Calculus

Applications of the Derivative

Applications of Derivative Logo

Related Rates

  • Suppose we have two quantities, which are connected to each other and both changing with time. A related rates problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

    Let the two variables be \(x\) and \(y.\) The relationship between them is expressed by a function \(y = f\left( x \right).\) The rates of change of the variables \(x\) and \(y\) are defined in terms of their derivatives \(\large{\frac{{dx}}{{dt}}}\normalsize\) and \(\large{\frac{{dy}}{{dt}}}\normalsize.\) If \(\large{\frac{{dx}}{{dt}}}\normalsize\) is known, we can determine \(\large{\frac{{dy}}{{dt}}}\normalsize\) (and vice versa).

    Any related rates problem can be solved as follows:

    1. Decide what are the two variables describing your system or process.
    2. Drawing a picture can often be useful.
    3. Write an equation that relates the variables. To derive the equation you may use a geometric fact (like the Pythagorean theorem or similar triangles), a trigonometric identity, or a physical law.
    4. Take the derivative \(\large{\frac{d}{{dt}}}\normalsize\) of both sides of the equation.
    5. Solve for the unknown rate of change.
    6. Substitute all known values to get the final answer.

    As an example, let’s consider the well-known sliding ladder problem.

    A ladder \(13\) feet long leans against a wall. The top of the ladder slides down at a constant rate of \(12\,\large{\frac{\text{ft}}{\text{sec}}}\normalsize.\) Find the velocity of the ladder when \(h = 15\,\text{ft}.\)

    sliding ladder
    Figure 1.

    By the Pythagorean equation:

    \[{{x^2} + {h^2} = {13^2}.}\]

    Differentiate both sides with respect to time \(t.\)

    \[{\frac{d}{{dt}}\left( {{x^2} + {h^2}} \right) = \frac{d}{{dt}}\left( {{{13}^2}} \right),\;\;} \Rightarrow {2x\frac{{dx}}{{dt}} + 2h\frac{{dh}}{{dt}} = 0,\;\;} \Rightarrow {x\frac{{dx}}{{dt}} = – h\frac{{dh}}{{dt}},\;\;} \Rightarrow {\frac{{dx}}{{dt}} = – \frac{h}{x}\frac{{dh}}{{dt}}.}\]

    As \(x = \sqrt {{{13}^2} – {h^2}} ,\) we have

    \[{\frac{{dx}}{{dt}} = – \frac{h}{{\sqrt {{{13}^2} – {h^2}} }}\frac{{dh}}{{dt}}.}\]

    Substitute the known values:

    \[{\frac{{dh}}{{dt}} = – 12\,\frac{\text{ft}}{\text{sec}}}\]

    (the minus sign denotes that the ladder is sliding down),

    \[{h = 15\,\text{ft}.}\]

    This yields:

    \[{\frac{{dx}}{{dt}} = – \frac{h}{{\sqrt {{{13}^2} – {h^2}} }}\frac{{dh}}{{dt}} }={ – \frac{5}{{\sqrt {{{13}^2} – {5^2}} }} \cdot \left( { – 12} \right) }={ \frac{5}{{12}} \cdot 12 }={ 5\,\frac{\text{ft}}{\text{sec}}}.\]


  • Solved Problems

    Click a problem to see the solution.

    Example 1

    Oil from an uncapped well is radiating outward in the form of a circular film on the surface of the water. If the radius of the circle is increasing at the rate of \(0.5\) meters per minute, how fast is the area of the oil film growing at the instant when the radius is \(100\,\text{m}?\)

    Example 2

    A triangle has two sides \(a = 1\,\text{cm}\) and \(b = 2\,\text{cm}.\) How fast is the third side \(c\) increasing when the angle \(\alpha\) between the given sides is \(60^\circ\) and is increasing at the rate of \(3^\circ\) per second (Figure \(2\))?

    Example 3

    A light is at the top of a \(16\) ft pole. A man \(6\) ft tall walks away from the pole at a rate of \(5\,\large{\frac{\text{ft}}{\text{sec}}}\normalsize\) (Figure \(3\)). How fast is the tip of his shadow moving when he is \(20\) ft from the pole?

    Example 4

    The volume of a cube is increasing at the rate of \(2\) cubic inches per minute. How fast is the surface of the cube increasing when the side is \(5\) inches?

    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.

