Calculus Review Worksheet with Related Rates and Optimization Problems

Let's solve each problem step by step from the Calculus Review Worksheet. These are all related rates problems, which involve using derivatives to find how quantities change with respect to time.

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(1) A 13 ft. ladder is leaning against a house...

Given:
- Ladder length = 13 ft (constant)
- Base sliding away at $ \frac{dx}{dt} = 5 $ ft/sec
- At a certain instant, base is 12 ft from wall → $ x = 12 $
- Find: Rate of change of area of triangle formed by ladder, wall, ground ($ \frac{dA}{dt} $)

Step 1: Use Pythagorean theorem:
$$
x^2 + y^2 = 13^2 = 169
$$
At $ x = 12 $, solve for $ y $:
$$
12^2 + y^2 = 169 \Rightarrow y^2 = 25 \Rightarrow y = 5
$$

Area of triangle:
$$
A = \frac{1}{2}xy
$$

Differentiate both sides w.r.t. $ t $:
$$
\frac{dA}{dt} = \frac{1}{2}\left(x \frac{dy}{dt} + y \frac{dx}{dt}\right)
$$

We know:
- $ x = 12 $
- $ y = 5 $
- $ \frac{dx}{dt} = 5 $
- Need $ \frac{dy}{dt} $

From $ x^2 + y^2 = 169 $, differentiate:
$$
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0
\Rightarrow x\frac{dx}{dt} + y\frac{dy}{dt} = 0
$$
Plug in values:
$$
12(5) + 5\frac{dy}{dt} = 0 \Rightarrow 60 + 5\frac{dy}{dt} = 0 \Rightarrow \frac{dy}{dt} = -12
$$

Now plug into area derivative:
$$
\frac{dA}{dt} = \frac{1}{2}(12(-12) + 5(5)) = \frac{1}{2}(-144 + 25) = \frac{1}{2}(-119) = -59.5
$$

Answer: The area is decreasing at 59.5 ft²/sec.

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(2) Rectangle dimensions changing...

Given:
- $ \frac{dl}{dt} = 5 $ in/sec (length increasing)
- $ \frac{dw}{dt} = -4 $ in/sec (width decreasing)
- At instant: $ w = 15 $ in, $ l = 20 $ in
- Find: $ \frac{dd}{dt} $ where $ d $ is diagonal

Diagonal:
$$
d = \sqrt{l^2 + w^2}
\Rightarrow d^2 = l^2 + w^2
$$

Differentiate:
$$
2d\frac{dd}{dt} = 2l\frac{dl}{dt} + 2w\frac{dw}{dt}
\Rightarrow d\frac{dd}{dt} = l\frac{dl}{dt} + w\frac{dw}{dt}
$$

Find $ d $ at that instant:
$$
d = \sqrt{20^2 + 15^2} = \sqrt{400 + 225} = \sqrt{625} = 25
$$

Now plug in:
$$
25\frac{dd}{dt} = 20(5) + 15(-4) = 100 - 60 = 40
\Rightarrow \frac{dd}{dt} = \frac{40}{25} = 1.6
$$

Answer: Diagonal is increasing at 1.6 in/sec.

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(3) Water tank – cone shape, vertex down...

Given:
- Cone: height = 10 ft, radius = 40 ft → so ratio $ \frac{r}{h} = \frac{40}{10} = 4 \Rightarrow r = 4h $
- Water flowing in at $ \frac{dV}{dt} = 7 $ ft³/min
- Find $ \frac{dh}{dt} $ when $ h = 8 $ ft

Volume of cone:
$$
V = \frac{1}{3}\pi r^2 h
$$
Substitute $ r = 4h $:
$$
V = \frac{1}{3}\pi (4h)^2 h = \frac{1}{3}\pi (16h^2)h = \frac{16}{3}\pi h^3
$$

Differentiate:
$$
\frac{dV}{dt} = \frac{16}{3}\pi \cdot 3h^2 \frac{dh}{dt} = 16\pi h^2 \frac{dh}{dt}
$$

Set $ \frac{dV}{dt} = 7 $, $ h = 8 $:
$$
7 = 16\pi (64) \frac{dh}{dt} = 1024\pi \frac{dh}{dt}
\Rightarrow \frac{dh}{dt} = \frac{7}{1024\pi}
$$

Answer: Depth rising at $ \boxed{\frac{7}{1024\pi}} $ ft/min ≈ 0.00216 ft/min

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(4) Trough with equilateral triangle ends...

