Related Rates

When two quantities are tied by an equation at every moment, their rates are tied by its derivative. Choose a scene, set the instant, and work the five steps yourself: the page puts the chain rule's answer beside a rate measured from the motion alone, and draws that rate over the whole range.

Related rates	$\frac{d y}{d t} = \frac{d y}{d x} \cdot \frac{d x}{d t}$

Height of the top, y (m)	4
Rate of the top, y′ (m/s)	−0.375
Top's speed as a multiple of the foot's	0.75
Angle with the floor, θ (degrees)	53.1
Time until the top reaches the floor (s)	4
Height at which that push reaches zero (m)	0.752

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

Related rates are the time derivatives of quantities tied together by an equation. If an equation such as x² + y² = L² holds at every moment, differentiating both sides with respect to t, by the chain rule, gives an equation between the rates x′ and y′, so one known rate fixes the other at any instant where x and y are known. The order matters: differentiate first and put in the instant's numbers afterwards.

What each symbol means

x′: dx/dt, the rate at which x changes with time
y*: the height at which a uniform ladder's top leaves the wall, (2a²L²/(3g))^⅓

Assumes: The ladder is rigid, its foot moves at a constant speed and its top is taken to stay on the wall; the cone is a perfect cone and the shadow is cast by a point of light. The wall's push is for a uniform ladder on a wall and floor without friction, with g = 9.81 m/s². The rates themselves use only the geometry.

The ladder at 3-4-5 3 steps
A 5 m ladder leans against a wall with its foot 3 m out, sliding away at 0.5 m/s. How fast is the top coming down, and how long until the foot reaches 5 m?

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1. The relation holds at every moment. At this one, it gives the height of the top:
  
  $x^{2} + y^{2} = 5^{2}, x = 3 \Rightarrow y = 25 - 9 = 4$
2. Differentiate with respect to t, then put in this instant’s numbers:
  
  $2 x x^{'} + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{x}{y} x^{'} = - \frac{3}{4} \times 0.5 = - 0.375 m/s$
3. The foot has 2 m left to go at a steady 0.5 m/s:
  
  $t = \frac{L - x}{x ^{'}} = \frac{5 - 3}{0 . 5} = 4 s$
Answer

The top comes down at 0.375 m/s, and the foot reaches 5 m after 4 s. The minus sign is the height falling. The answer is a rate at one instant: a second later the foot is at 3.5 m and the top is coming down faster.

$y^{'} = - 0.375 m/s, t = 4 s$
A cone filling at half depth 3 steps
Water runs at 2 m³/min into a cone 4 m deep whose rim has a radius of 2 m, point down. How fast is the level rising when the water is 2 m deep, and how does that compare with its speed at the brim?

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Show solution
1. Similar triangles tie the surface’s radius to the depth, which leaves the volume in one variable:
  
  $\frac{r}{h} = \frac{R}{H} = \frac{2}{4} \Rightarrow r = \frac{h}{2}, V = \frac{π}{3} r^{2} h = \frac{π}{12} h^{3}$
2. Differentiate. The factor in front of h′ is the area of the surface:
  
  $V^{'} = \frac{π}{4} h^{2} h^{'} = π r^{2} h^{'} \Rightarrow h^{'} = \frac{Q}{π r ^{2}} = \frac{2}{π \cdot 1 ^{2}} = 0.6366 m/min$
3. At the brim the surface has twice the radius:
  
  $h = 4, r = 2 : h^{'} = \frac{2}{π \cdot 2 ^{2}} = 0.1592 m/min$
Answer

The level rises at 2/π ≈ 0.6366 m/min, four times the 0.1592 m/min it manages at the brim. Eliminate r before differentiating. Differentiating V = πr²h/3 with r still in it brings in r′ as a second unknown rate.

$h^{'} = \frac{2}{π} \approx 0.6366 m/min, \frac{0 . 6366}{0 . 1592} = 4$
The tip of a shadow 3 steps
A lamp is 6 m up a post. Someone 1.8 m tall walks away from it at 1.5 m/s. When they are 4 m from the post, how long is their shadow, how fast is it growing, and how fast is its tip moving?

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Show solution
1. The ray past the head makes two similar triangles. Solve for the shadow’s length s:
  
  $\frac{s}{1 . 8} = \frac{d + s}{6} \Rightarrow 4.2 s = 1.8 d \Rightarrow s = \frac{1 . 8 \times 4}{4 . 2} = 1.714 m$
2. Differentiate. s is a fixed multiple of d, so its rate is the same multiple of d′:
  
  $s^{'} = \frac{1 . 8}{4 . 2} d^{'} = \frac{1 . 8 \times 1 . 5}{4 . 2} = 0.643 m/s$
3. The tip is d + s from the post:
  
  $(d + s)^{'} = d^{'} + s^{'} = 1.5 + 0.643 = 2.143 m/s$
Answer

The shadow is 1.714 m long and growing at 0.643 m/s, and its tip moves at 2.143 m/s. Neither rate depends on d: ask the same question at 16 m and the answers are the same.