    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 \(\large{\frac{{2\pi }}{3}}\normalsize?\)

    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 1.

    Oil from an uncapped well is radiating outward in the form of a circular film on the surface of the water. If the radius of the circle is increasing at the rate of \(0.5\) meters per minute, how fast is the area of the oil film growing at the instant when the radius is \(100\,\text{m}?\)

    Solution.

    Suppose that t is time in minutes, R and A are the radius and area of the circle, respectively.

    The rate of change of the area is giveb by the derivative \(\large{\frac{{dA}}{{dt}}}\normalsize,\) where

    \[A = \pi {R^2}.\]

    Differentiating the right-hand side of the relation by the chain rule, we get

    \[{\frac{{dA}}{{dt}} = \frac{d}{{dt}}\left( {\pi {R^2}} \right) }={ 2\pi R\frac{{dR}}{{dt}}.}\]

    It is known that \(\large{\frac{{dR}}{{dt}}}\normalsize = 0.5\,\large{\frac{\text{m}}{\text{min}}}\normalsize.\) Therefore, the oil spot is growing at the rate

    \[{\frac{{dA}}{{dt}} = 2\pi R\frac{{dR}}{{dt}} }={ 2\pi R \cdot 0.5 }={ \pi R.}\]

    For \(R = 100\,\text{m},\) we have

    \[\frac{{dA}}{{dt}} = 100\pi \approx 314\,\frac{{{\text{m}^2}}}{\text{min}}.\]

    Example 2.

    A triangle has two sides \(a = 1\,\text{cm}\) and \(b = 2\,\text{cm}.\) How fast is the third side \(c\) increasing when the angle \(\alpha\) between the given sides is \(60^\circ\) and is increasing at the rate of \(3^\circ\) per second (Figure \(2\))?

    Solution.

    A triangle with increasing side c.
    Figure 2.

    According to the law of cosines,

    \[{c^2} = {a^2} + {b^2} – 2ab\cos \alpha .\]

    We differentiate both sides of this equation with respect to time t:

    \[{\frac{d}{{dt}}\left( {{c^2}} \right) }={ \frac{d}{{dt}}\left( {{a^2} + {b^2} – 2ab\cos \alpha } \right),}\]

    \[{2c\frac{{dc}}{{dt}} }={ – 2ab\left( { – \sin \alpha } \right)\frac{{d\alpha }}{{dt}},}\]

    or

    \[\frac{{dc}}{{dt}} = \frac{{ab\sin \alpha }}{c}\frac{{d\alpha }}{{dt}}.\]

    Calculate the length of the side \(c:\)

    \[{c = \sqrt {{a^2} + {b^2} – 2ab\cos \alpha } }={ \sqrt {{1^2} + {2^2} – 2 \cdot 1 \cdot 2 \cdot \cos 60^\circ } }={ \sqrt {1 + 4 – 2} }={ \sqrt 3 .}\]

    Now we know all quantities to determine the rate of change \(\large{\frac{{dc}}{{dt}}}\normalsize:\)

    \[{\frac{{dc}}{{dt}} = \frac{{ab\sin \alpha }}{c}\frac{{d\alpha }}{{dt}} }={ \frac{{1 \cdot 2 \cdot \sin 60^\circ }}{{\sqrt 3 }}\frac{{d\alpha }}{{dt}} }={ \frac{{2\frac{{\sqrt 3 }}{2}}}{{\sqrt 3 }} \cdot 3 }={ 3\,\frac{\text{cm}}{\text{sec}}.}\]

    Example 3.

    A light is at the top of a \(16\) ft pole. A man \(6\) ft tall walks away from the pole at a rate of \(5\,\large{\frac{\text{ft}}{\text{sec}}}\normalsize\) (Figure \(3\)). How fast is the tip of his shadow moving when he is \(20\) ft from the pole?

    Solution.

    Man walking away from lamp
    Figure 3.

