\(P\) = Power measured in watts

\(I\) = Current in amperes.

\(P = 25 I^2\)

At \(t = t_1\),

\(I = 4\) amperes

\(\dfrac{dI}{dt} = 2\) amps/ min

\(\dfrac{dP}{dt} = ?\)

\(P = 25I^2\)

\(\dfrac{dP}{dt} = 50 * \dfrac{dI}{dt}\)

\(= 50 I \dfrac{dI}{dt}\)

At \(t_1\),

\(\dfrac{dP}{dt} = 50 * 4 * 2 = 400\) W/min

The power is increasing 400 W/min.

\(V =\) Volume of gas in cyclinder in cubic inches

\(P =\) Pressure of gas in kilopascals

\(P = \dfrac{3000}{V}\)

At \(t = t_1\),

\(V = 20\) cubic inches.

\(\dfrac{dV}{dt} = 3\) \(in^3/sec\)

\(\dfrac{dP}{dt} = ?\)

\(P = \dfrac{3000}{V}\)

\(P = 3000 V^{-1}\)

\(\dfrac{dP}{dt} = 3000 * (-1)V^{-2} \dfrac{dV}{dt}\)

\(\dfrac{dP}{dt} = 3000 * -1 * (20)^-2 * 3\)

\(= \dfrac{-3000}{20^2} * 3 = -22.5\) kp/sec.

The pressure is decreasing 22.5 kp/sec.

V = Volume of spherical bulb in inches.

r = radius of the baloon in inches.

\(\forall t. \dfrac{dv}{dt} = 5\) in³/s

At \(t = t_1\),

\(r = 3\)

\(\dfrac{dr}{dt} = ?\)

\(V = \dfrac{4}{3}\pi r^3\)

\(\dfrac{dV}{dt} = \dfrac{4}{3} \pi 3r^2 \dfrac{dr}{dt} = 4\pi r^2 \dfrac{dr}{dt}\)

At \(t_1\),

\(\dfrac{dV}{dt} = 4\pi r^2 \dfrac{dr}{dt}\)

\(5 = 4\pi (3^2) \dfrac{dr}{dt}\)

\(\dfrac{dr}{dt} = \dfrac{5}{4\pi 3^2} = \dfrac{5}{36 \pi}\) in/s

x = Length of Alice's shardow in feet

y = Distance between Alice and street light

We need to prove \(\dfrac{dx}{dt}\) is constant.

\(\dfrac{dy}{dt} = 2 ft/s\)

Using the property of similar triangle,

\(\dfrac{9}{y + x} = \dfrac{5}{x}\)

\(9x = 5y + 5x\)

\(4x = 5y\)

\(4 \dfrac{dx}{dt} = 5 \dfrac{dy}{dt}\)

\(\dfrac{dx}{dt} = \dfrac{5dy}{4dt} = \dfrac{5}{4} * 2 = 2.5\) ft/s

So, \(\dfrac{dx}{dt}\) is constant and it is getting longer as \(2.5 > 2\).

x = Alice's height in feets

\(\dfrac{dx}{dt} = -2f/min\)

y = Alice's shadow in feets

\(\dfrac{dy}{dt} = ?\)

Using the property of similar triangle,

\(\dfrac{9}{8 + y} = \dfrac{x}{y}\)

\(9y = 8x + xy\)

\(9\dfrac{dy}{dt} = 8\dfrac{dx}{dt} + x\dfrac{dy}{dt} + y \dfrac{dx}{dt}\)

\(\dfrac{dy}{dt}(9-x) = \dfrac{dx}{dx}(8+y)\)

At \(x=3\) feet,

\(\dfrac{dy}{dt}(9-x) = \dfrac{dx}{dt}(8+y)\)

\(\dfrac{dy}{dt} * 6 = -2 (8+y)\)

\(\dfrac{dx}{dt} = -\dfrac{8+y}{3}\)

We already know that

\(\dfrac{9}{8+y} = \dfrac{x}{y} = \dfrac{3}{y}\)

\(9y = 24 + 3y\)

\(6y = 24\)

\(y = 4\)

\(\dfrac{dx}{dt} = \dfrac{-(8+y)}{3} = \dfrac{-12}{3} = -4\) ft/min

The shadow is decreasing 4 feet per min.

