Lesson 19 (3.11)
Today:
- Related Rates (part 2)
Office Hours: MWF: 130pm – 230pm (MATH 842)
Announcements:
- Exam 2, Monday 10/19
- No classes on Monday 10/12, Tuesday 10/13
Lesson 19 (3.11)
Page 1
Page 2
Warmup:
Visual Description:
Diagram showing two similar triangles. The larger triangle on the left has sides labeled 6 on the left, 5 on the right, and 8 along the bottom. The smaller triangle on the right has corresponding sides labeled 'a' on the left, 'b' on the right, and 4 along the bottom. Both triangles display matching angle tick arcs indicating angle congruence: a single arc at the top vertex, a double arc at the bottom-left vertex, and a triple arc at the bottom-right vertex.
Find \(a\) & \(b\) if the two triangles are similar
\[ \frac{6}{8} = \frac{a}{4} \rightarrow a = \frac{6}{8} * 4 = 3 \] \[ \frac{5}{8} = \frac{b}{4} \rightarrow b = \frac{5}{8} * 4 = 2.5 \]
Page 3
Related Rates
- Read carefully
- Draw a picture + label variables
- Write down:
- Given
- Unknown
- Equation
- \(\frac{d}{dt}\), apply chain Rule
- plug in givens & solve for unknowns
Page 4
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.
Visual Description:
Geometric sketch depicting a street light on a vertical pole on the left and a 6-foot tall person on the ground to the right. A light ray extends from the lamp at the top of the pole past the top of the person's head down to the ground, creating a shadow. The distance from the base of the lamp post to the person is labeled x, with a red arrow pointing to the left showing movement towards the post. The person's height is labeled 6. The ground segment corresponding to the person's shadow is labeled s. A bracket on the left marks the height of the lamp post.
\(x(t) = \text{distance of person from post}\)
\(s(t) = \text{length of shadow}\)
Given: \(\frac{dx}{dt} = -5\,\text{ft/sec}\).
Find: \(\frac{ds}{dt}\) when \(x = 24\,\text{ft}\).
Equation:
\[ \frac{x+s}{20} = \frac{s}{6} \]
\[ 6x + 6s = 20s \]
\[ 6x = 14s \]
\[ 6 \cdot \frac{dx}{dt} = 14 \frac{ds}{dt} \]
\[ \frac{ds}{dt} = -\frac{6 \times 5}{14}\,\text{ft/sec} = -15/7\,\text{ft/sec} \]
Visual Description:
Right triangle diagram representing the full geometric profile from the lamp to the tip of the shadow, showing a vertical side labeled 20 and a horizontal base labeled x + s.
Visual Description:
Smaller similar right triangle diagram representing the person and the shadow, showing a vertical side labeled 6 and a horizontal base labeled s.
Length of shadow is decreasing at const. Rate \(15/7\,\text{ft/sec}\).
Page 5
Water is poured into an inverted conical tank at a rate of \(2\text{ ft}^3/\text{min}\). If the tank is \(6\text{ ft}\) tall and has radius \(2\text{ ft}\), how fast does the water level rise when the water is \(4\text{ ft}\) deep?
Visual Description:
Hand-drawn diagram of an inverted conical tank pointing downwards. The circular opening at the top has a marked radius of 2. A vertical centerline shows the total height of the cone labeled as 6. The bottom part of the tank is partially filled with water, represented by dashed lines and ripples, with its surface having a radius labeled r.
\(r(t) = \text{radius at the top of water level}\)
\(h(t) = \text{height of the water}\).
\(V(t) = \text{Volume of water}\)
Given: \[\frac{dV}{dt} = 2\text{ ft}^3/\text{min}\]
Find: \[\frac{dh}{dt} \quad \text{when } h = 4\text{ ft}.\]
\(\text{Volume of cone} = V(t) = \frac{1}{3}\pi r^2 h\)
use similar triangles to write \(r\) in terms of \(h\).
Page 6
Visual Description:
Right triangle diagram representing the cross-section of an inverted conical tank. The full vertical left side is labeled with height 6, and the horizontal top base is labeled with radius 2. A horizontal line segment of radius r is shown inside the triangle at a height indicated by a curly bracket labeled h from the bottom vertex, establishing similar triangles.
Similar Triangles
\[ \frac{2}{6} = \frac{r}{h} \implies r = \frac{h}{3} \]
Plug into volume formula
\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 \cdot h = \frac{1}{27}\pi h^3 \] \[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{1}{27}\pi h^3\right) = \frac{1}{27}\pi \cdot 3h^2 \cdot \frac{dh}{dt} \] \[ \frac{dV}{dt} = \frac{1}{9}\pi h^2 \frac{dh}{dt} \]
Given \( \frac{dV}{dt} = 2\text{ ft}^3/\text{min} \), find \( \frac{dh}{dt} \) when \( h = 4\text{ ft} \)
\[ 2\frac{\text{ft}^3}{\text{min}} = \frac{1}{9}\pi \cdot 16\text{ ft}^2 \cdot \frac{dh}{dt} \implies \frac{18}{16\pi}\text{ ft}/\text{min} = \frac{dh}{dt} \]
Page 7
A van travels towards an intersection from the north while a car travels away from the intersection going east. When the van is 0.6 mi north of the intersection and the car is 0.8 mi east of the intersection, the distance between them is increasing at 20mph. If at that instant the van is moving at 60mph, what is the speed of the car?
