Related rates

A thick rope stretches from a weathered dock toward a boat in the background — Photo by Bruno Mira on Unsplash.

Until now, we have usually described one variable explicitly as a function of another. Not every relationship takes that form. Several quantities may instead be connected by a single equation, and that equation may be much easier to work with before any variable is isolated.

As the quantities change, their changes must continue to satisfy the original relationship. Differentiating the relationship itself allows us to find its slope directly; moreover, when every quantity depends on time, the same process connects their rates at a particular moment. This gives us a way to study both implicitly defined curves and situations in which several quantities change together.

By the end of this lesson, you should be able to:

Differentiate relations without solving for $$ y $$ by working with differentials.
Use implicit differentiation to prove previously established differentiation rules.
Set up and solve related-rates problems, including interpreting the signs and units of rates.

In many physical systems, several changing quantities are constrained by one geometric relationship. When one length changes, the others must respond in a coordinated way. Knowing the rate of one change can therefore help us determine another.

A taut rope creates a particularly useful constraint: its length and the horizontal and vertical distances between its endpoints form a right triangle. We'll follow how those lengths change as a winch pulls a boat toward a dock. At one particular moment, the following information is known:

The rope is being pulled in at a constant rate of $$ 2 $$ feet per second.
The winch is $$ 7 $$ feet above the point where the rope is attached to the boat.
At this particular instant, the boat is $$ 24 $$ feet horizontally from the dock.

You wonder: As the rope gets pulled in, how quickly is the boat moving toward the dock at this moment?

We can start to get a feel for the situation by drawing a diagram.

Loading the graph…If it does not appear, check the Desmos API configuration.

Move the slider to pull in the rope. The diagram is not drawn to scale.

To keep track of everything, we can label the relevant quantities in the diagram:

Let $\mathhlblue{x}$ be the horizontal distance from the boat to the dock.
Let $\mathhlgreen{y}$ be the vertical distance from the boat's point of attachment to the winch.
Let $\mathhlred{\ell}$ be the length of the rope between the boat and the winch.

Based on the information we have, we know that at the current moment, $$ x= $$ feet and $$ y= $$ feet.

Furthermore, since $$ x $$ , $$ y $$ , and $\ell$ are the sides of , they're related by the Pythagorean Theorem:

$\displaystyle =\ell^2\text{.}$

We can use this to find the current length of the rope, which is $\ell=$ feet.

Now, we were asked to find how quickly the boat is moving horizontally toward the dock. In other words, we're looking for a rate: . To find this rate, we'll need to look at how the quantities $$ x $$ , $$ y $$ , and $\ell$ are changing in relation to each other.

Although $$ x $$ , $$ y $$ , and $\ell$ may change as the boat moves, the relationship ${x^2+y^2=\ell^2}$ remains true at every moment. Their changes must therefore respect this relationship, which means the of both sides must be equal as well: $\dd(x^2+y^2)=\dd(\ell^2)\text.$

Using the and the gives $\displaystyle {\dd{x}+}\)\(\displaystyle \dd{y}=2\ell\dd{\ell}\text.$

Then, since these quantities are changing with time, we divide each term by $\dd{t}$ to turn these differentials into rates: $2x\dv{x}{t}+2y\dv{y}{t}=2\ell\dv{\ell}{t}\text{.}$

Now that we have this equation relating the different rates, we can fill in the information we know:

We know from earlier that $$ x=24 $$ feet, $$ y=7 $$ feet, and $\ell=25$ feet.
Because the winch is pulling in the rope, its length is . Therefore, we have $\displaystyle \dv{\ell}{t}=$ feet per second.
The winch is always the same height above the point of attachment on the boat, so $$ y $$ is . Therefore, we have $\displaystyle \dv{y}{t}=$ feet per second.

We can now substitute everything we know into our rate equation: $\begin{aligned}2x\dv{x}{t}+2y\dv{y}{t}&=2\ell\dv{\ell}{t}\\ 2(24)\dv{x}{t}+2(7)(0)&=2(25)(-2)\end{aligned}$

Solving for the missing rate, we find that $\displaystyle \dv{x}{t}=$ feet per second.

Since $$ x $$ measures the distance from the boat to the dock, the negative value of $\displaystyle \dv{x}{t}$ means that this distance is , which means the boat is moving .

Implicit differentiation

An equation is not a magical incantation that produces a graph.

Rather, an equation is a relationship between quantities, and the graph consists of all points that satisfy that relationship.

For example, if we want an equation with a circle of radius $$ 5 $$ centered at the origin, then the distance from $$ (0,0) $$ to a point $$ (x,y) $$ on the circle should always be $$ 5 $$ : $\sqrt{(x-0)^2+(y-0)^2}=5$ If we square both sides, this can be represented in the simplified form $x^2+y^2=25\text.$

Now, what if we want the slope at any point of this circle?