x = distance between building and car in feets

y = distance between the road and the car in feets

We need to find \(\dfrac{dx}{dt}\) when \(y = 100\)

\(x^2 = 50^2 + y^2\)

\(2x\dfrac{dx}{dt} = 2y \dfrac{dy}{dt}\)

\(\dfrac{dx}{dt} = \dfrac{y}{x} * \dfrac{dy}{dt}\)

\(\dfrac{dy}{dt} = 30 ft/sec\)

\(\dfrac{dx}{dt} = \dfrac{y}{x} \dfrac{dy}{dt}\)

We know that

\(x^2 = 50^2 + y^2\)

\(x^2 = 50^2 + 100^2 = 2500 + 10000\)

\(x^2 = 12500\)

\(\dfrac{dx}{dt} = \dfrac{100}{12500} \dfrac{dy}{dt} = \dfrac{1}{125} * 30 = 30/125\)

\(\dfrac{dx}{dt} \approx 0.24\) ft/s

x = Distance between the bottom of the ladder and the wall in feets.

y = Distance between the top of the ladder and the wall in feets.

Length of ladder = 15 feet

$\dfrac{dx}{dt} = 2 $ ft/s

At \(y = 12\) feet, \(\dfrac{dy}{dt} = ?\)

\(15^2 = x^2 + y^2\)

\(0 = 2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt}\)

\(15^2 = x^2 + 12^2\)

\(x^2 = 15^2 - 12^2 = 81\)

\(x = 9\)

\(2x \dfrac{dx}{dt} + 2y \dfrac{dy}{dt} = 0\)

\(2*9*2 + 2*12*\dfrac{dy}{dt} = 0\)

\(\dfrac{dy}{dt} = \dfrac{-2*9*2}{2*12} = \dfrac{-2*9}{12} = \dfrac{-3}{2} = -1.5\) ft/s

The top of the ladder is sliding down at \(1.5\) feet per second.

Bottom of the ladder is fixed.

y = Distance between the top of the ladder and the wall in feets.

x = Distance between the ladder.

\(\dfrac{dx}{dt} = 2\) ft/s

At \(x = 5\) feet, \(\dfrac{dy}{dt} = ?\)

\(x^2 = 3^2 + y^2\)

\(5^2 = 3^2 + y^2\)

\(y^2 = 5^2 - 3^2 = 16\)

\(y = 4\)

\(x^2 = 3^2 + y^2\)

\(2x \dfrac{dx}{dt} = 2y\dfrac{dy}{dt}\)

\(2*5*2 = 2*4* \dfrac{dy}{dt}\)

\(\dfrac{dy}{dt} = \dfrac{2*5*2}{2*4} = \dfrac{5}{2} = 2.5\) ft/s

The ladder is sliding up the wall at \(2.5\) ft/s when the ladder is 5 feet long.

x = Distance between the rocket and ground in km.

\(\theta\) = Anble between the camera and rocket when focused.

When \(x=1\) km, \(\dfrac{d\theta}{dt} = 0.1\) rad/sec

\(\dfrac{dx}{dt} = ?\)

\(\tan \theta = \dfrac{x}{x}\)

\(\sec^2 \theta \dfrac{d\theta}{dt} = \dfrac{dx}{dt} * \dfrac{1}{2}\)

y = Distance between the camera and the top of the rocket.

\(y^2 = 2^2 + 1^2 = 5\)

\(y = \sqrt{5}\)

\(\cos \theta = \dfrac{2}{\sqrt{5}}\)

\(\cos^2 \theta = \dfrac{4}{5}\)

\(\sec^2 \theta = \dfrac{5}{4}\)

\(\sec^2 \theta \dfrac{d\theta}{dt} = \dfrac{dx}{dt} * \dfrac{1}{2}\)

\(\dfrac{5}{4} * 0.1 = \dfrac{1}{2} \dfrac{dx}{dt}\)

\(\dfrac{5}{4} * 0.1 = \dfrac{1}{2} \dfrac{dx}{dt}\)

\(\dfrac{5}{4} * 0.1 = \dfrac{1}{2} \dfrac{dx}{dt}\)

\(\dfrac{dx}{dt} = \dfrac{0.5}{4} * 2 = \dfrac{0.5}{2} = \dfrac{5}{20} = \dfrac{1}{4}\)

\(= 0.25\) km/s

\(\theta =\) Angle between the police and car in radians.

x = Distance between car and the shortedst distance between road and police.

