Related Rates Problems: These are problems where:
Related rates problems involve finding how fast one quantity is changing when we know how fast a related quantity is changing. The key is that both quantities depend on time, and they are connected by some equation.
Understanding Rates of Change:
For example, if a radius is "increasing at 2 cm/sec", this means \(\frac{dr}{dt} = 2\) cm/sec (positive value).
If a distance is "decreasing at 3 ft/sec", this means \(\frac{dx}{dt} = -3\) ft/sec (negative value).
Problem: How fast is the area of a circle changing if the radius of the circle is increasing at the rate of 2 cm/sec when the radius is 4 cm?
Diagram: Imagine a circle with radius \(r\) expanding outward. The radius increases uniformly in all directions.
Solution:
Step 1 & 2: Understand the problem and visualize it.
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating \(A\) and \(r\):
\[A = \pi r^2\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}\]Plug in known values when \(r = 4\) cm:
\[\frac{dA}{dt} = 2\pi (4 \text{ cm})(2 \text{ cm/sec}) = 16\pi \text{ cm}^2\text{/sec}\]Answer: The area is increasing at \(16\pi \text{ cm}^2\text{/sec}\) (approximately 50.27 cm²/sec).
Interpretation: Since \(\frac{dA}{dt} = 16\pi > 0\), the area is increasing, which makes sense because the radius is increasing.
Problem: How fast is the area of a circle changing if the circumference of the circle is increasing at the rate of 2 cm/sec when the radius is 4 cm?
Solution:
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating \(A\) and \(C\):
We know \(C = 2\pi r\), so \(r = \frac{C}{2\pi}\)
Substituting into \(A = \pi r^2\):
\[A = \pi r^2 = \pi \left(\frac{C}{2\pi}\right)^2 = \pi \cdot \frac{C^2}{4\pi^2} = \frac{C^2}{4\pi}\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{dA}{dt} = \frac{1}{4\pi} \cdot \frac{d(C^2)}{dt} = \frac{1}{4\pi} \cdot 2C \cdot \frac{dC}{dt} = \frac{C}{2\pi} \cdot \frac{dC}{dt}\]Plug in known values when \(r = 4\) cm (so \(C = 8\pi\) cm):
\[\frac{dA}{dt} = \frac{8\pi \text{ cm}}{2\pi} \cdot 2 \text{ cm/sec} = 4 \text{ cm} \cdot 2 \text{ cm/sec} = 8 \text{ cm}^2\text{/sec}\]Answer: The area is increasing at 8 cm²/sec.
Problem: A 13-foot ladder leans against a tall wall. If the bottom is pulled away from the wall at 3 ft/sec, how fast is the top of the ladder sliding down the wall when the bottom is 5 ft from the wall?
Diagram: A ladder of length 13 ft leans against a vertical wall. Let \(x\) be the distance from the wall to the bottom of the ladder (along the floor), and let \(y\) be the height of the top of the ladder (along the wall). As the bottom moves away from the wall (to the right), the top slides down the wall.
Solution:
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating \(x\) and \(y\):
By the Pythagorean theorem:
\[x^2 + y^2 = L^2 = 13^2 = 169\]First, find \(y\) when \(x = 5\):
\[5^2 + y^2 = 169\] \[25 + y^2 = 169\] \[y^2 = 144\] \[y = 12 \text{ ft}\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(169)\] \[2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0\]Plug in known values:
\[2(5)(3) + 2(12)\frac{dy}{dt} = 0\] \[30 + 24\frac{dy}{dt} = 0\] \[\frac{dy}{dt} = -\frac{30}{24} = -\frac{15}{12} = -\frac{5}{4} \text{ ft/sec}\]Answer: The top of the ladder is sliding down the wall at \(\frac{5}{4}\) ft/sec (or 1.25 ft/sec).
Interpretation: The negative sign indicates that \(y\) is decreasing, which makes sense because the distance between the top of the ladder and the floor is decreasing as the ladder slides down.
Problem: If the base of a right-angled triangle is decreasing at 2 cm/sec when the base is 5 cm and the height is 4 cm, how should the height change for the area of the triangle to remain unchanged?
Diagram: A right triangle with base \(b\) (horizontal) and height \(h\) (vertical). As the base decreases, we want to find how the height must change to keep the area constant.
Solution:
Step 3: Identify knowns and unknowns:
Step 4: Write equation relating base, height, and area:
\[A = \frac{1}{2}bh\]Step 5: Take \(\frac{d}{dt}\) of both sides:
\[\frac{dA}{dt} = \frac{1}{2}\left[\frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt}\right]\]Since the area remains unchanged, \(\frac{dA}{dt} = 0\):
\[0 = \frac{1}{2}\left[\frac{db}{dt} \cdot h + b \cdot \frac{dh}{dt}\right]\]Plug in known values:
\[0 = \frac{1}{2}\left[(-2)(4) + (5)\frac{dh}{dt}\right]\] \[0 = \frac{1}{2}\left[-8 + 5\frac{dh}{dt}\right]\] \[0 = -8 + 5\frac{dh}{dt}\] \[\frac{dh}{dt} = \frac{8}{5} \text{ cm/sec}\]Answer: The height should be increasing at \(\frac{8}{5}\) cm/sec (or 1.6 cm/sec) to keep the area constant.
Interpretation: As the base decreases, the height must increase to maintain the same area. This makes intuitive sense.
Important Tips:
Related rates problems involve finding the rate of change of one quantity in terms of the rate of change of another related quantity. The key strategy is:
With practice, these problems become a matter of following the systematic approach and being careful with the calculus.