \). A dashed vertical segment labeled \( h \) drops from the top vertex down to the dashed horizontal extension of the base, representing the height of the triangle.

Run through the checklist:

  1. Let \( \theta \) denote the interior angle of the fixed side lengths of the triangle and \( A \) the area of the triangle.
  2. \( A = \frac{1}{2} bh = \frac{1}{2}(4)(5\sin\theta) = 10\sin\theta \)
  3. K: \( \frac{d\theta}{dt} = 0.06 \)
  4. NTK: \( \frac{dA}{dt} \) when \( \theta = \frac{\pi}{3} \)
\[ \begin{gathered} A = 10\sin\theta \\[0.5em] \frac{dA}{dt} = 10\cos\theta\, \frac{d\theta}{dt} \\[0.5em] \frac{dA}{dt}\bigg|_{\substack{\theta=\pi/3 \\ \theta'=0.06}} = 10\cos\left(\frac{\pi}{3}\right)(0.06) \\[0.5em] = 10\left(\frac{1}{2}\right)\cdot\frac{6}{100} \\[0.5em] = \frac{3}{10} \end{gathered} \]

Therefore, the area of the triangle is increasing at a rate of \( \frac{3}{10} \) \( \frac{m^2}{s} \).