We could start by solving for $$ y $$ : $y=\pm\sqrt{25-x^2}\text{.}$

The two choices give two branches of the relation: the upper semicircle is described by $\displaystyle y=\sqrt{25-x^2}$ , while the lower semicircle is described by $\displaystyle y=-\sqrt{25-x^2}$ . Each branch defines $$ y $$ as a function of $$ x $$ , so each has its own derivative.

Using the on these two functions gives $\dv{y}{x}=\pm\frac12(25-x^2)^{-1/2}(-2x)\text{.}$

This is a somewhat “ugly” solution. First of all, we have two formulas to work with, one for each branch. Second, the square root made their derivatives relatively complicated. Neither difficulty is impossible, but it turns out that both are unnecessary because there's a better way to do it: differentiate the original relationship in place without solving for $$ y $$ first.

Example

Find the slope of the circle $$ x^2+y^2=25 $$ at the point $$ (3,4) $$ .

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The point

\( (3,4) \)

lies on the circle

\( x^2+y^2=25 \)

. Its coordinates and the radius form a right triangle.

Solution

$\begin{aligned}x^2+y^2&=25\\ \dd(x^2+y^2)&=\dd(25)\end{aligned}$

Start with the equation relating $$ x $$ and $$ y $$ , and take the differential of both sides.

$\begin{aligned}\dd(x^2+y^2)&=\dd(25)\\ \dd(x^2)+\dd(y^2)&=\dd(25)\\ {2x\dd{x}}+{2y\dd{y}}&=0\end{aligned}$

Apply the and . Since $$ 25 $$ is a constant, its differential is $$ 0 $$ .

$\begin{aligned}{2x\dd{x}}+{2y\dd{y}}&=0\\ 2x+2y\dv{y}{x}&=0\end{aligned}$

Divide every term by $\dd{x}$ , with $\displaystyle \frac{\dd{x}}{\dd{x}}=1$ .

$\begin{aligned}2x+2y\dv{y}{x}&=0\\[0.5 ex] 2y\dv{y}{x}&=-2x\\[0.5 ex] \dv{y}{x}&=-\frac{x}{y}\end{aligned}$

Solve for $\displaystyle \dv{y}{x}$ algebraically.

Where $y\ne0$ , the slope of the circle is given by $\displaystyle -\frac{x}{y}$ . In particular, the slope at $$ (3,4) $$ is $\displaystyle -\frac{3}{4}$ .

The restriction comes from dividing by $$ y $$ . At $$ (5,0) $$ and $$ (-5,0) $$ , the circle's locally matching line is vertical, so this calculation doesn't give a finite slope there.

You can check that the solution above is identical to what we would have gotten from solving for $$ y $$ first and then differentiating with the Chain Rule. However, in some situations it's not even possible to solve for $$ y $$ ; even so, we can still use implicit differentiation to find the slope of a curve.

Example

Find $\displaystyle \dv{y}{x}$ for the relation $x^5+4xy^3-3y^5=2\text{,}$ and then find the slope at the point $$ (1,1) $$ .

Solution

This is a fifth-degree equation in both $$ x $$ and $$ y $$ , so there's no practical way to solve for either variable first. Implicit differentiation instead lets us work with the equation as it's written.

$\begin{aligned}x^5+4xy^3-3y^5&=2\\ \dd(x^5+4xy^3-3y^5)&=\dd(2)\end{aligned}$

Take the differential of both sides.

$\begin{aligned}\dd(x^5+4xy^3-3y^5)&=\dd(2)\\ \dd(x^5)+\dd(4xy^3)-\dd(3y^5)&=\dd(2)\\ {5x^4\dd{x}}+{4\dd(xy^3)}-{15y^4\dd{y}}&=0\end{aligned}$

Apply the Sum Rule and Power Rule. The term $$ xy^3 $$ is a product of two changing quantities, so its differential still needs the .

$\begin{aligned}{5x^4\dd{x}}+\mathhlyellow{{4\dd(xy^3)}}-{15y^4\dd{y}}&=0\\ {5x^4\dd{x}}+\mathhlyellow{{4\qty(y^3\dd(x)+x\dd(y^3))}}-{15y^4\dd{y}}&=0\\ {5x^4\dd{x}}+\mathhlyellow{{4y^3\dd{x}}+{12xy^2\dd{y}}}-{15y^4\dd{y}}&=0\end{aligned}$

Use the Product Rule to compute $\dd(xy^3)$ .

$\begin{aligned}{5x^4\dd{x}}+{4y^3\dd{x}}+{12xy^2\dd{y}}-{15y^4\dd{y}}&=0\\ {5x^4}+{4y^3}+{12xy^2\dv{y}{x}}-{15y^4\dv{y}{x}}&=0\\ 5x^4+4y^3+(12xy^2-15y^4)\dv{y}{x}&=0\end{aligned}$

Divide every term by $\dd{x}$ and collect the terms containing $\displaystyle \dv{y}{x}$ .

