Introduction to

Applications of Derivative in Engineering

Application of derivative is defined as the change (increase or decrease) in the quantity such as motion represents derivative.

Differentiation and integration can help us solve many types of real-world problems.

We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.).

Derivatives are met in many engineering and science problems, especially when modelling the behavior of moving objects.

Our discussion begins with some general applications which we can then apply to specific problems.

1. Tangents and NormalWe often need to find tangents and normal to curves when we are analyzing forces acting on a moving body.

A tangent to a curve is a line that touches the curve at one point and has the same slope as the curve at that point.

A normal to a curve is a line perpendicular to a tangent to the curve.

Note 1: As we discussed before (in Slope of a Tangent to a Curve), we can find the slope of a tangent at any point (x, y) using d (x) / dy .

Note 2: To find the equation of a normal, recall the condition for two lines with slopes m1and m2 to be perpendicular (see Perpendicular Lines):

m1 × m2 = −1

2. Curvilinear Motion In The Derivative as an Instantaneous Rate of Change, we found out how to find the velocity from the displacement function using:

v=d (t)/ds

And the acceleration from the velocity function (or displacement function), using:

a=d (t)/dv =d(t2)d(2s)

These formulae are only appropriate for rectilinear motion (i.e. velocity and acceleration in a straight line). This is inadequate for most real situations, so we introduce here the concept of curvilinear motion, where an object is moving in a plane along a specified curved path.

We generally express the x and y components of the motion as functions of time. This form is called parametric form.

3. Curve Sketching There are now many tools for sketching functions (Mathcad,

Scientific Notebook, graphics calculators, etc). It is important in this section to learn the basic shapes of each curve that you meet. An understanding of the nature of each function is important for your future learning. Most mathematical modelling starts with a sketch.

You need to be able to sketch the curve, showing important features. Avoid drawing x-y boxes and just joining the dots.

We will be using calculus to help find important points on the curve.

The kinds of things we will be searching for in this section are:

x-interceptsUse y=0 NOTE: In many cases, finding x-intercepts is not so easy. If so, delete this step.

y-intercepts Use x=0

local maxima Use dx/dy = 0, sign of first derivative changes +→−

local minima Use dx/dy = 0, sign of first derivative changes −→+

points of inflection Use dx/2d2y = 0, and sign of dx2d2y changes

4. Finding Maxima and MinimaLocal maximum

A local maximum occurs when y′=0 and y′ changes sign from positive to negative (as we go left to right).

Local minimum

A local minimum occurs when y′=0andy′ changes sign from negative to positive.

5. Physical Applications of DerivativesMany physical quantities are derivatives of other quantities. One of the most basic examples is velocity (speed and direction), which is the derivative of position. For our discussion, we will only consider motion in one direction, although in general, motion may occur simultaneously in three independent directions, corresponding to the three dimensions of space. Let x (t) denote the position of an object at time t. typically, x is measured in meters and time in seconds. Then the velocity v (t) of the object is equal to the derivative of its position. In other words, we have

(11.5.1) v (t) = dx / dt

Furthermore, the acceleration a (t) of the object is defined as the derivative of its velocity, i.e. we have

(11.5.2) a (t) = dv / dt

Let us now return to our discussion from Section 1.10 of the law of falling bodies. We mentioned that if we drop an object from rest, the distance it falls after t seconds is equal to y (t) = 4.9t2 and its speed after t seconds is given by v (t) = 9.8t. We now want to generalize this law to the case of an object which is thrown directly up or down from an arbitrary height above the ground. We denote the height from which the object is thrown by y0 and the velocity at which it is thrown v0, which is positive if the object is thrown upward and

negative if it is thrown downward. Then the formula for the height of the object above the ground after t seconds is given by

(11.5.3) y(t) = y0 + v0t - 4.9 t2

Where as always, height is measured in meters and time in seconds. Taking the derivative of Equation (11.5.3), we find the velocity of the object is given by

(11.5.4) v (t) = d (y) / dt = v0 - 9.8 t.

Taking the derivative once again, we get the acceleration of the object, which is given by

(11.5.5) a (t) = dv / dt = -9.8.

Note that the acceleration of the object is constant; it is always equal to -9.8 meters per second per second. The fact that it is negative means the object is accelerating toward the ground. The rate at which object accelerate toward the ground during free-fall is a physical constant, which goes by the symbol g, which stands for the acceleration due to gravity. Thus, we have g = 9.8 meters per second per second. (Actually, g is not exactly constant, but varies slightly with latitude and altitude, but the maximum variation along the earth's surface is only about 1 part in 200.)