Calculus AP/Calc AB worksheet with related rates problems.

You’ve uploaded a worksheet titled “Calculus AP/IB Math Review 2017” with six related rates problems. Since I can’t view or process images directly, I’ll solve each problem based on the text you provided — which appears to be a clean transcription of the worksheet.

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## ✔ Problem 1:
> Each side *s* of a square is increasing at a constant rate of 3 in/sec. How fast is the area increasing when *s = 6* in?

Solution:

- Area of square:
\[ A = s^2 \]

- Differentiate both sides w.r.t. time *t*:
\[ \frac{dA}{dt} = 2s \cdot \frac{ds}{dt} \]

- Given:
\[ \frac{ds}{dt} = 3 \text{ in/sec}, \quad s = 6 \text{ in} \]

- Plug in:
\[ \frac{dA}{dt} = 2(6)(3) = 36 \text{ in}^2/\text{sec} \]

✔ Answer: The area is increasing at 36 in²/sec.

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## ✔ Problem 2:
> The radius of a balloon is increasing at the constant rate of 4 ft/min. Find the rate of change of the volume of the balloon at the instant when *r = 2* in, *V = \frac{4}{3}\pi r^3*

*(Note: Units are inconsistent — radius given in feet per minute, but evaluated at r=2 inches. We’ll assume it’s a typo and proceed with r=2 ft. If not, convert units.)*

Solution:

- Volume of sphere:
\[ V = \frac{4}{3}\pi r^3 \]

- Differentiate w.r.t. time *t*:
\[ \frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt} \]

- Given:
\[ \frac{dr}{dt} = 4 \text{ ft/min}, \quad r = 2 \text{ ft} \]

- Plug in:
\[ \frac{dV}{dt} = 4\pi (2)^2 (4) = 4\pi \cdot 4 \cdot 4 = 64\pi \text{ ft}^3/\text{min} \]

✔ Answer: The volume is increasing at \( 64\pi \) ft³/min.

*(If r=2 inches, convert to feet: 2 in = 1/6 ft → then dV/dt = 4π(1/6)²(4) = 4π(1/36)(4) = 16π/36 = 4π/9 ft³/min. But context suggests r=2 ft.)*

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## ✔ Problem 3:
> A 25-foot ladder leans against a vertical wall. If the base of the ladder is pulled away from the wall at the constant rate of 2 ft/sec, how fast is the top of the ladder sliding down the wall when the base of the ladder is 12 feet from the wall?

Solution:

Let:
- *x* = distance from wall to base of ladder (horizontal)
- *y* = height of ladder on wall (vertical)
- By Pythagoras:
\[ x^2 + y^2 = 25^2 = 625 \]

Differentiate w.r.t. time *t*:
\[ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \]
→
\[ x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \]

Given:
- \(\frac{dx}{dt} = 2\) ft/sec
- \(x = 12\) ft → find \(y\):
\[ y = \sqrt{625 - 144} = \sqrt{481} \approx 21.93 \text{ ft} \]

Now solve for \(\frac{dy}{dt}\):

\[ 12(2) + \sqrt{481} \cdot \frac{dy}{dt} = 0 \]
→
\[ \frac{dy}{dt} = -\frac{24}{\sqrt{481}} \approx -1.094 \text{ ft/sec} \]

✔ Answer: The top of the ladder is sliding down at approximately 1.094 ft/sec.

*(Negative sign indicates downward motion.)*

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## ✔ Problem 4:
> A right-circular cone for which *h = r* is used as a reservoir. If water is drained off at the constant rate of 8 cubic feet/min, how fast is the water level decreasing when the water is 12 feet deep? (*V = \frac{1}{3}\pi r^2 h*)

Solution:

Given: *h = r*, so substitute into volume formula:

\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h^2 \cdot h = \frac{1}{3}\pi h^3 \]

Differentiate w.r.t. time *t*:

\[ \frac{dV}{dt} = \pi h^2 \cdot \frac{dh}{dt} \]

Given:
- \(\frac{dV}{dt} = -8\) ft³/min (negative because draining)
- \(h = 12\) ft

Solve for \(\frac{dh}{dt}\):

\[ -8 = \pi (12)^2 \cdot \frac{dh}{dt} \]
→
\[ \frac{dh}{dt} = \frac{-8}{144\pi} = -\frac{1}{18\pi} \text{ ft/min} \]

✔ Answer: The water level is decreasing at \(\frac{1}{18\pi}\) ft/min ≈ 0.0177 ft/min.