$\begin{aligned}5x^4+4y^3+(12xy^2-15y^4)\dv{y}{x}&=0\\ (12xy^2-15y^4)\dv{y}{x}&=-5x^4-4y^3\\ \dv{y}{x}&=\frac{-5x^4-4y^3}{12xy^2-15y^4}\end{aligned}$

Solve for $\displaystyle \dv{y}{x}$ .

$\begin{aligned}\dv{y}{x}&=\frac{-5\mathhlred{x}^4-4\mathhlblue{y}^3}{12 \mathhlred{x}\mathhlblue{y}^2-15\mathhlblue{y}^4}\\ \eval{\dv{y}{x}}_{(\mathhlred{1},\mathhlblue{1})}&=\frac{-5(\mathhlred{1})^4-4(\mathhlblue{1})^3}{12(\mathhlred{1})(\mathhlblue{1})^2-15(\mathhlblue{1})^4}=3\end{aligned}$

The point $(\mathhlred{1},\mathhlblue{1})$ really is on the curve because $\mathhlred{1}^5+4(\mathhlred{1})(\mathhlblue{1})^3-3(\mathhlblue{1})^5=2$ . Substituting its coordinates into the slope formula shows that the slope there is $$ 3 $$ .

Implicit differentiation can also be useful even when a variable has already been isolated, if doing so makes the relation simpler. For example, we can use implicit differentiation to verify cases of the that up to this point we'd only assumed were true.

Example

Use implicit differentiation to find the derivative of $\displaystyle y=\frac{1}{x}$ .

Solution

The equation $\displaystyle y=\frac1x$ contains a quotient, but we can't use the Quotient Rule — we based our proof of the Quotient Rule on the derivative of $\displaystyle x^{-1}$ , so using the Quotient Rule would be circular reasoning.

Instead, we can multiply both sides by $$ x $$ to obtain a product:

$xy=1\text{.}$

Now we can use the :

$y\dd{x}+x\dd{y}=0\text{.}$

Next isolate $\dd{y}$ :

$\begin{aligned}x\dd{y}&=-y\dd{x}\\ \dd{y}&=-\frac{y}{x}\dd{x}\text.\end{aligned}$

Since $\displaystyle y=\frac{1}{x}$ , we can make that substitution:

$\begin{aligned}\dd{y}&=-\frac{1/x}{x}\dd{x}\\ &=-\frac{1}{x^2}\dd{x}\end{aligned}$

Finally, divide both sides by $\dd{x}$ :

$\begin{aligned}\dv{y}{x}&=-\frac{1}{x^2}\end{aligned}$

This is exactly the result the Power Rule predicts for the exponent $$ -1 $$ .

Related rates

Implicit differentiation relates changes in quantities without requiring us to solve for one of them first. A related rates problem uses this same idea when several quantities change with time.

In a related rates problem, we still have an equation that relates our variables. However, instead of dividing through by a differential of one of those variables, we treat all quantities as functions of $$ t $$ and divide through by $\dd{t}$ . This turns each differential into a rate with respect to time. From here, we can substitute the information we know and find the desired rate.

Example

An oil slick spreads across the surface of calm water in the shape of a circle. Its area is increasing at a rate of $12\pi$ square meters per minute. At the moment when the radius is $$ 6 $$ meters, how quickly is its radius increasing?

Loading the graph…If it does not appear, check the Desmos API configuration.

Move the slider to watch the slick spread. Equal movements of the slider represent equal time intervals and therefore add equal areas, so the radius grows more slowly as the slick gets larger.

Solution

$A=\pi r^2$

Let $$ A $$ be the area of the slick and $$ r $$ its radius. The area formula relates the changing quantities.

$\dd{A}=2\pi r\dd{r}$

Take the differential of both sides. Note that we have not yet substituted the radius or the given area rate, because $$ A $$ and $$ r $$ are both changing.

$\dv{A}{t}=2\pi r\dv{r}{t}$

Divide every term by $\dd{t}$ to express the relationship in terms of rates.

$12\pi=2\pi(6)\dv{r}{t}$

Now substitute $\dv{A}{t}=12\pi$ square meters per minute and $$ r=6 $$ meters, the information that describes the moment in question.

$\dv{r}{t}=1$

The radius of the oil slick is increasing at $$ 1 $$ meter per minute. The positive sign agrees with the fact that the slick is spreading outward.

We can summarize the method for solving a related-rates problem as follows.

Sometimes a model contains more than two changing quantities. In that case, we may be able to use an additional relationship to connect their rates.

Example

Water is poured into an inverted cone-shaped container at a rate of $4\pi$ cubic feet per minute. The container has a radius of $$ 5 $$ feet and a height of $$ 10 $$ feet. At the moment when the water is $$ 8 $$ feet deep, how quickly is the water level rising?

