NZ Level 8 (NZC) Level 3 (NCEA) [In development] Applications of differentiation of various functions
Lesson

Differentiation is used when the rate of change of a function at some point in the domain is required. This is equivalent to asking for the gradient of the graph of the function at the point in question.

For example, if a body accelerates at a varying rate, and its velocity changes continuously and smoothly with time according to some rule, we might ask when was the acceleration greatest. That is, at what point was the rate of change of the velocity function at a maximum.

At times, we will not only want the gradient of a function at a particular point but we will also want the equation of the tangent to the graph at the point. In the following diagram, a tangent is drawn to the parabola defined by $y=-\frac{x^2}{2}+1$y=x22+1 at the point on the graph where $x=1$x=1. The equation of the tangent is $y=\frac{3}{2}-x$y=32x. Gradients are found by differentiation and then the equation of a tangent line is found from the gradient and the point on the curve.

##### Example 1

The semicircle given by $y=\sqrt{9-x^2}$y=9x2 is drawn with a tangent at $x=-\frac{3}{2}$x=32. What is the equation of the tangent line?

We begin by finding the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$dydx

 $y$y $=$= $\sqrt{9-x^2}$√9−x2 $=$= $\left(9-x^2\right)^{\frac{1}{2}}$(9−x2)12​ $\frac{\mathrm{d}y}{\mathrm{d}x}$dydx​ $=$= $\frac{1}{2}\left(9-x^2\right)^{-\frac{1}{2}}.(-2x)$12​(9−x2)−12​.(−2x) $=$= $\frac{-x}{\sqrt{9-x^2}}$−x√9−x2​

At $x=-\frac{3}{2}$x=32, we have $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{3}{2}}{\sqrt{9-\frac{9}{4}}}=\frac{3}{\sqrt{27}}=\frac{1}{\sqrt{3}}$dydx=32994=327=13

The point of tangency on the semicircle is $\left(-\frac{3}{2},\frac{3}{2}\sqrt{3}\right)$(32,323). So, we can write the gradient of the tangent as $\frac{y-\frac{3}{2}\sqrt{3}}{x+\frac{3}{2}}=\frac{1}{\sqrt{3}}$y323x+32=13. After rearrangement, this is

$y=\frac{x}{\sqrt{3}}+2\sqrt{3}$y=x3+23 ##### Example 2

Consider the function given by $e^{\sin x}$esinx on the domain $\left(0,2\pi\right)$(0,2π). Find the points where the gradient is zero.

The derivative is $e^{\sin x}.\cos x$esinx.cosx. Since, $e^{\sin x}$esinx is never zero, the derivative can only be zero when $\cos x=0$cosx=0. This occurs when $x=\frac{\pi}{2}$x=π2 and $x=\frac{3\pi}{2}$x=3π2.

The function values at these points are respectively $e^{\sin\frac{\pi}{2}}=e$esinπ2=e and $e^{\sin\frac{3\pi}{2}}=e^{-1}$esin3π2=e1. #### Worked Examples

##### Question 1

Researchers have created a model to project the country’s population for the next $10$10 years, so that the population $t$t years from now is given by the function $P\left(t\right)=\frac{57460e^{\frac{t}{7}}}{t+13}$P(t)=57460et7t+13, where $P$P is the population (in thousands).

1. What is the current population of the country? Write your answer as an equation with $P\left(0\right)$P(0) as the subject.

2. According to the model, what is the current rate of growth of the population to the nearest thousand?

3. At what rate will the population be growing $7$7 months from now to the nearest thousand?

4. At what rate will the population be growing $10$10 years from now to the nearest thousand?

##### Question 2

Find the equation of the tangent to the curve $y=e^{\cos x}$y=ecosx at the point $x=\frac{3\pi}{2}$x=3π2.

##### Question 3

Consider the function $y=\frac{\ln3x}{e^{3x}}$y=ln3xe3x.

1. Determine the $x$x-value of the $x$x-intercept of the function.

2. Determine $\frac{dy}{dx}$dydx.

3. Show that the turning point of the graph occurs when $3x\ln3x-1=0$3xln3x1=0.

4. An approximation to the solution of $3x\ln3x-1=0$3xln3x1=0 is $x=0.58$x=0.58.

Complete the table of values.

 $x$x $\frac{dy}{dx}$dydx​ $0.01$0.01 $0.58$0.58 $1.58$1.58 $\editable{}$ $0$0 $\editable{}$
5. Hence determine whether there is a maximum or minimum turning point at $x=0.58$x=0.58.

Minimum

A

Neither

B

Maximum

C

Minimum

A

Neither

B

Maximum

C
6. Determine the coordinates of the turning point.

Give each coordinate to two decimal places.

### Outcomes

#### M8-11

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

#### 91578

Apply differentiation methods in solving problems