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## ✔ Problem 5:
> Sand poured onto the ground at the constant rate of 12 cubic feet/min forms a conical pile whose altitude is 2/3 of the radius of the base. How fast is the height of the pile increasing when the base radius is 6 ft? (*V = \frac{1}{3}\pi r^2 h*)

Solution:

Given:
- \(\frac{dV}{dt} = 12\) ft³/min
- \(h = \frac{2}{3}r\) → so \(r = \frac{3}{2}h\)

Substitute into volume formula:

\[ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{3}{2}h\right)^2 h = \frac{1}{3}\pi \cdot \frac{9}{4}h^2 \cdot h = \frac{3}{4}\pi h^3 \]

Differentiate w.r.t. *t*:

\[ \frac{dV}{dt} = \frac{9}{4}\pi h^2 \cdot \frac{dh}{dt} \]

We need \(\frac{dh}{dt}\) when *r = 6 ft* → since \(h = \frac{2}{3}r\), then \(h = \frac{2}{3}(6) = 4\) ft.

Plug in:

\[ 12 = \frac{9}{4}\pi (4)^2 \cdot \frac{dh}{dt} \]
→
\[ 12 = \frac{9}{4}\pi \cdot 16 \cdot \frac{dh}{dt} = 36\pi \cdot \frac{dh}{dt} \]
→
\[ \frac{dh}{dt} = \frac{12}{36\pi} = \frac{1}{3\pi} \text{ ft/min} \]

✔ Answer: The height is increasing at \(\frac{1}{3\pi}\) ft/min ≈ 0.106 ft/min.

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## ✔ Problem 6:
> Water leaking on a floor creates a circular puddle whose area increases at a rate of 5 square feet per minute. How fast is the radius of the puddle increasing when the radius is 10 inches?

Solution:

Area of circle:
\[ A = \pi r^2 \]

Differentiate w.r.t. time *t*:
\[ \frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} \]

Given:
- \(\frac{dA}{dt} = 5\) ft²/min
- \(r = 10\) inches → convert to feet: \(10/12 = \frac{5}{6}\) ft

Solve for \(\frac{dr}{dt}\):

\[ 5 = 2\pi \left(\frac{5}{6}\right) \cdot \frac{dr}{dt} \]
→
\[ \frac{dr}{dt} = \frac{5}{2\pi \cdot \frac{5}{6}} = \frac{5 \cdot 6}{2\pi \cdot 5} = \frac{6}{2\pi} = \frac{3}{\pi} \text{ ft/min} \]

✔ Answer: The radius is increasing at \(\frac{3}{\pi}\) ft/min ≈ 0.955 ft/min.

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## 📌 Summary of Answers:

1. 36 in²/sec
2. \(64\pi\) ft³/min *(assuming r=2 ft)*
3. ≈ -1.094 ft/sec (downward)
4. \(-\frac{1}{18\pi}\) ft/min (≈ -0.0177 ft/min)
5. \(\frac{1}{3\pi}\) ft/min (≈ 0.106 ft/min)
6. \(\frac{3}{\pi}\) ft/min (≈ 0.955 ft/min)

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## 💡 General Strategy for Related Rates:

1. Draw a diagram if possible.
2. Write down knowns and unknowns.
3. Write an equation relating variables.
4. Differentiate implicitly w.r.t. time.
5. Plug in known values and solve for desired rate.
6. Include units and interpret signs (positive = increasing, negative = decreasing).

Let me know if you’d like step-by-step diagrams or explanations for any specific problem!