$s^{'} = 0.643 m/s, (d + s)^{'} = 2.143 m/s$
How fast the angle closes 4 steps
A 6.5 m ladder leans against a wall, and its foot slides away at 0.4 m/s. When the foot is 2.5 m out, how fast is the top coming down? Where is the foot when both ends move at the same speed? And how fast is the angle between the ladder and the floor shrinking when the foot is 2.5 m out? Press Worked problem for the first two answers. The page does not print the third.

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Show solution
1. The relation, and the height of the top at this instant:
  
  $x^{2} + y^{2} = 6 . 5^{2}, x = 2.5 \Rightarrow y = 42.25 - 6.25 = 6$
2. Differentiate with respect to t and solve for y′:
  
  $2 x x^{'} + 2 y y^{'} = 0 \Rightarrow y^{'} = - \frac{x}{y} x^{'} = - \frac{2 . 5}{6} \times 0.4 = - 0.1667 m/s$
3. The ends move at the same speed when x/y = 1, that is when x = y:
  
  $\frac{x}{y} = 1 \Rightarrow x = y = \frac{6 . 5}{2} = 4.596 m$
4. For the angle, differentiate x = L cos θ instead, and notice that L sin θ is y:
  
  $x = L cos θ \Rightarrow x^{'} = - L sin θ θ^{'} = - y θ^{'} \Rightarrow θ^{'} = - \frac{x ^{'}}{y} = - \frac{0 . 4}{6} = - 0.0667 rad/s$
Answer

The top comes down at 0.1667 m/s, the speeds match when the foot is 4.596 m out, and the angle shrinks at 0.0667 rad/s, about 3.82° a second. θ′ = −x′/y has the same y underneath as the top's speed, so it too grows without limit as the ladder flattens.

$y^{'} = - 0.1667 m/s, x = 4.596 m, θ^{'} = - 0.0667 rad/s \approx - 3.8 2^{\circ} / s$

The formula sends the top of the ladder into the floor infinitely fast 🖖

y′ = −(x/y)·x′ says the top moves x/y times as fast as the foot. That ratio is below 1 while the ladder stands steeper than 45°, and past 45° it has no ceiling: as the top nears the floor, y goes to zero and the formula’s answer goes to infinity. Press The last 10 cm to watch it. The differentiation assumed the top stays on the wall, and that is what fails. The top needs the wall to push on it, and for a uniform ladder pulled at a steady speed the push reaches zero at y* = (2a²L²/(3g))^⅓, 0.752 m for this ladder. Lower down, the wall would have to pull, so the top comes away. To the right of the amber line, the blue curve is the formula carried on after a real ladder has left the wall.

The tip of your shadow keeps one speed, however far you walk 🖖

Similar triangles make the shadow a fixed fraction of your distance from the post, s = pd/(P − p), and a fixed fraction of d grows at the same fraction of d′. So where you stand does not enter the rate. Press Shadow at 4 m, then Shadow at 16 m: the shadow is four times as long, and both rates are unchanged, flat lines on the chart. The ladder’s rate runs away because x² + y² = L² is a curve. This relation is a straight line, and a straight line ties two rates by a constant.

A cone fills to half its depth in an eighth of the time 🖖

The inflow spreads over the water’s surface, so the level rises at h′ = Q/(πr²): quickly while the surface is small, slowly once it is wide. At half depth the surface has a quarter of the brim’s area, so the level climbs four times as fast as it will at the top. And because volume goes as the cube of depth, half the depth holds an eighth of the water and is reached in an eighth of the filling time. Neither ratio involves the cone’s size or the pour. Change them, and at half depth the page still prints 4 and 12.5%. Press Reset and Run to watch the level race and then crawl.

Learning path

References (1)

The five-step strategy the page's stepper follows (assign symbols, state what is given and what is wanted, find an equation relating the variables, differentiate both sides with respect to time by the chain rule, then substitute the known values and solve), with worked examples on a balloon, an aeroplane, a rocket launch and water draining from a cone-shaped funnel, and the ladder and lamppost-shadow problems among the exercises: G. Strang and E. Herman, Calculus Volume 1, OpenStax (2016), section 4.1, “Related Rates”.

By the 42Math team · Reviewed against published sources. Corrections welcome.