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Move the time slider to fill the tank, and drag the figure to rotate it. Because water enters at the constant rate

\dv{V}{t}=4\pi

, equal movements of the slider add equal volumes. The water level therefore rises quickly near the cone's tip and more slowly as the widening tank fills.

Solution

By drawing a vertical cross-section of the tank, we can see two similar right triangles: one formed by half of the full tank, and one formed by half of the water.

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Move the slider to change the water level in the tank. The triangle formed by the water is similar to the triangle formed by the full tank. Their radius-to-height ratios agree, so

\displaystyle \frac rh=\frac5{10}=\frac12

at every water depth.

Using this information, we can set up the related-rates problem.

$\begin{aligned}V&=\frac13\pi r^2h\\ r&=\frac12h\end{aligned}$

Write equations for the water's volume and the relationship between its radius and depth.

$\begin{aligned}\dd{V}&=\frac13\pi\qty(2rh\dd{r}+r^2\dd{h})\\ \dd{r}&=\frac12\dd{h}\end{aligned}$

Take differentials. The Product Rule accounts for changes in both $$ r^2 $$ and $$ h $$ .

$\begin{aligned}\dv{V}{t}&=\frac13\pi\qty(2rh\dv{r}{t}+r^2\dv{h}{t})\\ \dv{r}{t}&=\frac12\dv{h}{t}\end{aligned}$

Divide by $\dd{t}$ to turn the differentials into rates.

$\begin{aligned}4\pi&=\frac13\pi\qty(2(4)(8)\qty(\frac12\dv{h}{t})+4^2\dv{h}{t})\\ 4\pi&=16\pi\dv{h}{t}\end{aligned}$

Now substitute the known information. Remember that $\displaystyle r=\frac{h}{2}$ , so when $$ h=8 $$ , we also have $$ r=4 $$ .

$\dv{h}{t}=\frac14\text{ foot per minute}$

Solve for the desired rate and include its units. Since this rate is positive, the water level is rising at $\frac14$ foot per minute.

We also could have substituted $\displaystyle r=\frac12h$ into the volume formula before differentiating, because this relationship holds throughout the filling process. Substituting $$ h=8 $$ would be different: that measurement describes only the moment we're investigating. Keeping $$ r $$ and $$ h $$ in place here lets us track both changing dimensions and use the related rate $\displaystyle \dv{r}{t}=\frac12\dv{h}{t}$ .

For each relation, first use implicit differentiation to enter an expression for $\displaystyle \dv{y}{x}$ . Then use that expression to enter the slope at the indicated point.
1. $x^2+y^2=${c}$ at $$ (${x0},${y0}) $$
  
  $\displaystyle \dv{y}{x}=\)$\displaystyle \text{.}$
  
  $\displaystyle \eval{\dv{y}{x}}_{(${x0},${y0})}=$$\displaystyle \text{.}$
  
  Solution
  
  Start by taking the differential of both sides. The Sum Rule and Power Rule give
  
  $2x\dd{x}+2y\dd{y}=0\text{.}$
  
  Since we want the rate of change of $\( y $$ with respect to $$ x $$ , divide every term by $\dd{x}$ :
  
  $2x+2y\dv{y}{x}=0\text{.}$
  
  Now isolate $\displaystyle \dv{y}{x}$ to obtain a formula for the slope:
  
  $\dv{y}{x}=-\frac{x}{y}\text{.}$
  
  Finally, substitute both coordinates of the indicated point:
  
  $\eval{\dv{y}{x}}_{(${x0},${y0})}=-\frac{${x0}}{${y0}}=${slope}\text{.}$
  
  Therefore, the slope of the circle at $$ (${x0},${y0}) $$ is $$ ${slope} $$ .
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The point $$ (${x0},${y0}) $$ lies on the circle $x^2+y^2=${c}$ .
2. $x^2+${A}y^2=${c}$ at $$ (${x0},${y0}) $$
  
  $\displaystyle \dv{y}{x}=\)$\displaystyle \text{.}$
  
  $\displaystyle \eval{\dv{y}{x}}_{(${x0},${y0})}=$$\displaystyle \text{.}$
  
  Solution
  
  Take the differential of both sides:
  
  $2x\dd{x}+${2*A}y\dd{y}=0\text{.}$
  
  Divide by $\dd{x}$ , and then solve for $\displaystyle \dv{y}{x}$ to obtain a formula for the slope:
  
  $\begin{aligned}2x+${2*A}y\dv{y}{x}&=0\\ \dv{y}{x}&=-\frac{x}{${A}y}\text{.}\end{aligned}$
  
  Substitute the coordinates of the indicated point:
  
  $\eval{\dv{y}{x}}_{(${x0},${y0})}=-\frac{${x0}}{${A}(${y0})}=${slope}\text{.}$
  
  Thus the slope of the ellipse at $\( (${x0},${y0}) $$ is $$ ${slope} $$ .
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The point $$ (${x0},${y0}) $$ lies on the ellipse $x^2+${A}y^2=${c}$ .
3. $y^2=x^3${signed(A)}x${signed(B)}$ at $$ (${x0},${y0}) $$
  
  $\displaystyle \dv{y}{x}=\)$\displaystyle \text{.}$
  
  $\displaystyle \eval{\dv{y}{x}}_{(${x0},${y0})}=$$\displaystyle \text{.}$
  
  Solution
  
  This relation is an example of an elliptic curve; these curves are an important part of modern cryptography.
  
