Related Rates for Distances and Angles

Motion, Geometry, and Rates

A moving point can change its position, its distance from another point, and its direction. These quantities are connected by geometric relations. If the point’s coordinates depend on time, the chain rule turns those relations into equations for their rates of change.

The essential idea is to describe the geometry at a general time first. Then differentiate that relation with respect to time, and only after differentiating substitute the information about a particular instant. This preserves the way the quantities are related throughout the motion.

Definition: A related rate is the time derivative of a quantity whose value depends on other quantities that are changing with time. For example, if $r(t)$ is the distance between two moving points, its rate of change is $\frac{dr}{dt}$.

Distance from Coordinates

For a point $P=(x(t),y(t))$ in the plane, its distance $r(t)$ from the origin satisfies the Pythagorean relation $r^2=x^2+y^2$. Differentiating with respect to time gives a useful formula for the rate at which the distance changes.

Theorem: If $r(t)=\sqrt{x(t)^2+y(t)^2}$ and $r(t)>0$, then $$ \frac{dr}{dt} = \frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{r}. $$ For a point in space with coordinates $(x(t),y(t),z(t))$, its distance $r(t)>0$ from the origin satisfies $$ \frac{dr}{dt} = \frac{x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}}{r}. $$

These formulas also apply to the distance between two moving points. Subtract the coordinates of one point from those of the other to obtain the relative position, then apply the distance formula to that relative position. The rate uses the corresponding differences of their coordinate rates.

Why the Distance Formula Works

In the plane, begin with the Pythagorean relation, which holds at every time:

$$ r^2=x^2+y^2. $$

Differentiate both sides with respect to time. The chain rule gives a factor of the derivative of each time-dependent quantity:

2r\frac{dr}{dt} = 2x\frac{dx}{dt}+2y\frac{dy}{dt}.

Since $r>0$, divide by $2r$ to obtain the plane formula. In space, start instead from $r^2=x^2+y^2+z^2$. Differentiating yields $2r\frac{dr}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}+2z\frac{dz}{dt}$, and division by $2r$ gives the stated formula.

The condition $r>0$ matters: at $r=0$, division by $r$ is not permitted, and the distance function may fail to have a derivative at that instant. The original squared-distance relation may still be differentiated there, but it does not automatically determine $\frac{dr}{dt}$.

Worked Example: A Point’s Distance from the Origin

A point moves in the plane. At a certain instant, its position is $(x,y)=(3,4)$ metres, and its coordinate rates are $\frac{dx}{dt}=2$ metres per second and $\frac{dy}{dt}=-1$ metre per second. Find the rate of change of its distance $r$ from the origin.

First calculate the distance at the instant:

r=\sqrt{3^2+4^2} =\sqrt{9+16} =5\text{ m}.

Apply the distance-rate formula, retaining the coordinate values and rates at this instant:

\frac{dr}{dt} = \frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{r} = \frac{(3)(2)+(4)(-1)}{5} = \frac{6-4}{5} = \frac{2}{5}\text{ m/s}.

Check by differentiating $r^2=x^2+y^2$ at the instant. The left side is $2r\frac{dr}{dt}=2(5)(\frac25)=4$. The right side is $2x\frac{dx}{dt}+2y\frac{dy}{dt}=2(3)(2)+2(4)(-1)=12-8=4$. Both sides agree.

Rates of Angles in the Plane

A moving point’s direction from the origin can be described by its angle $\theta(t)$ from the positive $x$-axis. Locally, where the angle is defined smoothly and the point is not at the origin, its coordinates can be written as $x=r\cos\theta$ and $y=r\sin\theta$.

Differentiate both coordinate expressions. In the combination $x\frac{dy}{dt}-y\frac{dx}{dt}$, the terms involving $\frac{dr}{dt}$ cancel, leaving a relation for the angular rate:

x\frac{dy}{dt}-y\frac{dx}{dt} = r^2\frac{d\theta}{dt}.

Formula: For a moving point $(x(t),y(t))\ne(0,0)$, its angular rate is $$ \frac{d\theta}{dt} = \frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2+y^2}. $$ The angle is measured in radians, and the formula applies on an interval where a differentiable choice of angle $\theta$ is available.

Worked Example: The Direction of a Moving Point

A point is at $(4,3)$ metres at an instant, with coordinate rates $\frac{dx}{dt}=-1$ metre per second and $\frac{dy}{dt}=2$ metres per second. Find the rate at which its direction angle $\theta$ is changing.

Substitute the position and coordinate rates into the angular-rate formula:

\frac{d\theta}{dt} = \frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2+y^2} = \frac{(4)(2)-(3)(-1)}{4^2+3^2} = \frac{8+3}{16+9} = \frac{11}{25}\text{ rad/s}.

For a check, the differentiated relation requires $x\frac{dy}{dt}-y\frac{dx}{dt}=r^2\frac{d\theta}{dt}$. Its left side is $4(2)-3(-1)=11$. Its right side is $(16+9)(\frac{11}{25})=11$, as required.

Angles from Right-Triangle Geometry

Not every angle-rate problem asks for the direction of a point from the origin. An angle may instead be related to a horizontal distance and a height. For example, suppose an observer is at a fixed horizontal distance $x(t)>0$ from a vertical object of fixed height $h>0$, and $\theta(t)$ is the angle of elevation to its top. The right-triangle relation is $\tan\theta=h/x$.

Differentiate this relation with respect to time. Since $h$ is constant, the quotient on the right changes only through $x$:

\sec^2\theta\frac{d\theta}{dt} = -\frac{h}{x^2}\frac{dx}{dt}.

This equation determines $\frac{d\theta}{dt}$ whenever the quantities are defined and $\sec^2\theta\ne0$, which holds for the acute angle of elevation in this setting.