Given:
- Ends are equilateral triangles, vertex down
- Trough length = 9 ft
- Water depth = 2 ft
- $ \frac{dh}{dt} = \frac{1}{2} $ ft/min
- Find $ \frac{dV}{dt} $

Let’s find volume $ V $ as function of $ h $. Since cross-section is triangle, and water fills up to height $ h $, we need area of triangular cross-section at height $ h $.

For an equilateral triangle, side length $ s $, height $ H = \frac{\sqrt{3}}{2}s $. But here, since it's pointed downward, and water fills from bottom up, we can use similar triangles.

Assume full height of triangle is $ H $. But not given — however, since it's equilateral, and we’re told depth is 2 ft, we can express width of water surface at depth $ h $.

Let’s suppose the full height of the triangle is $ H $. But since no value is given, maybe we assume it's large enough? Wait — actually, we don’t need full height because we can use proportionality.

In equilateral triangle, width at height $ h $ from vertex is proportional to $ h $.

Let’s define: at height $ h $ from vertex, the width $ w(h) $ is proportional to $ h $.

Since triangle is equilateral, the width at height $ h $ is:
$$
w(h) = k h
$$
But we can find relation.

Let’s say full height = $ H $, then at $ h = H $, width = $ s $, side of triangle.

But since it's equilateral, the width increases linearly from vertex.

So at any height $ h $, the width $ w = \frac{s}{H} h $

But we don’t know $ s $ or $ H $. However, we can use geometry: in an equilateral triangle, the width at a distance $ h $ from the tip (vertex) is proportional to $ h $, and the ratio is constant.

Let’s denote: at height $ h $, the horizontal width is $ w $. By similar triangles:
$$
\frac{w}{h} = \frac{W}{H}
$$
where $ W $ is full width, $ H $ is full height.

But since we don’t have $ W $ or $ H $, let’s instead compute area of water cross-section.

Cross-sectional area of water at depth $ h $ is:
$$
A = \frac{1}{2} \cdot \text{base} \cdot \text{height}
$$
But base varies with $ h $. Let’s call the width of water surface at depth $ h $ as $ b(h) $.

Because of similarity: at height $ h $, the width $ b(h) $ is proportional to $ h $.

Let’s suppose the full height of triangle is $ H $. Then at $ h = H $, width = $ b(H) $. But since equilateral, $ b(H) = s $, side length.

But we don’t know $ s $. However, we can express everything in terms of $ h $, and the ratio will cancel.

Actually, better: in equilateral triangle, the width at height $ h $ from vertex is:
$$
b(h) = \frac{2}{\sqrt{3}} h \cdot \tan(60^\circ)? \quad \text{No.}
$$

Wait — simpler: in equilateral triangle, the width at height $ h $ from vertex is:
$$
b(h) = \frac{2}{\sqrt{3}} h \cdot \sin(60^\circ)? \quad \text{Too messy.}
$$

Better: consider that in an equilateral triangle, the width increases linearly from 0 at vertex to $ s $ at height $ H $, so $ b(h) = \frac{s}{H} h $

But again, unknowns.

But we don't need absolute size — we just need to find $ \frac{dV}{dt} $, and we know $ \frac{dh}{dt} $, so if we can write $ V $ in terms of $ h $, we can differentiate.

But we need a relationship between width and height.