  Take the differential of both sides:
  
  $2y\dd{y}=(3x^2${signed(A)})\dd{x}\text{.}$
  
  Divide by $\dd{x}$ and then by $\( 2y $$ to obtain a formula for the slope:
  
  $\dv{y}{x}=\frac{3x^2${signed(A)}}{2y}\text{.}$
  
  At $$ (${x0},${y0}) $$ , this becomes
  
  $\eval{\dv{y}{x}}_{(${x0},${y0})}=\frac{3(${parenthesize_negative(x0)})^2${signed(A)}}{2(${y0})}=${slope}\text{.}$
  
  Therefore, the slope of the elliptic curve at $$ (${x0},${y0}) $$ is $$ ${slope} $$ .
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The point $$ (${x0},${y0}) $$ lies on the elliptic curve $y^2=x^3${signed(A)}x${signed(B)}$ .
4. $x^2+${A}xy+${B}y^2=${c}$ at $$ (${x0},${y0}) $$
  
  $\displaystyle \dv{y}{x}=\)$\displaystyle \text{.}$
  
  $\displaystyle \eval{\dv{y}{x}}_{(${x0},${y0})}=$$\displaystyle \text{.}$
  
  Solution
  
  The middle term is a product of two changing quantities, so the Product Rule gives $\dd(xy)=x\dd{y}+y\dd{x}$ . Taking the differential of the entire relation therefore gives
  
  $2x\dd{x}+${A}x\dd{y}+${A}y\dd{x}+${2*B}y\dd{y}=0\text{.}$
  
  Divide by $\dd{x}$ and collect the terms containing $\displaystyle \dv{y}{x}$ :
  
  $(2x+${A}y)+(${A}x+${2*B}y)\dv{y}{x}=0\text{.}$
  
  Solving for $\displaystyle \dv{y}{x}$ gives the slope formula
  
  $\dv{y}{x}=-\frac{2x+${A}y}{${A}x+${2*B}y}\text{.}$
  
  Now substitute both coordinates of the specified point:
  
  $\eval{\dv{y}{x}}_{(${x0},${y0})}=-\frac{2(${x0})+${A}(${y0})}{${A}(${x0})+${2*B}(${y0})}=${slope}\text{.}$
  
  Therefore, the slope of the relation at $\( (${x0},${y0}) $$ is $$ ${slope} $$ .
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The mixed term $$ ${A}xy $$ produces an ellipse whose axes are tilted relative to the coordinate axes; $$ (${x0},${y0}) $$ is the indicated point.
The curve $x^3+y^3=${K}xy$ is a Folium of Descartes. Complete the following investigation of its slope.
1. Find an expression for $\displaystyle \dv{y}{x}$ in terms of $$ x $$ and $$ y $$ .
  
  $\displaystyle \dv{y}{x}=\)$\displaystyle \text{.}$
  
  Solution
  
  Take the differential of both sides. On the right, the constant $\( ${K} $$ remains in place while the Product Rule is applied to $$ xy $$ :
  
  $3x^2\dd{x}+3y^2\dd{y}=${K}x\dd{y}+${K}y\dd{x}\text{.}$
  
  Divide every term by $\dd{x}$ . The terms containing $\dd{y}$ become terms containing $\displaystyle \dv{y}{x}$ :
  
  $3x^2+3y^2\dv{y}{x}=${K}x\dv{y}{x}+${K}y\text{.}$
  
  Put all terms containing $\displaystyle \dv{y}{x}$ on the left and all remaining terms on the right by subtracting $\displaystyle ${K}x\dv{y}{x}$ and $$ 3x^2 $$ from both sides:
  
  $3y^2\dv{y}{x}-${K}x\dv{y}{x}=${K}y-3x^2\text{.}$
  
  Factor $\displaystyle \dv{y}{x}$ from the two terms on the left:
  
  $(3y^2-${K}x)\dv{y}{x}=${K}y-3x^2\text{.}$
  
  Finally, divide by its coefficient:
  
  $\dv{y}{x}=\frac{${K}y-3x^2}{3y^2-${K}x}\text{.}$
2. Find the slope at the point $$ (${q},${q}) $$ .
  
