# Related Rates

## Solved Problems

Click or tap a problem to see the solution.

### Example 5

An object is moving along the curve y = x² − 2x + 3. Find a point on the curve at which the y-coordinate is changing 4 times faster than the x-coordinate.

### Example 6

A water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. The height of the cone is 8 feet and the radius is 4 feet (Figures 5, 6). Find the rate of change of the water level when the depth is 6 feet.

### Example 7

A kite $$160$$ feet above the ground moves in a direction parallel to the ground at the rate of $$5$$ feet per second (Figure $$7$$). How fast is the string unwinding when the length of string already let out is $$200$$ feet?

### Example 8

Ship A is 60 miles north of point O and moving in the north direction at $$20$$ miles per hour. Ship B is 80 miles east of point O and moving west at 25 miles per hour (Figure $$8$$). How fast is the distance between the ships changing at this moment?

### Example 9

The legs of an isosceles triangle are increasing at the rate of $$0.5$$ feet per minute, and the vertical angle $$\alpha$$ is increasing at the rate of $$1$$ radian per minute (Figure $$9$$). At what rate is the area $$A$$ of the triangle changing when the leg $$b$$ is 2 feet and the angle $$\alpha$$ is $$\frac{{2\pi }}{3}?$$

### Example 10

Find all points on the ellipse $$9{x^2} + 16{y^2} = 400,$$ at which the $$y-$$coordinate is decreasing and the $$x-$$coordinate is increasing at the same rate (Figure $$10$$).

### Example 5.

An object is moving along the curve $$y = {x^2} - 2x + 3.$$ Find a point on the curve at which the $$y-$$coordinate is changing $$4$$ times faster than the $$x-$$coordinate.

Solution.

First, we differentiate both sides of the curve equation with respect to time $$t:$$

$\frac{{dy}}{{dt}} = \frac{d}{{dt}}\left( {{x^2} - 2x + 3} \right),\;\; \Rightarrow \frac{{dy}}{{dt}} = \left( {2x - 2} \right)\frac{{dx}}{{dt}}.$

In the problem, we are told that

$\frac{{dy}}{{dt}} = 4\frac{{dx}}{{dt}}.$

Therefore, we get the relationship

$2x - 2 = 4,$

so $$x=3.$$

The $$y-$$coordinate is given by

$y = {3^3} - 2 \cdot 3 + 3 = 6.$

Thus, the answer is $$\left( {x,y} \right) = \left( {3,6} \right).$$

### Example 6.

A water tank in the form of an inverted cone is being emptied at the rate of $$2$$ cubic feet per second. The height of the cone is $$8$$ feet and the radius is $$4$$ feet (Figures $$5, 6$$). Find the rate of change of the water level when the depth is 6 feet.

Solution.

Let $$h$$ and $$r$$ be the height and the radius of the water. The volume of the water (in the form of right circular cone) is given by

$V = \frac{1}{3}\pi {r^2}h.$

We can use similar triangles to get a relationship between $$r$$ and $$h:$$

$\frac{h}{r} = \frac{8}{4} = 2.$

Hence,

${r = \frac{h}{2}.}$

Plug it into the formula for the volume:

$V = \frac{1}{3}\pi {r^2}h = \frac{1}{3}\pi {\left( {\frac{h}{2}} \right)^2}h = \frac{1}{{12}}\pi {h^3}.$

Differentiate this formula with respect to time $$t:$$

$\frac{{dV}}{{dt}} = \frac{d}{{dt}}\left( {\frac{1}{{12}}\pi {h^3}} \right) = \frac{1}{{12}}\pi \cdot 3{h^2}\frac{{dh}}{{dt}} = \frac{1}{4}\pi {h^2}\frac{{dh}}{{dt}}.$

Solve the last equation for $$\frac{{dh}}{{dt}} :$$

$\frac{{dh}}{{dt}} = \frac{4}{{\pi {h^2}}}\frac{{dV}}{{dt}}.$

Now we can substitute known values to compute the rate of change of the water level:

$\frac{{dh}}{{dt}} = \frac{4}{{\pi {h^2}}}\frac{{dV}}{{dt}} = \frac{4}{{\pi \cdot {6^2}}} \cdot 2 = \frac{2}{{9\pi }}\frac{\text{ft}}{\text{sec}}.$

### Example 7.

A kite $$160$$ feet above the ground moves in a direction parallel to the ground at the rate of $$5$$ feet per second (Figure $$7$$). How fast is the string unwinding when the length of string already let out is $$200$$ feet?

Solution.

We denote the length of string by $$s.$$ By the Pythagorean theorem,

${s^2} = {x^2} + {h^2}.$

Take the derivatives of both sides with respect to time $$t$$ assuming the height $$h$$ remains constant.

$\frac{d}{{dt}}\left( {{s^2}} \right) = \frac{d}{{dt}}\left( {{x^2} + {h^2}} \right),\;\; \Rightarrow 2s\frac{{ds}}{{dt}} = 2x\frac{{dx}}{{dt}},\;\; \Rightarrow \frac{{ds}}{{dt}} = \frac{x}{s}\frac{{dx}}{{dt}}.$

Write $$x$$ in terms of $$s$$ and $$h:$$

$x = \sqrt {{s^2} - {h^2}} .$

Hence,

$\frac{{ds}}{{dt}} = \frac{{\sqrt {{s^2} - {h^2}} }}{s}\frac{{dx}}{{dt}}.$

The speed of the kite $$\frac{{dx}}{{dt}}$$ is known (it is equal to $$5\,\frac{\text{ft}}{\text{sec}}$$). Substituting the values of $$s$$ and $$h,$$ we find the rate $$\frac{{ds}}{{dt}}:$$

$\frac{{ds}}{{dt}} = \frac{{\sqrt {{{200}^2} - {{160}^2}} }}{{200}} \cdot 5 = 3\,\frac{\text{ft}}{\text{sec}}.$

### Example 8.

Ship A is 60 miles north of point O and moving in the north direction at $$20$$ miles per hour. Ship B is 80 miles east of point O and moving west at 25 miles per hour (Figure $$8$$). How fast is the distance between the ships changing at this moment?

