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4. Related Rates

by M. Bourne

If 2 variables both vary with respect to time and have a relation between them, we can express the rate of change of one in terms of the other.

We need to differentiate both sides w.r.t. (with respect to) time.

That is, we'll be finding `(df)/(dt)` for some function `f(t)`.

Important!

Recall from implicit differentiation the following for some function `x` of `t`:

`d/(dt)x^2=2x(dx)/(dt)`

`d/(dt)x^3=3x^2(dx)/(dt)`

`d/(dt)x^4=4x^3(dx)/(dt)`

`d/(dt)x^5=5x^4(dx)/(dt)`

We use this concept throughout this section on related rates.

Example 1

A `20\ "m"` ladder leans against a wall. The top slides down at a rate of 4 ms-1. How fast is the bottom of the ladder moving when it is 16 m from the wall?

Steps:

  1. Make a sketch of the problem
  2. Identify constant and variable quantities
  3. Establish relationship between quantities.
  4. Differentiate w.r.t time.
  5. Evaluate at point of interest.

Answer

Here's the sketch of the situation. The variables `x` and `y` vary as time varies.

20 m x y
`(dy)/(dt) = -4" ms"^-1`
`(dx)/(dt) =\ ?`

Ladder sliding down a wall

(We regard UP as being the POSITIVE direction.)

Now the relation between x and y is:

x2 + y2 = 202

Differentiating throughout with respect to time (since the value of x and y depends on t):

`d/(dt)x^2+d/(dt)y^2=0`

`2x(dx)/(dt)+2y(dy)/(dt)=0`

That is:

`x(dx)/(dt)+y(dy)/(dt)=0`

Now, we know

`(dy)/(dt)=-4`

and we need to know the horizontal velocity (`dx/(dt)`) when

`x = 16`.

The only other unknown is y, which we obtain using Pythagoras' Theorem:

`y=sqrt(20^2-16^2)=sqrt(144)=12`

So

`(16)(dx)/(dt)+(12)(-4)=0`

gives

`(dx)/(dt)=3\ "ms"^-1`.

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Example 2

A stone is dropped into a pond, the ripples forming concentric circles which expand. At what rate is the area of one of these circles increasing when the radius is `4\ "m"` and increasing at the rate of 0.5 ms-1?

Answer

Ripples on a pond

The area of a circle with radius `r` is:

`A=pi r^2`

Differentiate w.r.t. time, and then substitute known values:

`(dA)/(dt)=d/(dt)(pir^2)`

`=2pir(dr)/(dt)`

`=2pi(4)(0.5)`

`~~12.56 "m"^2"s"^-1`

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Example 3

An earth satellite moves in a path that can be described by

`x^2/72.5+y^2/71.5=1`

where x and y are in thousands of kilometres.

If `dx/dt = 12900\ "km/h"` for `x = 3200\ "km"` and `y > 0`, find `dy/dt`.

Answer

Here is the path of the satellite. It is an ellipse, but is very nearly circular.

Elliptical path of the satellite.

We can see from the above that `(dy)/(dt)` should be a negative value (as it is going "down" in this reference frame).

We differentiate the expression with respect to t:

`x^2/72.5+y^2/71.5=1`

`(2x(dx)/(dt))/72.5+(2y(dy)/(dt))/71.5=0`

We need to find y. We do this by substituting `x = 3.2` (because `x` is in 1000s of kilometers) into the original expression:

`3.2^2/72.5+y^2/71.5=1`

Solving this gives:

`(71.5)(3.2)^2+72.5y^2=(71.5)(72.5)`

`732.16+72.5y^2=5183.75`

`y^2=(5183.75-732.16)/(72.5)`

`y^2=61.401`

`y=+-7.836`

The question tells us to take the positive value only. Substituting our known values gives:

`(2(3.2)(12.9))/72.5+(2(7.836)(dy)/dt)/71.5=0`

`(71.5)(2)(3.2)(12.9)+` `(72.5)(2)(7.836)(dy)/(dt)=0`

`(dy)/(dt)=-5.195`

This means the velocity in the y-direction is −5195 km/h (since the units are 1000s of kilometers).

Example 4

The tuning frequency f of an electronic tuner is inversely proportional to the square root of the capacitance `C` in the circuit.

If f = 920 kHz for C = 3.5 pF, find how fast f is changing at this frequency if `(dC)/(dt) =0.3\ "pF/s"`.

Answer

Now

`f=k/sqrtC`

and substituting our given values, we have that

`920000=k/sqrt(3.5xx10^-12`

and this gives `k = 1.721`.

So

`f=1.721/sqrtC=1.721C^(-1"/"2)`

We need to find `(df)/(dt)`.

`(df)/(dt)=-1.721/2C^(-3"/"2)(dC)/(dt)`

We are told that `C = 3.5\ "pF"` and `(dC)/(dt) =0.3\ "pF/s"`.

So

`(df)/(dt)=-1.721/2(3.5xx10^-12)^(-3"/"2)xx` `(0.3xx10^-12)`

`=-39424.8\ "Hz s"^-1`

So the frequency of the electronic tuner is decreasing at the rate of 39.4 kHz s-1.

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