Diagram: Two triangles with sides labeled. First triangle has sides 3, 6, and 8. Second triangle has sides \(a\), 4, and \(b\).
Solution:
When two triangles are similar, the lengths of their corresponding sides are proportional.
Setting up the proportion:
\[\frac{3}{6} = \frac{a}{4}\]Solving for \(a\):
\[a = \frac{3 \cdot 4}{6} = \frac{12}{6} = 2\]Similarly:
\[\frac{8}{6} = \frac{b}{4}\]Solving for \(b\):
\[b = \frac{8 \cdot 4}{6} = \frac{32}{6} = \frac{16}{3}\]Answer: \(a = 2\) and \(b = \frac{16}{3}\)
Problem: A 6 ft tall man walks towards a street light on a post 20 ft above the ground at a rate of 5 ft/sec. Find the rate of change of the length of his shadow when he is 24 ft from the base of the lamp post.
Diagram: A vertical lamp post 20 ft tall on the left. A 6 ft tall man stands at distance \(x\) from the base of the lamp post. His shadow extends to the right with length \(L\). Light rays from the top of the lamp post pass over the man's head to the tip of his shadow, forming similar triangles.
Solution:
Step 1 & 2: Define variables and understand the setup.
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating \(x\) and \(L\):
Using similar triangles, we compare the large triangle (from lamp top to shadow tip) with the small triangle (from man's head to shadow tip):
The large triangle has height 20 ft and base \(x + L\).
The small triangle has height 6 ft and base \(L\).
By similar triangles:
\[\frac{x + L}{20} = \frac{L}{6}\]Cross-multiplying:
\[6(x + L) = 20L\] \[6x + 6L = 20L\] \[6x = 14L\] \[L = \frac{6x}{14} = \frac{3x}{7}\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{dL}{dt} = \frac{3}{7} \cdot \frac{dx}{dt}\]Plug in known values:
\[\frac{dL}{dt} = \frac{3}{7} \cdot (-5) = -\frac{15}{7} \text{ ft/sec}\]Answer: The length of the shadow is decreasing at \(\frac{15}{7}\) ft/sec (approximately 2.14 ft/sec).
Note: The negative sign indicates the shadow is getting shorter, which makes sense as the man walks toward the lamp post.
Problem: Water is poured into an inverted conical tank at a rate of \(\frac{2}{3}\) ft³/min. If the tank is 6 ft tall and has radius 2 ft at the top, how fast does the water level rise when the water is 4 ft deep?
Diagram: An inverted cone (point down) with total height 6 ft and top radius 2 ft. Water fills the cone to height \(h\) from the bottom (the vertex), forming a smaller cone with radius \(r\) at the water surface.
Solution:
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating variables:
The volume of a cone is:
\[V = \frac{1}{3}\pi r^2 h\]However, we have two variables (\(r\) and \(h\)). We need to relate them using similar triangles.
The full tank has height 6 ft and radius 2 ft, so:
\[\frac{r}{h} = \frac{2}{6} = \frac{1}{3}\]Therefore:
\[r = \frac{h}{3}\]Substituting into the volume formula:
\[V = \frac{1}{3}\pi \left(\frac{h}{3}\right)^2 h = \frac{1}{3}\pi \cdot \frac{h^2}{9} \cdot h = \frac{\pi h^3}{27}\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{dV}{dt} = \frac{\pi}{27} \cdot 3h^2 \cdot \frac{dh}{dt} = \frac{\pi h^2}{9} \cdot \frac{dh}{dt}\]Plug in known values when \(h = 4\) ft:
\[\frac{2}{3} = \frac{\pi (4)^2}{9} \cdot \frac{dh}{dt}\] \[\frac{2}{3} = \frac{16\pi}{9} \cdot \frac{dh}{dt}\]Solving for \(\frac{dh}{dt}\):
\[\frac{dh}{dt} = \frac{2}{3} \cdot \frac{9}{16\pi} = \frac{18}{48\pi} = \frac{3}{8\pi} \text{ ft/min}\]Answer: The water level is rising at \(\frac{3}{8\pi}\) ft/min (approximately 0.12 ft/min or about 1.4 inches/min).
Problem: 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 20 mph. If at that instant the van is moving at 60 mph, what is the speed of the car?
Diagram: An intersection forming a coordinate system. The van is on the vertical axis (north) at distance \(V\) from the intersection, moving downward (toward the intersection). The car is on the horizontal axis (east) at distance \(C\) from the intersection, moving to the right (away from the intersection). The straight-line distance between them is \(D\).
Solution:
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating \(C\), \(V\), and \(D\):
By the Pythagorean theorem:
\[C^2 + V^2 = D^2\]First, find \(D\) when \(C = 0.8\) and \(V = 0.6\):
\[D^2 = (0.8)^2 + (0.6)^2 = 0.64 + 0.36 = 1.0\] \[D = 1 \text{ mi}\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{d}{dt}(C^2 + V^2) = \frac{d}{dt}(D^2)\] \[2C\frac{dC}{dt} + 2V\frac{dV}{dt} = 2D\frac{dD}{dt}\]Dividing by 2:
\[C\frac{dC}{dt} + V\frac{dV}{dt} = D\frac{dD}{dt}\]Plug in known values:
\[(0.8)\frac{dC}{dt} + (0.6)(-60) = (1)(20)\] \[0.8\frac{dC}{dt} - 36 = 20\] \[0.8\frac{dC}{dt} = 56\] \[\frac{dC}{dt} = \frac{56}{0.8} = 70 \text{ mph}\]Answer: The car is traveling at 70 mph.
Interpretation: The positive value confirms the car is moving away from the intersection (in the positive direction).
Important Problem-Solving Tips:
Common Mistakes to Avoid:
Related rates problems in Part 2 involve more complex geometric relationships: