Calculus

Applications of the Derivative

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Related Rates

Solved Problems

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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.

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

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

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.

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 \(\frac{{dx}}{{dt}}\) 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.

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:

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

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.

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

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

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.

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

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

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:

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