  $\displaystyle \eval{\dv{y}{x}}_{(${q},${q})}=\)$\displaystyle \text{.}$
  
  Solution
  
  Substitute $\( x=${q} $$ and $$ y=${q} $$ into the slope formula from part (a):
  
  $\eval{\dv{y}{x}}_{(${q},${q})}=\frac{${K}(${q})-3(${q})^2}{3(${q})^2-${K}(${q})}=-1\text{.}$
  
  Thus the slope of the folium at $$ (${q},${q}) $$ is $$ -1 $$ .
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  The Folium of Descartes $x^3+y^3=${K}xy$ forms a loop in the first quadrant and has branches extending beyond the viewing window.
3. Find the point in the first quadrant where the curve has a horizontal slope.
  
  Enter your answer as an ordered pair. You can type cbrt to produce a cube root.
  
  $\displaystyle (x,y)=$
  
  Solution
  
  A horizontal slope occurs when the numerator of $\displaystyle \dv{y}{x}$ is zero. Therefore,
  
  $\begin{aligned}${K}y-3x^2&=0\\ y&=\frac{3x^2}{${K}}\text{.}\end{aligned}$
  
  Substitute this expression for $$ y $$ into the original equation:
  
  $\begin{aligned}x^3+\left(\frac{3x^2}{${K}}\right)^3&=${K}x\left(\frac{3x^2}{${K}}\right)\\ x^3+\frac{27x^6}{${K}^3}&=3x^3\text{.}\end{aligned}$
  
  Because the desired point is in the first quadrant, $$ x>0 $$ , so we may divide by $$ x^3 $$ and solve:
  
  $\begin{aligned}1+\frac{27x^3}{${K}^3}&=3\\ x^3&=\frac{2(${K})^3}{27}\\ x&=${radical_coefficient}\sqrt[3]{2}\text{.}\end{aligned}$
  
  Using $\displaystyle y=3x^2/${K}$ then gives $\displaystyle y=${radical_coefficient}\sqrt[3]{4}$ . Thus the point is
  
  $\left(${radical_coefficient}\sqrt[3]{2},${radical_coefficient}\sqrt[3]{4}\right)\text{.}$
  
  The denominator of $\displaystyle \dv{y}{x}$ is nonzero at this point, so the slope is indeed zero.
4. Find the point in the first quadrant where the curve has a vertical slope.
  
  Enter your answer as an ordered pair.
  
  $\displaystyle (x,y)=$
  
  Solution
  
  A vertical slope occurs when the denominator of $\displaystyle \dv{y}{x}$ is zero. Therefore,
  
  $\begin{aligned}3y^2-${K}x&=0\\ x&=\frac{3y^2}{${K}}\text{.}\end{aligned}$
  
  Substitute this expression for $$ x $$ into the original equation:
  
  $\begin{aligned}\left(\frac{3y^2}{${K}}\right)^3+y^3&=${K}\left(\frac{3y^2}{${K}}\right)y\\ \frac{27y^6}{${K}^3}+y^3&=3y^3\text{.}\end{aligned}$
  
  Since $$ y>0 $$ in the first quadrant, divide by $$ y^3 $$ and solve:
  
  $\begin{aligned}\frac{27y^3}{${K}^3}+1&=3\\ y^3&=\frac{2(${K})^3}{27}\\ y&=${radical_coefficient}\sqrt[3]{2}\text{.}\end{aligned}$
  
  Using $\displaystyle x=3y^2/${K}$ gives $\displaystyle x=${radical_coefficient}\sqrt[3]{4}$ . Thus the point is
  
  $\left(${radical_coefficient}\sqrt[3]{4},${radical_coefficient}\sqrt[3]{2}\right)\text{.}$
  
  These are the coordinates from part (c) in the opposite order. This reflects the folium's symmetry across the line $$ y=x $$ : interchanging $$ x $$ and $$ y $$ leaves its equation unchanged.
Solve each related-rates problem. Draw and label a diagram when appropriate, form the rate equation before substituting numerical values, and interpret the sign and units of the result.
1. After a rock is thrown into still water, a circular ripple expands outward. Its radius is increasing at $$ ${dr} $$ ${plural(dr, "foot", "feet")} per second. How quickly is the circumference of the ripple increasing?
  
  Answer:
  
  Solution
  
  Let $$ C $$ be the circumference of the ripple and $$ r $$ its radius. These changing quantities are related by the circumference formula
  
  $C=2\pi r\text{.}$
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  Move the slider to watch the ripple expand. The circumference $$ C $$ is its outer edge, and $$ r $$ measures its radius.
  
  Take the differential of both sides, and then divide by $\dd{t}$ :
  
  $\begin{aligned}\dd{C}&=2\pi\dd{r}\\ \dv{C}{t}&=2\pi\dv{r}{t}\text{.}\end{aligned}$
  
  Now substitute the given rate $\displaystyle \dv{r}{t}=${dr}$ ${plural(dr, "foot", "feet")} per second:
  
  $\dv{C}{t}=2\pi(${dr})=${dC}\pi\text{ feet per second.}$
  
  The circumference of the ripple is increasing at $${dC}\pi$ feet per second. Notice that the rate does not depend on the ripple's current radius, so we did not need to know how large the ripple was at a particular moment.
2. Air is blown into a spherical balloon so that its volume increases at a constant rate of $${Q}\pi$ cubic centimeters per second. How quickly is its radius increasing when the radius is $$ ${R} $$ centimeters?
  
