Lesson 18 (3.11)

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Page 1

Lesson 18 (3.11)

Today:

  • Related Rates – part 1

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Related Rates

Find rate at which one thing (volume, Area, cost..) is changing when rate at which something else is changing is given (side length, radius, --)

  1. Read the problem very very very carefully.
  2. Draw a schematic picture + Label variable
  3. Write:
    • What is given/known
    • what is to be found/unknown
    • equation relating known & unknown
  4. take derivative w.r.t time, using chain Rule
  5. plug in knowns, find/solve for unknown.

Page 3

How fast is the area of a circle changing if the radius of the circle is increasing at the rate of 2cm/sec when the radius is 4 cm.

A hand-drawn sketch of a circle with green hatch lines covering its interior to indicate area, labeled with an 'A' near the bottom. A center point is marked with a blue line extending to the circumference, labeled with the radius 'r'.
Visual Description: A hand-drawn sketch of a circle with green hatch lines covering its interior to indicate area, labeled with an 'A' near the bottom. A center point is marked with a blue line extending to the circumference, labeled with the radius 'r'.

Know rate: \( \frac{dr}{dt} = 2\text{ cm/sec} \) when \( r = 4\text{ cm} \)

Unknown / find: \( \frac{dA}{dt} \)

Equation relating \( A \) & \( r \):

\[ A = \pi r^2 \]

Take derivative w.r.t time:

\[ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) \] \[ \frac{dA}{dt} = \pi \cdot 2r \frac{dr}{dt} \]

Plug in \( r = 4\text{ cm} \), \( \frac{dr}{dt} = 2\text{ cm/sec} \implies \)

\[ \frac{dA}{dt} = \pi \cdot 2 \cdot 4\text{ cm} \times 2\frac{\text{cm}}{\text{sec}} = \frac{16\pi \text{ cm}^2}{\text{sec}} \]

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How fast is the area of a circle changing if the circumference of the circle is increasing at the rate of 2cm/sec when the radius is 4 cm.

known:

\[ \frac{dC}{dt} = 2\,\text{cm/sec} \quad \text{when } r = 4\,\text{cm} \]

\(C = 8\pi\,\text{cm}\) when \(r = 4\,\text{cm}\).

unknown / find:

\[ \frac{dA}{dt} \]

use: \(A = \pi r^2\), \(C = 2\pi r\) to write a relation b/w \(A\) & \(C\)

\[ C = 2\pi r \Rightarrow r = \frac{C}{2\pi} \quad \text{plug in} \quad A = \pi r^2 \Rightarrow A = \pi \cdot \frac{C^2}{4\pi^2} = \frac{C^2}{4\pi} \]

take \(\frac{d}{dt}\):

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

find \(\frac{dA}{dt}\) when \(\frac{dC}{dt} = 2\,\text{cm/sec}\), \(C = 8\pi\,\text{cm}\).

\[ \frac{dA}{dt} = \frac{1}{2\pi} * 8\pi\,\text{cm} * 2\,\text{cm/sec} = 8\,\text{cm}^2/\text{sec} \]

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\[ A = \pi r^2, \quad C = 2\pi r \]

\[ \frac{dC}{dt} = 2\pi \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{2\pi} \frac{dC}{dt} \]

\[ = \frac{1}{2\pi} * 2\text{ cm/sec} \]

\[ = \frac{1}{\pi}\text{ cm/sec} \]

at \( r = 4\text{cm} \)

\[ A = \pi r^2 \rightsquigarrow \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \]

\[ = 2\pi * 4\text{ cm} * \frac{1}{\pi}\text{ cm/sec} = 8\text{ cm}^2/\text{sec} \]

Page 6

A 13 foot ladder leans against a tall wall. If the bottom is pulled away from the wall at 3ft/sec, how fast is the top of the ladder sliding down the ball when the bottom is 5 ft from the wall.

Diagram of a ladder leaning against a vertical wall, forming a right triangle with the horizontal ground. The ladder serves as the hypotenuse and is labeled '13 ft = L'. The vertical wall is hatched along the left, with the vertical distance from the floor to the top of the ladder labeled 'y'. The horizontal distance along the ground from the wall to the base of the ladder is labeled 'x'. A red arrow at the base of the ladder points away from the wall to the right, indicating the direction of movement.
Visual Description: Diagram of a ladder leaning against a vertical wall, forming a right triangle with the horizontal ground. The ladder serves as the hypotenuse and is labeled '13 ft = L'. The vertical wall is hatched along the left, with the vertical distance from the floor to the top of the ladder labeled 'y'. The horizontal distance along the ground from the wall to the base of the ladder is labeled 'x'. A red arrow at the base of the ladder points away from the wall to the right, indicating the direction of movement.