At \(\theta = \dfrac{\pi}{3}\), \(\dfrac{d\theta}{dt} = \dfrac{1}{20}\) rad/s

Speed limit of road = 65mph

Since sum of angle inside the triangle is 180,

\(\dfrac{\pi}{3} + \dfrac{\pi}{2} + a = 180\)

\(\dfrac{2\pi + 3\pi}{6} + a = 2\pi\)

\(a = 2\pi - \dfrac{5\pi}{6} = \dfrac{12\pi}{6} - \dfrac{5\pi}{6}\)

\(a = \dfrac{7\pi}{6}\)

\(\tan \dfrac{7\pi}{6} = \dfrac{1/10}{x} = \dfrac{1}{10x}\)

\(x = \dfrac{1}{10 \tan (7\pi/6)} = \dfrac{\sqrt{3}}{10}\)

We know that,

\(\tan(\pi - \dfrac{\pi}{2} - \theta) = \dfrac{1/10}{x}\)

\(\tan (\dfrac{\pi}{2} - \theta) = \dfrac{1}{10x}\)

\(\cot \theta = \dfrac{1}{10x}\)

\(\cot \theta = \dfrac{1}{10}x^{-1}\)

\(-\csc^2 \theta \dfrac{d\theta}{dt} = -\dfrac{1}{10}x^{-2}\dfrac{dx}{dt}\)

\(-\csc^2 \theta \dfrac{d\theta}{dt} = -\dfrac{1}{10} (\dfrac{\sqrt{3}}{10})^{-2}\dfrac{dx}{dt}\)

\(-(\dfrac{2}{\sqrt{3}})^2 \dfrac{1}{20} = -\dfrac{1}{10} (\dfrac{10}{\sqrt{3}})^{2}\dfrac{dx}{dt}\)

\(-\dfrac{4}{3} \dfrac{1}{20} = -\dfrac{1}{10} * \dfrac{10^2}{3} \dfrac{dx}{dt}\)

\(-\dfrac{1}{15} = \dfrac{-10}{3} \dfrac{dx}{dt}\)

\(\dfrac{dx}{dt} = \dfrac{1}{5*10} = \dfrac{1}{50}\) m/s

\(= \dfrac{1}{50} * 3600\) m/h

\(= 72\) m/h

So, yes the driver should be given a speeding ticket!

V = Volume of sand in the cone measured in \(ft^3\)

h = height of the sand pile

r = Radius of the base

\(\dfrac{dr}{dt} = 2\) ft³/sec

\(h = r\)

At \(h = 5\), \(\dfrac{dh}{dt} = ?\)

\(V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \pi h^3\)

\(\dfrac{dV}{dt} = \dfrac{3h^2}{3} \pi \dfrac{dh}{dt} = h^2 \pi \dfrac{dh}{dt}\)

\(2 = 5^2 \pi \dfrac{dh}{dt}\)

\(\dfrac{dh}{dt} = \dfrac{2}{5^2 \pi} = 0.025\) feet/sec

y = Distance between traffic intersection and the van.

x = Distance between traffic intersection and the police car.

At \(t = t_1\),

At \(x = 2\) miles,

\(\dfrac{dx}{dt} = 80\) mph

\(y = 3\) miles

\(\dfrac{dy}{dt} = 60\) mph

z = Distance between the van and the police vehicle.

\(z^2 = x^2 + y^2\)

We need to find \(\dfrac{dz}{dt}\) at \(t = t_1\)

\(2z \dfrac{dz}{dt} = 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt}\)

\(z^2 = 2^2 + 3^2 = 13\)

\(z = \sqrt{13}\)

\(2\sqrt{13} \dfrac{dz}{dt} = 2*2*80 + 2*3*60\)

\(\dfrac{dz}{dt} = \dfrac{680}{2\sqrt{13}} = 94.29\) mph

Note: Best to learn a little about baseball before solving this!

x = Distance between the first base and the runner.

y = Distance between the home plate and the point of a runner.

Reference: https://www.youtube.com/watch?v=z0lxZJYqBAE

a = Distance between the person and the base of the ferris wheel.