Let’s suppose the full height of the triangle is $ H $. Then at height $ h $, the width $ w $ satisfies:
$$
\frac{w}{h} = \frac{W}{H}
\Rightarrow w = \frac{W}{H} h
$$

Then cross-sectional area of water:
$$
A(h) = \frac{1}{2} \cdot w \cdot h = \frac{1}{2} \cdot \left(\frac{W}{H} h\right) \cdot h = \frac{W}{2H} h^2
$$

Then volume:
$$
V = \text{length} \times A = 9 \cdot \frac{W}{2H} h^2 = C h^2
$$
where $ C = \frac{9W}{2H} $

Then:
$$
\frac{dV}{dt} = 2C h \frac{dh}{dt}
$$

But we still don’t know $ C $. But wait — we can find $ w $ at $ h = 2 $ ft using geometry.

In equilateral triangle, the width at height $ h $ from vertex is related to the angle.

The triangle has 60° angles. From vertex, two lines at 60° to vertical?

Actually, easier: in equilateral triangle, the width at height $ h $ is:
$$
w = \frac{2}{\sqrt{3}} h \cdot \tan(60^\circ)? \quad \text{No.}
$$

Let me draw: from vertex, two sides go out at 60° to each other. So the horizontal width at height $ h $ is:

The slope of side: from vertex to base, over height $ H $, the horizontal run is $ \frac{s}{2} $. So slope = $ \frac{s/2}{H} $

Thus, at height $ h $, the half-width is $ \frac{s}{2H} h $, so total width $ w = \frac{s}{H} h $

Now, area of water cross-section:
$$
A = \frac{1}{2} \cdot w \cdot h = \frac{1}{2} \cdot \left(\frac{s}{H} h\right) \cdot h = \frac{s}{2H} h^2
$$

So $ V = 9 \cdot \frac{s}{2H} h^2 = K h^2 $, $ K = \frac{9s}{2H} $

But $ \frac{s}{H} = \frac{s}{(\sqrt{3}/2)s} = \frac{2}{\sqrt{3}} $, since height of equilateral triangle is $ H = \frac{\sqrt{3}}{2}s $

So $ \frac{s}{H} = \frac{2}{\sqrt{3}} $

Therefore:
$$
A = \frac{1}{2} \cdot \left(\frac{2}{\sqrt{3}} h\right) \cdot h = \frac{1}{\sqrt{3}} h^2
$$

Then volume:
$$
V = 9 \cdot \frac{1}{\sqrt{3}} h^2 = \frac{9}{\sqrt{3}} h^2 = 3\sqrt{3} h^2
$$

Now differentiate:
$$
\frac{dV}{dt} = 3\sqrt{3} \cdot 2h \frac{dh}{dt} = 6\sqrt{3} h \frac{dh}{dt}
$$

At $ h = 2 $, $ \frac{dh}{dt} = \frac{1}{2} $:
$$
\frac{dV}{dt} = 6\sqrt{3} \cdot 2 \cdot \frac{1}{2} = 6\sqrt{3}
$$

Answer: Water is flowing into trough at $ \boxed{6\sqrt{3}} $ ft³/min ≈ 10.39 ft³/min

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(5) Right circular cone — radius increasing, height decreasing...

Given:
- Radius $ r = 4 $ in, $ \frac{dr}{dt} = 5 $ in/sec
- Height $ h = 3 $ in, $ \frac{dh}{dt} = -6 $ in/sec
- Find $ \frac{dV}{dt} $

Volume of cone:
$$
V = \frac{1}{3}\pi r^2 h
$$

Differentiate:
$$
\frac{dV}{dt} = \frac{1}{3}\pi \left(2r\frac{dr}{dt} h + r^2 \frac{dh}{dt}\right)
$$

Plug in:
$$
= \frac{1}{3}\pi \left(2(4)(5)(3) + (4)^2 (-6)\right)
= \frac{1}{3}\pi \left(120 - 96\right) = \frac{1}{3}\pi (24) = 8\pi
$$

Answer: Volume is increasing at $ \boxed{8\pi} $ in³/sec

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(6) Clock hands — approaching each other at 4:00

Given:
- Hour hand: 3 ft
- Minute hand: 5 ft
- Find rate at which tips are approaching at exactly 4:00

At 4:00:
- Hour hand points at 4 → 120° from 12 o'clock
- Minute hand points at 12 → 0°
- Angle between them: $ \theta = 120^\circ = \frac{2\pi}{3} $ radians

Let $ \theta(t) $ be angle between hands.

Let $ r_1 = 3 $, $ r_2 = 5 $

Distance between tips:
$$
D^2 = r_1^2 + r_2^2 - 2r_1 r_2 \cos\theta
\Rightarrow D^2 = 9 + 25 - 2(3)(5)\cos\theta = 34 - 30\cos\theta
$$

Differentiate:
$$
2D \frac{dD}{dt} = 30 \sin\theta \frac{d\theta}{dt}
\Rightarrow \frac{dD}{dt} = \frac{15 \sin\theta}{D} \frac{d\theta}{dt}
$$

Now find $ \frac{d\theta}{dt} $: rate of change of angle between hands.

Minute hand: $ 360^\circ $ per hour → $ 6^\circ $/min = $ \frac{\pi}{30} $ rad/sec
Hour hand: $ 30^\circ $ per hour → $ 0.5^\circ $/min = $ \frac{\pi}{360} $ rad/sec

So relative angular speed:
$$
\frac{d\theta}{dt} = \frac{\pi}{30} - \frac{\pi}{360} = \frac{12\pi - \pi}{360} = \frac{11\pi}{360} \text{ rad/sec}
$$

At 4:00, $ \theta = 120^\circ = \frac{2\pi}{3} $, so $ \sin\theta = \sin(120^\circ) = \frac{\sqrt{3}}{2} $

$ D^2 = 34 - 30\cos(120^\circ) = 34 - 30(-0.5) = 34 + 15 = 49 \Rightarrow D = 7 $

Now:
$$
\frac{dD}{dt} = \frac{15 \cdot \frac{\sqrt{3}}{2}}{7} \cdot \frac{11\pi}{360}
= \frac{15\sqrt{3}}{14} \cdot \frac{11\pi}{360}
= \frac{165\sqrt{3}\pi}{5040} = \frac{11\sqrt{3}\pi}{336}
$$

But this is positive, meaning distance is increasing? Wait — but we want approaching, so sign?

Wait: $ \frac{d\theta}{dt} > 0 $, meaning angle is increasing (minute hand catching up), so distance may be decreasing?

Wait — at 4:00, minute hand is behind hour hand, moving faster, so angle between them is decreasing? No!

At 4:00, hour hand at 120°, minute hand at 0°, so angle is 120°.

As time passes, minute hand moves forward (increasing angle), hour hand moves slowly forward. So angle between them decreases until they meet.

So $ \frac{d\theta}{dt} $ should be negative?

Wait: define $ \theta $ as angle from minute to hour hand? Or absolute?

Actually, if $ \theta $ is the angle between them, and minute hand moves faster than hour hand, the angle decreases.

So $ \frac{d\theta}{dt} = \omega_m - \omega_h = \frac{\pi}{30} - \frac{\pi}{360} = \frac{11\pi}{360} $ — positive? That would mean angle is increasing.

But from 4:00 to 4:05, minute hand goes to 30°, hour hand to ~122.5°, so angle becomes ~92.5°, so decreasing.

So $ \theta $ is decreasing → $ \frac{d\theta}{dt} < 0 $

So we must define $ \theta $ as the angle between them, decreasing.

So $ \frac{d\theta}{dt} = -\left(\frac{11\pi}{360}\right) $ rad/sec

Then:
$$
\frac{dD}{dt} = \frac{15 \sin\theta}{D} \cdot \left(-\frac{11\pi}{360}\right)
= \frac{15 \cdot \frac{\sqrt{3}}{2}}{7} \cdot \left(-\frac{11\pi}{360}\right)
= \frac{15\sqrt{3}}{14} \cdot \left(-\frac{11\pi}{360}\right)
= -\frac{165\sqrt{3}\pi}{5040} = -\frac{11\sqrt{3}\pi}{336}
$$

So the distance is decreasing at $ \frac{11\sqrt{3}\pi}{336} $ ft/sec

But the question asks: "How fast are the tips approaching?" → so answer is magnitude.

Answer: $ \boxed{\frac{11\sqrt{3}\pi}{336}} $ ft/sec ≈ 0.055 ft/sec

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(7) Jessica and Karen running at 120° angle...

Given:
- Jessica: 6 mph
- Karen: 9 mph
- Initial angle between paths: 120°
- After 1 hour, how fast is distance between them changing?

Let $ x(t) = 6t $ miles (Jessica)
$ y(t) = 9t $ miles (Karen)

Angle between paths: $ \theta = 120^\circ $

Distance between them:
$$
D^2 = x^2 + y^2 - 2xy\cos\theta
$$

$ \cos(120^\circ) = -\frac{1}{2} $

So:
$$
D^2 = x^2 + y^2 - 2xy(-1/2) = x^2 + y^2 + xy
$$

Differentiate:
$$
2D \frac{dD}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} + \left(y\frac{dx}{dt} + x\frac{dy}{dt}\right)
$$

At $ t = 1 $:
- $ x = 6 $, $ y = 9 $
- $ \frac{dx}{dt} = 6 $, $ \frac{dy}{dt} = 9 $

Compute $ D $:
$$
D^2 = 6^2 + 9^2 + 6*9 = 36 + 81 + 54 = 171 \Rightarrow D = \sqrt{171} = 3\sqrt{19}
$$

Now:
$$
2D \frac{dD}{dt} = 2(6)(6) + 2(9)(9) + (9)(6) + (6)(9)
= 72 + 162 + 54 + 54 = 342
$$

So:
$$
\frac{dD}{dt} = \frac{342}{2D} = \frac{342}{2\sqrt{171}} = \frac{171}{\sqrt{171}} = \sqrt{171}
$$

Wait: $ \frac{171}{\sqrt{171}} = \sqrt{171} $? Yes, because $ \frac{a}{\sqrt{a}} = \sqrt{a} $

So $ \frac{dD}{dt} = \sqrt{171} $ mph

But $ \sqrt{171} = \sqrt{9 \cdot 19} = 3\sqrt{19} $

Answer: Distance is increasing at $ \boxed{3\sqrt{19}} $ mph ≈ 13.08 mph

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(8) Elf walking away from lamppost...

Given:
- Lamppost: 22 ft tall
- Elf: 4 ft tall
- Walking at 4 ft/sec
- When elf is 30 ft from base, find:
(a) How fast is shadow length changing?
(b) How fast is tip of shadow changing?

Let:
- $ x $ = distance from elf to lamppost
- $ s $ = length of shadow
- Tip of shadow is at $ x + s $ from lamppost

By similar triangles:
$$
\frac{22}{x + s} = \frac{4}{s}
\Rightarrow 22s = 4(x + s) = 4x + 4s
\Rightarrow 18s = 4x \Rightarrow s = \frac{2}{9}x
$$

(a) $ \frac{ds}{dt} = \frac{2}{9} \frac{dx}{dt} = \frac{2}{9}(4) = \frac{8}{9} $ ft/sec

(b) Tip of shadow: $ T = x + s = x + \frac{2}{9}x = \frac{11}{9}x $

So $ \frac{dT}{dt} = \frac{11}{9} \cdot 4 = \frac{44}{9} $ ft/sec

Answers:
- Shadow length changing at $ \boxed{\frac{8}{9}} $ ft/sec
- Tip of shadow changing at $ \boxed{\frac{44}{9}} $ ft/sec

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## ✔ Final Answers Summary:

1. Area changing at –59.5 ft²/sec (decreasing)
2. Diagonal changing at 1.6 in/sec (increasing)
3. Depth rising at $ \frac{7}{1024\pi} $ ft/min
4. Water flowing in at $ 6\sqrt{3} $ ft³/min
5. Volume increasing at $ 8\pi $ in³/sec
6. Tips approaching at $ \frac{11\sqrt{3}\pi}{336} $ ft/sec
7. Distance changing at $ 3\sqrt{19} $ mph
8. Shadow length: $ \frac{8}{9} $ ft/sec, tip: $ \frac{44}{9} $ ft/sec

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