  Answer:
  
  Solution
  
  Let $$ V $$ be the balloon's volume and $$ r $$ its radius. The volume of a sphere is
  
  $V=\frac43\pi r^3\text{.}$
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  Move the slider to inflate the balloon at a constant volume rate. Its changing radius is labeled $$ r $$ .
  
  Take a differential before substituting the current radius, then divide by $\dd{t}$ :
  
  $\begin{aligned}\dd{V}&=4\pi r^2\dd{r}\\ \dv{V}{t}&=4\pi r^2\dv{r}{t}\text{.}\end{aligned}$
  
  At the instant described, $$ r=${R} $$ centimeters and $\displaystyle \dv{V}{t}=${Q}\pi$ cubic centimeters per second. Therefore,
  
  $${Q}\pi=4\pi(${R})^2\dv{r}{t}\text{.}$
  
  Solving gives $\displaystyle \dv{r}{t}=${dr}$ ${plural(dr, "centimeter", "centimeters")} per second. The positive result means the balloon's radius is increasing.
3. A $$ ${L} $$ -foot ladder rests against a vertical wall. Its base slides away from the wall at $$ ${dx} $$ ${plural(dx, "foot", "feet")} per second. How quickly is the top sliding downward when the base is $$ ${x0} $$ feet from the wall?
  
  Answer:
  
  Solution
  
  Let $$ x $$ be the horizontal distance from the wall to the base of the ladder, and let $$ y $$ be the height of the top of the ladder. The ladder has constant length, so the Pythagorean Theorem gives
  
  $x^2+y^2=${L}^2\text{.}$
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  Move the slider to watch the ladder slide. The changing legs are labeled $$ x $$ and $$ y $$ , while the ladder's length $$ L $$ remains constant.
  
  Take the differential of both sides. Since the ladder's length is constant, the right side has differential $$ 0 $$ . Dividing by $\dd{t}$ then gives
  
  $\begin{aligned}2x\dd{x}+2y\dd{y}&=0\\ 2x\dv{x}{t}+2y\dv{y}{t}&=0\text{.}\end{aligned}$
  
  When $$ x=${x0} $$ , the equation $${x0}^2+y^2=${L}^2$ gives $$ y=${y0} $$ . The base is moving away from the wall, so $\displaystyle \dv{x}{t}=${dx}$ . Substitute this information into the rate equation:
  
  $2(${x0})(${dx})+2(${y0})\dv{y}{t}=0\text{.}$
  
  Solving gives $\displaystyle \dv{y}{t}=${dy}$ ${plural(dy, "foot", "feet")} per second. Because $$ y $$ measures height, the negative sign means the top of the ladder is sliding downward.
4. One car is $$ ${x0} $$ miles east of an intersection and traveling east at $$ ${dx} $$ miles per hour. At the same moment, a second car is $$ ${y0} $$ miles north of the intersection and traveling north at $$ ${dy} $$ miles per hour. How quickly is the distance between the cars changing?
  
  Answer:
  
  Solution
  
  Let $$ x $$ and $$ y $$ be the cars' distances from the intersection, and let $$ s $$ be the distance between the cars. Their positions form a right triangle, so
  
  $x^2+y^2=s^2\text{.}$
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  Move the slider to watch both cars travel away from the intersection. Their distances are labeled $$ x $$ and $$ y $$ , and their separation is $$ s $$ .
  
  Take the differential of both sides and divide by $\dd{t}$ :
  
  $\begin{aligned}2x\dd{x}+2y\dd{y}&=2s\dd{s}\\ x\dv{x}{t}+y\dv{y}{t}&=s\dv{s}{t}\text{.}\end{aligned}$
  
  At this moment, the distance between the cars is $s=\sqrt{${x0}^2+${y0}^2}=${s0}$ miles. Since each car is moving away from the intersection, both given rates are positive. Substitution gives
  
  $${x0}(${dx})+${y0}(${dy})=${s0}\dv{s}{t}\text{.}$
  
  Therefore, $\displaystyle \dv{s}{t}=${ds}$ miles per hour. The positive result means the distance between the cars is increasing.
5. A $$ ${person} $$ -foot-tall person walks directly away from a $$ ${post} $$ -foot lamppost at $$ ${speed} $$ feet per second. How quickly is the length of the person's shadow increasing?
  