    By similar triangles,

    \[\frac{L}{{16}} = \frac{{L – x}}{6}.\]

    Solving for \(L,\) we have:

    \[{6L = 16\left( {L – x} \right),\;\;} \Rightarrow {6L = 16L – 16x,\;\;} \Rightarrow {L = \frac{8}{5}x.}\]

    Differentiating both sides with respect to time \(t\) yields:

    \[\frac{{dL}}{{dt}} = \frac{8}{5}\frac{{dx}}{{dt}}.\]

    We are told that the rate \(\frac{{dx}}{{dt}} = 5\,\large{\frac{\text{ft}}{\text{sec}}}\normalsize.\) Therefore

    \[\frac{{dL}}{{dt}} = \frac{8}{5} \cdot 5 = 8\,\frac{\text{ft}}{\text{sec}}.\]

    Thus, the tip of the shadow is moving at the rate of \(8\,\large{\frac{\text{ft}}{\text{sec}}}\normalsize.\)

    Example 4.

    The volume of a cube is increasing at the rate of \(2\) cubic inches per minute. How fast is the surface of the cube increasing when the side is \(5\) inches?

    Solution.

    The volume of the cube of side \(a\) is given by

    \[V = {a^3}.\]

    We differentiate both sides of the equation to find the relation between the rates of change:

    \[\frac{{dV}}{{dt}} = 3{a^2}\frac{{da}}{{dt}}.\]

    Solve for \(\large{\frac{{da}}{{dt}}}\normalsize:\)

    \[\frac{{da}}{{dt}} = \frac{1}{{3{a^2}}}\frac{{dV}}{{dt}}.\]

    Recall that the surface area of the cube is

    \[A = 6{a^2}.\]

    Similarly, take the derivatives of both sides:

    \[\frac{{dA}}{{dt}} = 12a\frac{{da}}{{dt}}.\]

    Now we substitute the expression for \(\large{\frac{{da}}{{dt}}}\normalsize\) from the first equation. This yields

    \[{\frac{{dA}}{{dt}} = 12a\frac{{da}}{{dt}} }={ 12a \cdot \frac{1}{{3{a^2}}}\frac{{dV}}{{dt}} }={ \frac{4}{a}\frac{{dV}}{{dt}}.}\]

    As the volume of the cube is increasing at the rate of \(2\,\large{\frac{{{\text{in}^3}}}{\text{min}}}\normalsize,\) we can write

    \[{\frac{{dA}}{{dt}} = \frac{4}{a}\frac{{dV}}{{dt}} }={ \frac{4}{a} \cdot 2 }={ \frac{8}{a}.}\]

    Finally, we substitute the side length of the cube, which is equal to \(2\) inches. As a result, we find the rate of change of the area:

    \[{\frac{{dA}}{{dt}} = \frac{8}{a} }={ \frac{8}{8} }={ 1\,\frac{{{\text{in}^2}}}{\text{min}}.}\]

    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.

    A point moving along the curve y=x^2-2x+3
    Figure 4.

    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.

    water tank related rates problem
    Figure 5.
    water tank cone section
    Figure 6.

    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 \(\large{\frac{{dh}}{{dt}}}\normalsize :\)

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

    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.

    flying kite
    Figure 7.

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

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

    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.

    Two ships related rates problem
    Figure 8.

    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:

    \[{2s\frac{{ds}}{{dt}} = 2x\frac{{dx}}{{dt}} + 2y\frac{{dy}}{{dt}},\;\;} \kern0pt {\frac{{ds}}{{dt}} = \frac{{x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}}}}{s} }={ \frac{{x\frac{{dx}}{{dt}} + y\frac{{dy}}{{dt}}}}{{\sqrt {{x^2} + {y^2}} }}.}\]

    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 \(\large{\frac{{2\pi }}{3}}\normalsize?\)

    Solution.

    An isosceles triangle with increasing legs and vertical angle.
    Figure 9.

    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 \(\large{\frac{{dA}}{{dt}}}\normalsize:\)

    \[{\frac{{dA}}{{dt}} }={ 2 \cdot \sin \frac{{2\pi }}{3} \cdot 0,5 }\kern0pt{+ \frac{1}{2} \cdot {2^2} \cdot \cos \frac{{2\pi }}{3} \cdot 1 }={ \frac{{\sqrt 3 }}{2} – 1 \approx – 0.134\frac{{{\text{ft}^2}}}{\text{min}}}.\]

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

    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.

    Ellipse related rate problem
    Figure 10.

    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

    \[{9x = 16y,\;\;}\kern0pt{\text{or}\;\;x = \frac{{16}}{9}y.}\]

    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:

    \[{A\left( {\frac{{16}}{3},3} \right),\;}\kern0pt{B\left( { – \frac{{16}}{3}, – 3} \right)}.\]