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Differentiating a relation with respect to time, choosing the right relation, and substituting the instant last.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to set up a related-rates problem by naming every changing quantity and finding a relation that connects the rate you have with the rate you want, eliminate a variable you have no rate for before differentiating, differentiate the relation with respect to time so that every changing quantity contributes its rate, substitute the instant's values only after differentiating and say what goes wrong if you do not, carry signs through for shrinking quantities, and keep both terms when a product has two changing factors.
Two quantities connected by an equation have connected rates. Differentiate the equation with respect to time and the connection between the rates falls out.
The mechanism is the chain rule. If $A = x^2$ and $x$ is a function of $t$, then $$\frac{dA}{dt} = \frac{dA}{dx}\cdot\frac{dx}{dt} = 2x\,\frac{dx}{dt}.$$ The factor $\dfrac{dx}{dt}$ is the chain rule's inner derivative — the same factor that was easy to forget three lessons ago, and here it is the entire point of the exercise.
The one procedural rule that matters: substitute the instant last.
A related-rates problem gives you values that hold at one moment. "The side is $5$ cm" is not a fact about the side; it is a fact about the side now. Substituting it before differentiating replaces a function of time with a constant, and constants have rate zero. Differentiate first, with every changing quantity still a symbol; substitute afterwards.
Signs carry meaning. A shrinking quantity has a negative rate. Put the minus into the given rate at the start rather than reasoning it back in at the end.
Another way: picture
A balloon inflating. The radius and the volume are two numbers changing together, tied by $V = \tfrac43\pi r^3$. Knowing how fast either changes fixes the other, because the relation holds at every instant — which is precisely what makes it differentiable in time.
Another way: steps
Most of the difficulty is upstream of the calculus. The relation has to connect the rate you have with the rate you want, and introduce nothing else that varies.
For a ladder sliding down a wall: $x$ is the foot's distance from the wall, $y$ the top's height, $L$ the ladder's length. Then $x^2 + y^2 = L^2$ with $L$ constant, and differentiating gives $$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0.$$ Two rates, one equation — solvable.
Choosing the triangle's area instead would give $A = \tfrac12 xy$, which relates three rates when you only know one. The equation is true and useless.
Constants must be constants. The ladder's length does not change, so it contributes no term. Mistaking a constant for a variable adds a phantom rate; mistaking a variable for a constant loses a real one, and that is the substitution error wearing a different hat.
Eliminate variables you have no rates for, using the geometry, before differentiating. A cone draining with a fixed shape has $r$ proportional to $h$, so $r$ can be written in terms of $h$ and the problem reduced to one changing length.
Substituting the instant's values too early. The defining error, and it gives an answer of zero.
Dropping a rate term. If two quantities in a product are both changing, the product rule gives two terms and both count.
Losing a sign. "Shrinking at $3$ cm/s" is $\dfrac{dx}{dt} = -3$. Reading the magnitude and fixing the sign afterwards works until two effects compete, and then it does not.
Using a relation with an unavailable rate. Count the rates in the differentiated equation: if there is more than one you cannot supply, the relation was the wrong one.
Skipping the units check. $\dfrac{dV}{dt}$ in cm³ per second, $\dfrac{dr}{dt}$ in cm per second — the factor between them must be an area. That check catches a surprising share of algebraic slips.
"The side is $5$ cm" and "the side is $5t$ cm" look like the same kind of statement and are not. The first holds at one instant; the second describes the motion. A related-rates problem gives you the first, and the temptation to use it as if it were the second — by substituting it into the relation straight away — is what produces the zero answer.
A useful habit: write the relation with every changing quantity as a letter and no numbers at all, differentiate it, and only then open the brackets and put the instant's values in. If a number appears in your working before the differentiation, check that it is genuinely a constant.
A $10$ m ladder's foot slides away at $2$ m/s. How fast is the top falling when the foot is $6$ m out? Relation: $x^2 + y^2 = 100$.
$L$ is constant and contributes nothing.
Differentiate: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$, so $\dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt}$.
Differentiate before substituting anything.
At the instant: $x = 6$, so $y = 8$ from the relation, and $\dfrac{dy}{dt} = -\dfrac{6}{8}(2) = -1.5$ m/s. Negative: the top is falling. As $y \to 0$ the rate runs to infinity, which is the model failing rather than the ladder accelerating.
The answer's limit is worth reading too.
Water drains from a cone of radius $3$ m and height $6$ m at $2$ m³/min. How fast is the depth falling when the water is $4$ m deep? $V = \tfrac13\pi r^2 h$ has two changing lengths and you have a rate for neither.
Stop before differentiating.
The water's surface is a similar triangle, so $\dfrac{r}{h} = \dfrac{3}{6}$, giving $r = h/2$. Substitute: $V = \tfrac13\pi\dfrac{h^2}{4}h = \dfrac{\pi h^3}{12}$.
Eliminate first, differentiate second.
$\dfrac{dV}{dt} = \dfrac{\pi h^2}{4}\dfrac{dh}{dt}$, and with $\dfrac{dV}{dt} = -2$ and $h = 4$: $\dfrac{dh}{dt} = \dfrac{-8}{16\pi} = -\dfrac{1}{2\pi}$ m/min.
$A = \pi r^2$, and both $A$ and $r$ are functions of time.
Relation first.
$\dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt}$.
Differentiate, nothing substituted.
Now substitute: $2\pi(5)(3) = 30\pi$ cm² per second.
Match each related-rates step to its purpose.
| Models the quantities at every moment | $dA/dt=2\pi r\,dr/dt$ | Evaluates the general rate at the requested moment | Square units per unit time | |
|---|---|---|---|---|
| For a circle, $A=\pi r^2$ | ||||
| Differentiate $A=\pi r^2$ with respect to $t$ | ||||
| Use $r=3$ after differentiation | ||||
| A rate of area change |
Match each related-rates move to the reason it belongs in that order.
| Links the requested rate to the given changing quantity | Preserves the rate of the still-changing radius | Evaluates the general rate at the named instant | Uses square units per unit time and a negative sign for shrinking | |
|---|---|---|---|---|
| For a circle, start with $A=\pi r^2$ | ||||
| Differentiate to $dA/dt=2\pi r\,dr/dt$ | ||||
| Then use $r=3$ and $dr/dt=-2$ | ||||
| Report the area rate |
A square's side grows at $3$ cm per second. How fast is its area growing when the side is $7$ cm?
Answer:
A student writes "$A = x^2$, and $x = 8$, so $A = 64$, so $\dfrac{dA}{dt} = 0$". What went wrong?
A cube's edge is shrinking at $2$ cm per second. How fast is its surface area changing when the edge is $5$ cm? (Give a signed rate, in cm² per second.)
Answer:
A rectangle's width grows at $1$ cm/s while its height shrinks at $3$ cm/s. How fast is the area changing when the width is $7$ cm and the height is $8$ cm?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A ladder leans against a wall. Its foot slides away at a known rate and you want the rate at which the top slides down. Which relation should be differentiated?
You can set up and solve a related-rates problem and say why the substitution comes last. Say in your own words why substituting first gives an answer of zero. Next: linear approximation, where the tangent line is used in place of the curve.
8. Your turn: a circle's radius grows at $3$ cm/s. How fast is the area growing when $r = 5$?, step 3