\( x(t) = \text{distance of bottom from wall} \)

\( y(t) = \text{distance of top from floor} \)

\( L = 13\text{ ft} \)

\( \frac{dx}{dt} = 3\text{ ft/sec} \quad \text{when } x = 5\text{ ft} \)

Find \( \frac{dy}{dt} \)

Equation:

\[ (x(t))^2 + (y(t))^2 = (13)^2 \]

\[ \frac{d}{dt}\left((x(t))^2 + (y(t))^2\right) = \frac{d}{dt}(13)^2 = 0 \]

\[ 2x(t)\frac{dx}{dt} + 2y(t)\frac{dy}{dt} = 0 \]

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find \( \frac{dy}{dt} \) when \( x(t) = 5 \), \( \frac{dx}{dt} = 3 \)

but first find \( y(t) \) when \( x(t) = 5 \)

Right triangle diagram showing a vertical leg labeled y, a horizontal base labeled x(t) = 5, and a hypotenuse labeled 13.
Visual Description: Right triangle diagram showing a vertical leg labeled y, a horizontal base labeled x(t) = 5, and a hypotenuse labeled 13.

\[ 5^2 + y^2 = 13^2 \Rightarrow y^2 = 144 \] \[ y = 12 \] \[ 2x(t)\frac{dx}{dt} + 2y(t)\frac{dy}{dt} = 0 \] \[ 2 * 5 * 3 + 2 * 12 * \frac{dy}{dt} = 0 \] \[ \frac{dy}{dt} = \frac{-30}{24}\text{ ft/sec} \]

-ve Represents \( y \) is decreasing.

Page 8

If the base of a right angled triangle is decreasing at 2cm/sec when the base is 5cm and the height is 4cm. How should the height change for the area of the triangle to remain unchanged.

Sketch of a right-angled triangle. The vertical leg on the left is labeled 'h', the horizontal bottom base is labeled 'b', and the interior of the triangle is shaded with green diagonal hatch lines and labeled 'A' for area.
Visual Description: Sketch of a right-angled triangle. The vertical leg on the left is labeled 'h', the horizontal bottom base is labeled 'b', and the interior of the triangle is shaded with green diagonal hatch lines and labeled 'A' for area.

\( \frac{db}{dt} = -2\text{cm/sec} \), when \( b = 5\text{cm} \), \( h = 4\text{cm} \)

find \( \frac{dh}{dt} \) such that \( \frac{dA}{dt} = 0 \)

Equation:

\[ A(t) = \frac{1}{2} b(t) h(t) \] \[ \frac{d}{dt} A(t)^{\to 0} = \frac{1}{2} \frac{d}{dt} [b(t) h(t)] \] \[ 0 = \frac{1}{2} \left[ \frac{db}{dt} \cdot h(t) + b(t) \frac{dh}{dt} \right] \] \[ 0 = \frac{1}{2} \left[ -2 * 4 + 5 \frac{dh}{dt} \right] \implies \frac{dh}{dt} = 8/5\text{ cm/sec} \]

Page 9

A 6ft tall man walks towards a street light on a post 20ft above the ground at a rate of 5ft/sec. Find the rate of change of the length of his shadow when he is 24ft from the base of the lamp post.

Geometric diagram of the related rates problem illustrating similar right triangles. A vertical street light post on the left has a height labeled 20ft with a light at the top. The ground forms the horizontal base of the large right triangle. A stick figure representing a 6ft tall man stands vertically between the lamp post and the triangle's tip. The horizontal distance from the base of the lamp post to the man is labeled x, accompanied by a red arrow pointing left toward the post to indicate the man walking towards it. The horizontal distance from the man to the tip of the triangle on the ground is labeled s = Shadow. A line of light extends from the lamp at the top of the 20ft post, grazing the top of the man's 6ft frame, to intersect the ground, forming two similar right triangles.
Visual Description: Geometric diagram of the related rates problem illustrating similar right triangles. A vertical street light post on the left has a height labeled 20ft with a light at the top. The ground forms the horizontal base of the large right triangle. A stick figure representing a 6ft tall man stands vertically between the lamp post and the triangle's tip. The horizontal distance from the base of the lamp post to the man is labeled x, accompanied by a red arrow pointing left toward the post to indicate the man walking towards it. The horizontal distance from the man to the tip of the triangle on the ground is labeled s = Shadow. A line of light extends from the lamp at the top of the 20ft post, grazing the top of the man's 6ft frame, to intersect the ground, forming two similar right triangles.

\(x = \text{distance of man from light pole}\)

\(s = \text{length of shadow}\)

\[\frac{dx}{dt} = -5\text{ ft/sec}, \quad \text{when } x(t) = 24\text{ ft}\]

\[\text{Find } \frac{ds}{dt}\]