Solution.

Let us denote the distance between the two ships as $$s.$$ By the Pythagorean theorem,

${s^2} = {x^2} + {y^2},$

where $$y$$ is the coordinate of point $$A$$ and $$x$$ is the coordinate of point $$B.$$ Take the derivatives of both sides:

$2s\frac{{ds}}{{dt}} = 2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}},\;\;\frac{{ds}}{{dt}} = \frac{{x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}}}}{s} = \frac{{x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}}}}{{\sqrt {{x^2} + {y^2}} }}.$

Substitute the known values to get the rate of change of the distance between the ships:

$\frac{{ds}}{{dt}} = \frac{{80 \cdot \left( { - 25} \right) + 60 \cdot 20}}{{\sqrt {{{80}^2} + {{60}^2}} }} = \frac{{ - 800}}{{100}} = - 8\frac{\text{mi}}{\text{h}}.$

The minus sign means that the distance is decreasing at this moment.

### Example 9.

The legs of an isosceles triangle are increasing at the rate of $$0.5$$ feet per minute, and the vertical angle $$\alpha$$ is increasing at the rate of $$1$$ radian per minute (Figure $$9$$). At what rate is the area $$A$$ of the triangle changing when the leg $$b$$ is 2 feet and the angle $$\alpha$$ is $$\frac{{2\pi }}{3}?$$

Solution.

The area of an isosceles triangle is given by

$A = \frac{{{b^2}}}{2}\sin \alpha .$

Differentiate both sides of the equation with respect to time $$t.$$ The right-hand side contains two variables that change with time: $$b$$ and $$\alpha.$$ Therefore, we differentiate the right-hand side combining the product and chain rules.

$\frac{{dA}}{{dt}} = \frac{d}{{dt}}\left( {\frac{{{b^2}}}{2}\sin \alpha } \right) = \frac{{\sin \alpha }}{2} \cdot 2b\frac{{db}}{{dt}} + \frac{{{b^2}}}{2} \cdot \cos \alpha \frac{{d\alpha }}{{dt}} = b\sin \alpha \frac{{db}}{{dt}} + \frac{1}{2}{b^2}\cos \alpha \frac{{d\alpha }}{{dt}}.$

Substitute the known values to calculate $$\frac{{dA}}{{dt}}:$$

$\frac{{dA}}{{dt}} = 2 \cdot \sin \frac{{2\pi }}{3} \cdot 0,5 + \frac{1}{2} \cdot {2^2} \cdot \cos \frac{{2\pi }}{3} \cdot 1 = \frac{{\sqrt 3 }}{2} - 1 \approx - 0.134\frac{{{\text{ft}^2}}}{\text{min}}.$

The area of the triangle is decreasing at the approximate rate of $$0.134\,\frac{{{\text{ft}^2}}}{\text{min}}.$$

### Example 10.

Find all points on the ellipse $$9{x^2} + 16{y^2} = 400,$$ at which the $$y-$$coordinate is decreasing and the $$x-$$coordinate is increasing at the same rate (Figure $$10$$).

Solution.

We differentiate both sides of the ellipse equation with respect to $$t:$$

$\frac{d}{{dt}}\left( {9{x^2} + 16{y^2}} \right) = \frac{d}{{dt}}\left( {400} \right),$
$18x\frac{{dx}}{{dt}} + 32y\frac{{dy}}{{dt}} = 0,$

or

$9x\frac{{dx}}{{dt}} + 16y\frac{{dy}}{{dt}} = 0.$

We are asked to find the points where

$\frac{{dx}}{{dt}} = - \frac{{dy}}{{dt}},$

so we can write the system of equations:

$\left\{ \begin{array}{l} 9x\frac{{dx}}{{dt}} + 16y\frac{{dy}}{{dt}} = 0\\ \frac{{dx}}{{dt}} = - \frac{{dy}}{{dt}} \end{array} \right..$

Hence, the points on the ellipse must satisfy the condition

$9x = 16y,\;\;\text{or}\;\;x = \frac{{16}}{9}y.$

Substitute this in the ellipse equation to calculate the $$y-$$coordinates:

$9{x^2} + 16{y^2} = 400,\;\; \Rightarrow 9 \cdot {\left( {\frac{{16}}{9}y} \right)^2} + 16{y^2} = 400,\;\; \Rightarrow \frac{{256}}{9}{y^2} + 16{y^2} = 400,\;\; \Rightarrow \frac{{400}}{9}{y^2} = 400,\;\; \Rightarrow {y^2} = 9,\;\; \Rightarrow {y_1} = 3,{y_2} = - 3.$

The $$x-$$coordinates are given by

${x_1} = \frac{{16}}{9}{y_1} = \frac{{16}}{9} \cdot 3 = \frac{{16}}{3},$
${x_2} = \frac{{16}}{9}{y_2} = \frac{{16}}{9} \cdot \left( { - 3} \right) = - \frac{{16}}{3}.$

We've got two points on the ellipse:

$A\left( {\frac{{16}}{3},3} \right),\;B\left( { - \frac{{16}}{3}, - 3} \right).$