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We understand the derivative in two main ways:
f′(x) gives the slope of f at x.
dxdy gives the rate of change of y with respect to x.
Related rates describe the rates of change of two or more quantities that we know to be related. For instance, given a square of side length s, we know that if the side length is increasing with respect to time by dtds, then the area of the same square would also increase with respect to time. Given an expression for dtds and utilizing the known fact A=s2, we could find an expression for dtdA.
In general, if we take the derivative of y=f(x) with respect to time, we get dtdy=dtdx⋅f′(x), which can be rewritten dtdx⋅dxdy, as a result of chain rule. Therefore, to solve for dtdA, we would find dsdA and multiply it by dtds.
Related Rates
Given three variables x, y, and z,
Hence, given dzdx, we can find an expression for dzdy by calculating dxdy.
Discussion
The radius r of a sphere is changing at a rate of dtdr=0.2m/s.
Find drdV using the equation for the volume of a sphere.
Hence, state dtdV, the rate of change of the volume of the sphere with respect to time.
Part (a)
The volume of a sphere of radius r is
To find how V changes as r changes, we differentiate with respect to r:
Hence
Part (b)
We know that V depends on r, which in turn depends on t. By the chain-rule,
From part (a) we have drdV=4πr2, and we’re given dtdr=0.2m/s. Hence,
So the volume is increasing at the rate
Volume related rates
Given the time rate of change of radius, length, height, or width of a three dimensional object, you may find the time rate of change of volume by taking the derivative of the volume equation.
Checkpoint
A radius r of a 5cm tall cylinder is changing at a rate of dtdr=0.1cm/s.
Find dtdV.
Select the correct option
Discussion
Let L be the distance from the origin of a point with coordinates (x,y).
Find L at (3,4).
Show that 2LdtdL=2xdtdx+2ydtdy.
It is given that dtdx=0.3m/s and dtdy=0.1m/s.
Use your answers from (a) and (b) to find dtdL at
Part (a)
We want the distance from (0,0) to the point (3,4). Think of the horizontal shift of 3 and the vertical shift of 4 as the legs of a right‐angled triangle, with the distance L as the hypotenuse. By the Pythagorean theorem:
Part (b)
We start from the definition
and notice that squaring both sides removes the root:
Now we differentiate each side with respect to t. On the left we have a function of L, so we get
On the right we differentiate x2+y2, remembering that x and y both depend on t:
Putting these together gives exactly
as required.
Part (c)
We start from the result of (b):
Divide both sides by 2L:
At the point (x,y)=(3,4) we have L=5, dtdx=0.3 and dtdy=0.1. Hence
Distance related rates
Let L be the distance from the origin of a point with coordinates (x,y). Then, given dtdx and dtdy, we can find dtdL at a given point (x,y).
Checkpoint
Given dtdx=0.7m/s and dtdy=0.4m/s, find dtdL at (−5,12).
Select the correct option
Discussion
Consider the angle θ in the following right angled triangle:
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Find an expression for θ in terms of x.
It is given that dtdx=0.1m/s.
Find dtdθ when x=3m.
Part (a)
We label the side opposite θ as x and the side adjacent to θ as 3. By definition of the tangent ratio,
Inverting this gives
Part (b)
We know from part (a) that
Differentiate both sides with respect to t. Recall
so with u=x/3 and using the chain rule,
Since dtdx=0.1m/s,
Finally, when x=3,
So dtdθ=601 rad/s when x=3.
Angle related rates
Using a given rate of change dtdx and trigonometry, we can calculate dxdθ, which can be used to find dtdθ.
Recall the previously learned skill of implicit differentiation:
Implicit Differentiation
Implicit differentiation is when we differentiate both sides of an equation. It is helpful when we have an equation that cannot be simplified to y=f(x).
Discussion
Use implicit differentiation to find dxdy of the curve 2xy=ex−1.
Given that dtdx=3 when x=1, find dtdy.
Part (a)
We want to differentiate implicitly the relation
with respect to x, remembering that y is really y(x).
First, apply the product rule on the left and the chain rule on the right:
On the left, pull out the 2 and use dxd(x⋅y)=dxdxy+xdxdy:
On the right, the derivative of ex−1 is itself:
So we get
Expand the left side:
Now isolate dxdy:
Part (b)
We have from part (a)
and in general
We are given dx/dt=3 when x=1. To find dy/dt at that instant we need y at x=1. From the original curve
substitute x=1:
Now evaluate dy/dx at x=1,y=21:
Hence
Related rates with implicit differentiation
Since dtdy=dxdy⋅dtdx, you may be asked to use implicit differentiation to find dxdy, then with a given dtdx and point, you can find dtdy.
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