4. Applications: Derivatives of Trigonometric Functions

by M. Bourne

We can now use derivatives of trigonometric and inverse trigonometric functions to solve various types of problems.

Example 1

Find the equation of the normal to the curve of `y=tan^-1(x/2)` at `x=3`.

Answer

If you need a refresher, see the section on Tangent and Normals.

The derivative of `y=arctan(u)` is given by:

`(dy)/(du)=1/(1+(u)^2)`

In this example, we have `u=x/2`, so the derivative is:

`(dy)/(dx)=1/(1+(x/2)^2)(1/2)`

When `x = 3`, this expression is equal to: `0.153846`

So the slope of the tangent at `x = 3` is `0.153846`.

The slope of the normal at `x = 3` is given by:

`(-1)/0.153846=-6.5`

So the equation of the normal is given by:

(When `x = 3`, `y = 0.9828`)

`y − 0.9828 = -6.5(x − 3)`

That is,

`y = -6.5x + 20.483`

Here is the graph of our situation:

The graph of `y=arctan(x/2)` showing the tangent and the normal at `x=3.`

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Example 2

The apparent power Pa of an electric circuit whose power is P and whose impedance phase angle is θ, is given by

`P_a = P sec θ`.

Given that P is constant at 12 W, find the time rate of change of Pa if θ is changing at the rate of 0.050 rad/min, when θ = 40°.

Answer

Using chain rule, we have:

`(dP_a)/(dt)=(dP_a)/(d theta)(d theta)/(dt)`

Now `P_a= P sec θ = 12 sec θ` (since `P = 12\ "W"`)

`(dP_a)/(d theta)=12 sec theta tan theta`

We are told

`(d theta)/(dt)=0.050`

So

`(dP_a)/(dt)=(dP_a)/(d theta)(d theta)/(dt)`

`=(12 sec theta tan theta)(0.050)`

`=0.6 sec theta tan theta`

When θ = 40°, this expression is equal to: 0.657 W/min.

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Example 3

A machine is programmed to move an etching tool such that the position of the tool is given by x = 2 cos 3t and y = cos 2t, where the dimensions are in cm and time is in s. Find the velocity of the tool for t = 4.1 s.

Answer

`v_x=(dx)/(dt)=-6 sin 3t`

`v_y=(dy)/(dt)=-2 sin 2t`

At `t = 4.1`, `v_x = 1.579` and `v_y = -1.88`.

So

`v=sqrt((v_x)^2+(v_y)^2)`

`=sqrt((1.579)^2+(-1.88)^2)`

`=2.46\ "cm"//s`

For velocity, we need to also indicate direction. First, we find the appropriate acute angle (the "reference" angle):

`alpha=tan^-1(1.88 / 1.579)=50^@`

So since we are in the 4th quadrant when `v_x` is positive and `v_y` is negative, the required angle is `360^@ - 50^@ = 310^@`.

(See the following for background on how to find this angle: Trigonometric Functions of any Angle.)

Example 4

The television screen at a sports arena is vertical and 2.4 m high. The lower edge is 8.5 m above an observer's eye level. If the best view of the screen is obtained when the angle subtended by the screen at eye level is a maximum, how far from directly below the screen must the observer be?

Answer

(Diagam not to scale)

math expression

We define θ1 and θ2 as shown in the diagram such that:

`θ = θ_2 - θ_1`.

Let x be the distance from directly under the screen to the observer. To maximise `θ,` we will need to find

`dy/dx`

and then set it to 0.

We note that

`tan theta_1=8.5/x`, and

`tan theta_2=10.9/x`

This gives:

`theta_1=tan^-1(8.5)/x ` and

`theta_2=tan^(-1) 10.9/x`

Now since `θ = θ_2 - θ_1`,

`theta=tan^-1 10.9/x-tan^-1 8.5/x`

We have a function of a function in each term.

Now, in the first term, if we let

`u=10.9/x=10.9x^-1`

then

`(du)/(dx)=d/(dx)(10.9x^-1)`

`=-10.9x^-2`

`=-10.9/x^2`

Similarly for the second term, we will have:

`d/(dx)(8.5x^-1)`

`=-8.5x^-2`

`=-8.5/x^2`

So we have:

`(d theta)/(dx)=1/(1+((10.9)/x)^2)(-10.9/x^2)` `-1/(1+((8.5)/x)^2)(-8.5/x^2)`

`=1/x^2((-10.9)/(1+(10.9^2)/(x^2))+8.5/(1+(8.5^2)/(x^2)))`

Next, we multiply the x2 in the denominator (bottom) of the first fraction by the denominators of the 2 fractions in brackets, giving:

`=(-10.9)/(x^2+10.9^2)+8.5/(x^2+8.5^2)`

`=(-10.9(x^2+8.5^2)+8.5(x^2+10.9^2))/((x^2+10.9^2)(x^2+8.5^2))`

`=(-2.4x^2+222.36)/((x^2+10.9^2)(x^2+8.5^2))`

To find when this equals `0`, we need only determine when the numerator (the top) is `0`.

That is

`-2.4x^2 + 222.36 = 0`

This occurs when `x = 9.63` (we take positive case only)

So the observer must be `9.63\ "m"` from directly below the screen to get the best view.

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Example 5

A winch on a loading dock is used to drag a container along the ground. The winch winds the cable in at 2ms-1 and is 5 m above the ground. At what rate is the angle θ between the cable and the ground changing when 10 m of cable is out?

Answer

Scalar Diagram (involving distances only):

math expression

We can see that:

`sin theta=5/x`, so

`theta=sin^-1(5/x)`

Vector Diagram (involving velocities):

Related rates winch problem

[See the section on Vector concepts for more on vectors and scalars.]

We are also given that:

`(dx)/(dt)=-2\ "ms"^-1`

Since we need `(d theta)/(dt)`, we use:

`(d theta)/(dt)=(d theta)/(dx)(dx)/(dt)`

Because

`theta=sin^-1(5/x)`

we have:

`(d theta)/(dx)=1/(sqrt(1-(5/x)^2))(d(5/x))/(dx)`

`=(-5)/(x^2sqrt(1-(5/x)^2))`

So

`(d theta)/(dt)=(d theta)/(dx)(dx)/(dt)`

`=(-5)/(x^2sqrt(1-(5/x)^2))(-2)`

`=10/(x^2sqrt(1-(5/x)^2))`

We want to know the rate of change of θ when x = 10 m, so we substitute, as follows:

`(d theta)/(dt)=10/(10^2sqrt(1-(5/10)^2))`

`=0.1155\ "rad"//"s"`