\(a = \theta(t)\)

Using law of cosines,

\(a^2 = 60^2 + 60^2 - 2.60.60 \cos \theta\)

\(a^2 = 2.60^2 - 2.60^2 \cos \theta\)

\(a^2 = 60^2(2-2\cos \theta)\)

\(a = 60\sqrt{2-2\cos \theta}\)

Differentitating it,

\(\dfrac{da}{dt} = 60(2-2\cos \theta)^{-1/2} (-2 * -\sin \theta \dfrac{d\theta}{dt})\)

\(= \dfrac{60}{\sqrt{2-2\cos \theta}} * 2\sin \theta \dfrac{d\theta}{dt}\)

\(\dfrac{da}{dt} = \dfrac{120\sin \theta}{\sqrt{2-2\cos \theta}} \dfrac{d\theta}{dt}\)

It takes 30 minutes to complete one revolution. So to cover \(2\pi\) radians, it took 30 minutes.

\(\dfrac{d\theta}{dt} = \dfrac{2\pi}{30} = \dfrac{\pi}{15}\) radians/minutes

When the passenger is 90 meteres above the base of the wheel, we need to find \(\theta\).

\(\sin \alpha = \dfrac{30}{60} = \dfrac{1}{2}\)

\(\alpha = \dfrac{\pi}{6}\) radians

\(\theta = \dfrac{\pi}{6} + \dfrac{\pi}{2} = \dfrac{2\pi}{3}\)

\(\sin (\dfrac{2\pi}{3}) = \dfrac{\sqrt{3}}{2}\)

\(\cos (\dfrac{2\pi}{3}) = \dfrac{-1}{2}\)

\(\dfrac{da}{dt} = \dfrac{120. \sqrt{3}/2}{\sqrt{2+ \dfrac{1}{2}}} . \dfrac{\pi}{15}\)

\(= \dfrac{60\sqrt{3}\pi \sqrt{2}\sqrt{2}}{\sqrt{5}.15} = \dfrac{60.2.\sqrt{3}\pi}{15 \sqrt{5}}\)

\(= \dfrac{8\sqrt{3}\pi}{\sqrt{5}}\) m/min

\(\theta\) = Angle between the camera and the statue measured in radians.

x = Distance between the bus and the instersection.

\(\dfrac{dx}{dt} = 40 ft/s\)

y = Distance between the camera and the status

\(y^2 = x^2 + 100^2\)

Differentitating it,

\(2y \dfrac{dy}{dt} = 2x \dfrac{dx}{dt}\)

\(2y \dfrac{dy}{dt} = 2x . 40 = 80x\)

\(\dfrac{dy}{dt} = 40\dfrac{x}{y}\)

At \(x=50\),

\(y^2 = 50^2 + 100^2 = 12500\)

\(y = \sqrt{12500}\)

We need to find \(\dfrac{d\theta}{dt}\) at \(x=50\)

\(\cos \theta = \dfrac{x}{x}\)

\(y \cos \theta = x\)

Differentitating it,

\(y * -\sin \theta \dfrac{d\theta}{dt} + cos \theta \dfrac{dy}{dt} = \dfrac{dx}{dt}\)

\(-y\sin \theta \dfrac{d\theta}{dt} + \cos \theta \dfrac{dy}{dt} = \dfrac{dx}{dt}\)

At \(x = 50\),

\(\dfrac{dx}{dt} = 40\) ft/s

\(\dfrac{dy}{dt} = \dfrac{40x}{y} = \dfrac{40.50}{\sqrt{12500}} = \dfrac{40.50}{10.\sqrt{125}}\)

\(= \dfrac{40.50}{10.5\sqrt{5}} = \dfrac{40}{\sqrt{5}}\)

\(\cos \theta = \dfrac{50}{\sqrt{12500}} = \dfrac{50}{10.5\sqrt{5}} = \dfrac{1}{\sqrt{5}}\)

\(\sin \theta = \dfrac{2}{\sqrt{5}}\)

\(50 \sqrt{5} \dfrac{2}{\sqrt{5}} \dfrac{d\theta}{dt} + \dfrac{1}{\sqrt{5}}\dfrac{40}{\sqrt{5}} = 40\)

\(100\dfrac{d\theta}{dt} \theta + \dfrac{40}{5} = 40\)

\(100 \dfrac{d\theta}{dt} = 32\)

\(\dfrac{d\theta}{dt} = 0.32\) radians/s

I found this youtube video helpful to come with the above diagram.

x = Distance between the boat and the dock

y = Distance between the bow of the boat and the man's hand.