Visual Description:
Compass coordinate diagram centered at intersection origin 0 with axes labeled N (North), S (South), E (East), and W (West). A point labeled V representing the van is located on the North axis at distance v from the intersection, with a downward arrow indicating motion toward the intersection. A point labeled C representing the car is located on the East axis at distance c from the intersection, with a rightward arrow indicating motion away from the intersection. A line of length d connects V and C, forming the hypotenuse of a right triangle with legs v and c.
\( c(t) = \text{distance of car from intersection} \)
\( v(t) = \text{distance of van from intersection} \)
\( d(t) = \text{distance b/w car \& van} \)
at \( v = 0.6\text{mi} \), \( c = 0.8\text{mi} \),
\[ \frac{dd}{dt} = 20\text{mph}, \quad \frac{dv}{dt} = -60\text{mph} \]
Equation:
\[ (c(t))^2 + (v(t))^2 = (d(t))^2 \]
take \( \frac{d}{dt} \)
Page 8
\[ 2 c(t) \frac{dc}{dt} + 2 v(t) \frac{dv}{dt} = 2 d(t) \frac{dd}{dt} \]
find \( \frac{dc}{dt} \) when \( c = 0.6 \), \( v = 0.8 \), \( \frac{dv}{dt} = -60 \), \( \frac{dd}{dt} = 20 \)
Visual Description:
Right triangle sketch with vertical leg of length 0.8 on the left, horizontal base of length 0.6 at the bottom, and hypotenuse labeled d.
\( d \) when \( c = 0.6 \), \( v = 0.8 \) \( \rightsquigarrow \)
\[ d^2 = (0.6)^2 + (0.8)^2 = 1 \implies d = 1 \] \[ \cancel{2} * 0.6 \frac{dc}{dt} - \cancel{2} * 0.8 * 60 = \cancel{2} * 1 * 20 \] \[ 0.6 \frac{dc}{dt} = 68 \implies \frac{dc}{dt} = \frac{68}{0.6} \underline{\underline{\text{mph}}} \]
Page 9
A baseball player runs from 2nd to 3rd base 24ft/sec. Let \(\theta\) be the angle between the 3rd base line and the runner when viewed from home plate. How fast is \(\theta\) changing when the runner is 30ft from 3rd base?
Visual Description:
Diagram of a baseball diamond with vertices labeled Home, 1st, 2nd, and 3rd. The side from Home to 3rd base is labeled 90ft, and the side from 2nd to 1st base is labeled 90ft. A point along the path from 2nd to 3rd base indicates the runner, with the distance from the runner to 3rd base labeled x, and the segment between 2nd base and the runner labeled 90ft. A line connects Home plate to the runner, forming an angle labeled theta with the third base line.
\(x(t) = \text{distance btw player \& 3rd base}\)
\[ \frac{dx}{dt} = -24\,\text{ft/sec} \]
What is \(\frac{d\theta}{dt}\) when \(x = 30\,\text{ft}\)
equation:
\[ \tan\theta = \frac{x}{90} \]
\[ \frac{d}{dt}\tan\theta = \frac{d}{dt}\left(\frac{x}{90}\right) \]
\[ \sec^2\theta \frac{d\theta}{dt} = \frac{1}{90}\cdot\frac{dx}{dt} \]
Visual Description:
Right triangle diagram representing the geometric snapshot when x = 30. The vertical side opposite angle theta is labeled 30, the horizontal adjacent base is labeled 90, and the hypotenuse is labeled h. The angle between the hypotenuse and the base is labeled theta.
\[ \hookrightarrow h^2 = 30^2 + 90^2 = 9000 \implies h = \sqrt{9000} = 30\sqrt{10} \]
Page 10
\[ \sec\theta = \frac{h}{\text{adj}} = \frac{h}{90} = \frac{30\sqrt{10}}{90} = \frac{\sqrt{10}}{3}. \]
\[ \frac{dx}{dt} = -24 \]
\[ \frac{10}{9} \frac{d\theta}{dt} = \frac{1}{90} * -24 \]
\[ \frac{d\theta}{dt} = \frac{9}{10} * \frac{1}{90} * -24 \text{ Rad/sec}. \]
\[ \frac{d\theta}{dt} = -\frac{24}{100} \text{ Rad/sec}. \]