Worked Example: An Angle of Elevation

An observer is $5$ metres horizontally from the base of a $12$-metre vertical object. The observer moves toward it at $2$ metres per second, so $\frac{dx}{dt}=-2$ metres per second. Find the rate of change of the angle of elevation $\theta$ at that instant.

The triangle gives $\tan\theta=h/x=12/5$. Therefore, using $\sec^2\theta=1+\tan^2\theta$,

\sec^2\theta = 1+\left(\frac{12}{5}\right)^2 = 1+\frac{144}{25} = \frac{169}{25}.

Differentiate $\tan\theta=12/x$ and substitute the known values:

\sec^2\theta\frac{d\theta}{dt} = -\frac{12}{5^2}(-2) = \frac{24}{25}.

Now solve for the angular rate:

\frac{d\theta}{dt} = \frac{\frac{24}{25}}{\frac{169}{25}} = \frac{24}{169}\text{ rad/s}.

Check the differentiated relation: its left side is $(\frac{169}{25})(\frac{24}{169})=\frac{24}{25}$. Its right side is $-\frac{12}{25}(-2)=\frac{24}{25}$. The two sides agree.

Separation Between Moving Points in Space

To find the separation rate of two points in three dimensions, use their relative position rather than treating either point as stationary. If $A(t)=(x_A,y_A,z_A)$ and $B(t)=(x_B,y_B,z_B)$, the separation $r$ satisfies

$$ r^2=(x_B-x_A)^2+(y_B-y_A)^2+(z_B-z_A)^2. $$

Differentiation shows that the rate depends on the relative position and relative velocity. In particular, if $\mathbf{d}=B-A$ is the relative position and $\mathbf{v}_{\mathrm{rel}}=B'-A'$ is the relative velocity, then $r'=(\mathbf{d}\cdot\mathbf{v}_{\mathrm{rel}})/r$, provided $r>0$. In coordinates, the dot product here means the sum of the products of corresponding components.

Worked Example: Separation of Two Moving Points in Space

Let the positions, in metres, of two points be $A(t)=(t,0,0)$ and $B(t)=(2+2t,-1+2t,2-t)$, with time measured in seconds. Find the rate at which their separation changes at $t=0$.

At $t=0$, their positions are $A(0)=(0,0,0)$ and $B(0)=(2,-1,2)$. The relative position is therefore $(2,-1,2)$, and the separation is

r=\sqrt{2^2+(-1)^2+2^2} =\sqrt{4+1+4} =3\text{ m}.

Differentiate each position to obtain the velocities: $A'(t)=(1,0,0)$ and $B'(t)=(2,2,-1)$. Thus the relative velocity, $B'-A'$, is $(2-1,2-0,-1-0)=(1,2,-1)$ metres per second.

Apply the distance-rate formula to the relative position and velocity:

\frac{dr}{dt} = \frac{(2)(1)+(-1)(2)+(2)(-1)}{3} = \frac{2-2-2}{3} = -\frac{2}{3}\text{ m/s}.

For a direct check, the derivative of the squared separation must equal $2r\frac{dr}{dt}$. The dot-product calculation gives $2[(2)(1)+(-1)(2)+(2)(-1)]=2(-2)=-4\text{ m}^2/\text{s}$. The computed rate gives $2(3)(-\frac23)=-4\text{ m}^2/\text{s}$, so it satisfies the differentiated relation.

A Common Mistake: Substituting Before Differentiating

A common error is to insert the values for one instant into a geometric relation and then treat the resulting numbers as if they describe how the quantities change. For instance, at $x=3$, $y=4$, the relation $r^2=x^2+y^2$ becomes $r^2=25$. Differentiating that numerical equation gives $2r\frac{dr}{dt}=0$, which incorrectly suggests $\frac{dr}{dt}=0$. The equation $r^2=25$ describes only the values at that instant; it is not the relation as the point moves.

The correction is to differentiate the time-dependent identity before substituting the instant’s values:

r(t)^2=x(t)^2+y(t)^2 \quad\Longrightarrow\quad 2r\frac{dr}{dt} = 2x\frac{dx}{dt}+2y\frac{dy}{dt}.

For the first worked example, the right side at the instant is $2(3)(2)+2(4)(-1)=4$, not zero. The point’s coordinates and their rates must both appear in the differentiated equation.

A Reliable Geometric Setup

Before calculating a related rate, identify what geometric relation connects the changing quantities. For distances, this is often a Pythagorean relation. For angles in a right triangle, it may be a tangent, sine, or cosine relation. Then differentiate the relation with respect to time, keeping the variables as functions of time. Substitute the values and known rates only after differentiation, solve for the requested rate, and check the result in the differentiated equation.

This approach works because the chain rule tracks how each changing quantity contributes to the whole geometric relation. It applies to moving points in a plane or in space, provided the relation is valid at the instant and the relevant derivatives exist. A distance formula that has been divided by the distance additionally requires that the distance be nonzero.

Check Your Understanding

Use the geometric relation first, and delay substitution of instant-specific values until after differentiation.

A point has coordinates $(6,8)$ metres, with coordinate rates $\frac{dx}{dt}=-1$ metre per second and $\frac{dy}{dt}=2$ metres per second. What is its distance rate from the origin?
For a plane point $(x(t),y(t))$, write the formula for its angular rate $\frac{d\theta}{dt}$ in terms of $x,y,\frac{dx}{dt},\frac{dy}{dt}$.
A right triangle relates a fixed height $h$, a changing horizontal distance $x(t)$, and an angle $\theta(t)$ by $\tan\theta=h/x$. What equation results from differentiating this relation?
When calculating the separation rate of two moving points, which velocity should be used: one point’s velocity or the difference of their velocities?
Why must a time-dependent distance relation be differentiated before inserting the values for a particular instant?