  Answer:
  
  Solution
  
  Let $$ x $$ be the person's distance from the lamppost and $$ s $$ be the length of the shadow. The large triangle has height $$ ${post} $$ and base $$ x+s $$ , while the smaller triangle has height $$ ${person} $$ and base $$ s $$ . Similar triangles give
  
  $\frac{${post}}{x+s}=\frac{${person}}s\text{.}$
  
  Loading the graph…If it does not appear, check the Desmos API configuration.
  
  Move the slider to watch the person walk away. The lamppost and the full shadow form the large triangle; the person and the shadow of length $$ s $$ form the smaller triangle.
  
  Before differentiating, simplify this relationship:
  
  $\begin{aligned}${post}s&=${person}(x+s)\\ ${post-person}s&=${person}x\text{.}\end{aligned}$
  
  Now take the differential of both sides and divide by $\dd{t}$ :
  
  $${post-person}\dv{s}{t}=${person}\dv{x}{t}\text{.}$
  
  The person walks away from the lamppost, so $\displaystyle \dv{x}{t}=${speed}$ feet per second. Substituting and solving gives
  
  $\dv{s}{t}=\frac{${person}}{${post-person}}(${speed})=${shadow_rate}\text{ feet per second.}$
  
  The positive rate means that the shadow is getting longer.
In the Learn section, we showed how to find the derivatives of $\displaystyle \frac 1x$ and $\sqrt x$ ; in this problem, you'll extend those arguments further. In each problem, make sure to begin with the suggested relationship rather than using the rule you are trying to establish.
1. For a positive integer $$ n $$ , let $y=x^{-n}$ . Rewrite this as $$ x^ny=1 $$ and derive a formula for $\displaystyle \dv{y}{x}$ .
  
  Solution
  
  Multiplying $y=x^{-n}=1/x^n$ by $$ x^n $$ gives the implicit relationship
  
  $x^ny=1\text{.}$
  
  Use the Product Rule when taking its differential:
  
  ${x^n\dd{y}}+{nyx^{n-1}\dd{x}}=0\text{.}$
  
  Now solve for $\dd{y}$ and divide by $\dd{x}$ :
  
  $\dv{y}{x}=-\frac{nyx^{n-1}}{x^n}=-\frac{ny}{x}\text{.}$
  
  Because $y=x^{-n}$ , substitute that relationship back into the result:
  
  $\dv{y}{x}=-n x^{-n-1}\text{.}$
  
  The exponent decreases from $$ -n $$ to $$ -n-1 $$ , so this is exactly what the Power Rule predicts. We have now extended the rule to negative integer exponents.
2. Let $\displaystyle h=\frac{f}{g}$ , where $g\ne0$ . Rewrite this as $$ gh=f $$ and use differentials to rederive the Quotient Rule.
  
  Solution
  
  Multiplying $\displaystyle h=\frac{f}{g}$ by $$ g $$ removes the quotient and gives
  
  $gh=f\text{.}$
  
  Since both factors on the left can change, the Product Rule gives
  
  $g\dd{h}+h\dd{g}=\dd{f}\text{.}$
  
  Our goal is a formula for the differential of $$ h $$ , so isolate $\dd{h}$ :
  
  $\dd{h}=\frac{\dd{f}-h\dd{g}}{g}\text{.}$
  
  The right side still contains $$ h $$ . Substitute the original relationship $\displaystyle h=\frac{f}{g}$ and combine the terms over a common denominator:
  
  $\begin{aligned}\dd{h}&=\frac{\dd{f}-\frac fg\dd{g}}{g}\\ &=\frac{g\dd{f}-f\dd{g}}{g^2}\text{.}\end{aligned}$
  
  Since $\displaystyle h=\frac{f}{g}$ , this establishes the Quotient Rule in differential form:
  
  $\dd(\frac fg)=\frac{g\dd{f}-f\dd{g}}{g^2}\text{.}$
3. For positive $$ x $$ and positive integers $$ m $$ and $$ n $$ , let $y=x^{m/n}$ . Rewrite this as $$ y^n=x^m $$ and derive a formula for $\displaystyle \dv{y}{x}$ .
  
  Solution
  
  Raising both sides of $y=x^{m/n}$ to the $$ n $$ th power gives a relationship involving only positive integer powers:
  
  $y^n=x^m\text{.}$
  
  Take the differential of both sides:
  
  $ny^{n-1}\dd{y}=mx^{m-1}\dd{x}\text{.}$
  
  Divide by $\dd{x}$ and isolate the derivative:
  
  $\dv{y}{x}=\frac{mx^{m-1}}{ny^{n-1}}\text{.}$
  
  To express the result using $$ x $$ alone, substitute $y=x^{m/n}$ and simplify the exponent:
  
  $\begin{aligned}\dv{y}{x}&=\frac{mx^{m-1}}{n(x^{m/n})^{n-1}}\\ &=\frac mn x^{(m/n)-1}\text{.}\end{aligned}$
  
  Hence the familiar Power Rule also works for rational exponents (as long as $$ x>0 $$ ).