Calculus of Trigonometric Functions

Lesson

As can be done with other kinds of differentiable functions, we can use calculus to find the equations of tangents and normals at points on the trigonometric functions.

Differentiation allows us to find the gradient of the tangent to a curve at a given point. The coordinates of the point and the value of the gradient are enough information from which to deduce the equation of the tangent line.

To find the equation of the normal line at a point on a curve, we make use of the fact that the gradient of the normal is $\frac{-1}{m}$−1`m` times the gradient $m$`m`of the tangent at the point. We can show that this is true by constructing a pair of mutually perpendicular lines, as in the following diagram. One line can be taken to be the tangent and the other the normal at the point where they intersect.

The lines $PB$`P``B` and $AC$`A``C` have been constructed so that they are perpendicular to one another. The segment $PB$`P``B` has length $1$1 unit and segments $AB$`A``B` and $BC$`B``C` have lengths $m$`m` and $n$`n` respectively.

Because they have the same angles, triangles $PAB$`P``A``B` and $CPB$`C``P``B` are similar. Therefore, we can form equal ratios, $\frac{m}{1}=\frac{1}{n}$`m`1=1`n`. But the gradients of the lines are $m$`m` and $-n$−`n`. So, we see that each gradient is the negative reciprocal of the other.

Find the equation of the tangent line to the function given by $y=1+\tan x$`y`=1+`t``a``n``x` at $x=\frac{\pi}{4}$`x`=π4.

When $x=\frac{\pi}{4}$`x`=π4, $y=1+1=2$`y`=1+1=2. So, the point on the curve has coordinates $\left(\frac{\pi}{4},\ 2\right)$(π4, 2).

The gradient function for $1+\tan x$1+`t``a``n``x` is $\sec^2x$`s``e``c`2`x`. So, at $x=\frac{\pi}{4}$`x`=π4, the gradient is $\sec^2\frac{\pi}{4}=2$`s``e``c`2π4=2.

The equation for a line can be written $m=\frac{y-y_1}{x-x_1}$`m`=`y`−`y`1`x`−`x`1 where $m$`m` is the gradient and $\left(x_1,\ y_1\right)$(`x`1, `y`1) are the coordinates of a point on the line.

After making the appropriate substitutions, we have $2=\frac{y-2}{x-\frac{\pi}{4}}$2=`y`−2`x`−π4, and this can be re-arranged to $y=2x+2-\frac{\pi}{2}$`y`=2`x`+2−π2.

Find the equation of the normal at $x=\frac{\pi}{3}$`x`=π3 to the curve given by $y=\sin x+\tan x$`y`=`s``i``n``x`+`t``a``n``x`. Where dos the normal cross the $x$`x`-axis?

At $x=\frac{\pi}{3}$`x`=π3, $y=\sin\frac{\pi}{3}+\tan\frac{\pi}{3}=\frac{\sqrt{3}}{2}+\sqrt{3}=\frac{3}{2}\sqrt{3}$`y`=`s``i``n`π3+`t``a``n`π3=√32+√3=32√3.

On differentiating the function we find $y'=\cos x+\sec^2x$`y`′=`c``o``s``x`+`s``e``c`2`x`. At $x=\frac{\pi}{3}$`x`=π3, this has the value $\frac{1}{2}+4=\frac{9}{4}$12+4=94. So, the normal at this point must have gradient $-\frac{4}{9}$−49.

As before, using the equation $m=\frac{y-y_1}{x-x_1}$`m`=`y`−`y`1`x`−`x`1 for the normal line, we have $-\frac{4}{9}=\frac{y-\frac{3}{2}\sqrt{3}}{x-\frac{\pi}{3}}$−49=`y`−32√3`x`−π3. This can be re-arranged to$y=-\frac{4}{9}x+\frac{4\pi}{27}+\frac{3\sqrt{3}}{2}$`y`=−49`x`+4π27+3√32.

To find where the normal crosses the $x$`x`-axis, we set $y=0$`y`=0. Then, $x=\frac{9}{4}\left(\frac{4\pi}{27}+\frac{3\sqrt{3}}{2}\right)$`x`=94(4π27+3√32)

Consider the function $f\left(x\right)=\tan\left(\frac{x}{5}\right)$`f`(`x`)=`t``a``n`(`x`5).

Find the equation of the tangent to the curve at $x=\frac{5\pi}{4}$

`x`=5π4.Find the equation of the normal to the curve at $x=\frac{5\pi}{4}$

`x`=5π4.

Consider the function $y=\frac{\sin x}{1+\cos x}$`y`=`s``i``n``x`1+`c``o``s``x`.

Find $\frac{dy}{dx}$

`d``y``d``x`.How many turning points does $y$

`y`have?

The displacement of an object is a measure of how far it is from a fixed point. It can be negative or positive depending on which side of the fixed point the object is on.

The displacement of an object $x$`x` about a fixed position $x=0$`x`=0, at time $t$`t` seconds, is given by $x\left(t\right)=-5\cos2t$`x`(`t`)=−5`c``o``s`2`t`. Its velocity at time $t$`t` seconds is given by $v\left(t\right)$`v`(`t`) and its acceleration at time $t$`t` seconds is given by $a\left(t\right)$`a`(`t`).

Show that $a\left(t\right)=-4x\left(t\right)$

`a`(`t`)=−4`x`(`t`).Determine the acceleration of the object when it is at the origin $x=0$

`x`=0.

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems