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# 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.

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.

### 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°.

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.

### 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.

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?

(Diagam not to scale) 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.

### 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?

Scalar Diagram (involving distances only): We can see that:

sin theta=5/x, so

theta=sin^-1(5/x)

Vector Diagram (involving velocities): [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"

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