\(\dfrac{dy}{dt} = -4 ft/s\)

At \(x=12\), \(\dfrac{dx}{dt} = ?\)

\(y^2 = x^2 + 25\)

\(2y \dfrac{dy}{dt} = 2x\dfrac{dx}{dt}\)

\(2y * - 4 = 2x\dfrac{dx}{dt}\)

\(\dfrac{dx}{dt} = -\dfrac{8y}{2x} = \dfrac{-4y}{x}\)

\(y^2 = x^2 + 25\)

\(y^2 = 12^2 + 25 = 169\)

\(y = \sqrt{169} = 13\)

\(\dfrac{dy}{dt} = -4 * \dfrac{13}{12} = \dfrac{-13}{3} \approx -4.33\) ft/s

Components involved:

• Light bulb
• Table top
• Horizontal board

Board has a circular hole with 3 inch radius.

Distance between the table top and the light bulb is 20 inches.

x = Distance between light bulb and the board.

$\dfrac{dx}{dt} = -5 $ inches/sec

\(A =\) Area of the lighted region

At \(x = 10\), \(\dfrac{dA}{dt} = ?\)

The area depends on two parameters: x and r where r is the radius of the lighted region.

Note that \(r\) also changes with time and depends on \(x\).

By similar triangle law,

\(\dfrac{3}{x} = \dfrac{r}{20}\)

\(A = \pi {r^2}\)

\(= \pi * \dfrac{60^2}{x^2} = 60^2 \pi x^{-2}\)

\(A = 60^2 \pi x^{-2}\)

\(\dfrac{dA}{dt} = 60^2 \pi * -2x^{-3} \dfrac{dx}{dt}\)

\(= -7200 \pi x^{-3} \dfrac{dx}{dt}\)

\(= -7200 \pi 10^{-3} * -5\)

\(= -\dfrac{72}{10} \pi * -5\)

\(= \dfrac{-72}{2}\pi = 36 \pi \approx 113.097\) in²/sec

Trough dimensions:

• (w) Width: 2 feet
• (h) Height: 3 feet
• (l) Length: 5 feet

At noon, the water in trough was

1.5 deep

h = Height of the trough

\(\dfrac{dh}{dt} = 0.1\) ft/hr

Height of pyramid = 455 feet

Base is square that is 756 feet

V = Volume of the pyramid

\(\dfrac{dV}{dt} = 10\) ft³/min

h = Depth of the pyramid

At \(h=400\) feet, \(\dfrac{dh}{dt} = ?\)

Reference: mathisfun: Pyramids

Volume = \(\dfrac{1}{3} *\) base area * height

\(V = \dfrac{1}{3}S^2h\)

\(S =\) length of square.

We need to elimiate \(S\) from the above equation.

Using similar triangle property,

\(\dfrac{S}{h} = \dfrac{756}{455}\)

\(V = \dfrac{1}{3}(\dfrac{756}{455})^2 * h^3\)

\(\dfrac{dV}{dt} = (\dfrac{756}{455})^2h^2\dfrac{dh}{dt}\)

\(\dfrac{dV}{dt} = (\dfrac{756}{455})^2(406)^2*\dfrac{dh}{dt}\)

\(\dfrac{dh}{dt} = 10*(\dfrac{455}{756})^2*\dfrac{1}{400^2}\)

\(= \dfrac{2070250}{91445760000}\)

\(= 2.26 * 10^{-5}\)

\(\beta = \theta + \alpha\)

\(\tan \theta = \tan (\beta - \alpha) = \dfrac{\tan \beta - \tan \alpha}{1 - \tan \alpha \tan \beta}\)

x = Distance between the person and the wall

\(\tan \beta = \dfrac{4}{x}\)

\(\tan \alpha = \dfrac{1}{x}\)

\(\tan \theta = \dfrac{4/x - 1/x}{1 - 4/x.1/x} = \dfrac{(4-1)/x}{(x^2-4)/x^2}\)

\(\tan \theta = \dfrac{(4-1)x}{x^2-4} = \dfrac{-3x}{x^2-4}\)

\(\dfrac{dx}{dt} = 3\) ft/s