Tangent Line, Velocity and Other Rates of Changes

Related calculator: Tangent Line Calculator

Now we will talk about problems that lead to the concept of derivative.

The Tangent Line

Suppose we are given curve `y=f(x)` and point on curve `P(a,f(a))`. We want to find equation of tangent line at point `P`.

Word "tangent" means "touching", so we need to find line that touches the curve and is parallel (in some sense) to curve at point of intersection.

But this is verbal description. We need to state formal definition.

So, how do we find equation of tangent line at point `P(a,f(a))`?

To find equation of any line we need two points. Here we have only one point, namely, `(a,f(a))`. However, we can take point Q and find equation of line passing through points P and Q.

Let's see how this is done.tangent line

Let point Q has coordinates `(b,f(b))` and equation of tangent line is `y=mx+c` where `m` and `c` must be determined.

Then slope line that passes through points `P` and `Q` is `m=(f(b)-f(a))/(b-a)`.

If we move point `Q` towards point `P` along curve `y=f(x)` P then we will obtain tangent line as `Q` becomes very close to `P`. We say that slope of tangent line is limit of slopes of all lines through P and Q when Q approaches P. In other words slope of tangent line is `m=lim_(b->a)(f(b)-f(a))/(b-a)`.

It doesn't matter whether point `Q` approaches `P` from the left or from the right.

If we now set `b=a+h` then as `b->a` we have that `h->0`.

Now, slope of tangent line can be rewritten as `m=lim_(h->0)(f(a+h)-f(a))/h`.

But this is derivative at point `a`. So, `m=f'(a)`.

Thus, we have the following fact.

Fact. Equation of tangent line to the curve `y=f(x)` at point `a` is `y=f(a)+f'(a)(x-a)`.

Now, let's see how tangent line is shown on practice.

example of tangent lineExample 1. Find equation of tangent line to `y=2x-x^3+5` at point `P=(1,6)`.

We have that `f(x)=2x-x^3+5`.

Therefore `f'(x)=2-3x^2` and `f'(1)=2-3*1^2=-1`.

Thus, equation of tangent line is `y=6+(-1)(x-1)` or `y=-x+7`.

The Velocity Problem
Consider the object that is dropped from height (we neglect air resistance).

Galileo discovered that distance traveled by this object is `s=g/2 t^2` , where `g=9.81 m/s^2` is constant.

First consider case when we want to find average velocity. It is known that average velocity is distance travelled divided by time needed to travel that distance.

For example, let's find average velocity from `t=1` to `t=5` seconds: `v_(ave)=(s(5)-s(1))/(5-1)=(9.81/2*5^2-9.81/2*1^2)/4=3*9.81=29.43 m/s`.

But what if we want to find velocity at some particular time? What if we want to find instantaneous velocity?

In this case we don't have time interval and thus we can't find average velocity. However, let's take interval `[t,t+h]`: average velocity is `v=(s(t+h)-s(t))/(t+h-t)=(s(t+h)-s(t))/h`. If we now let `h->0` we will make interval `[t,t+h]` very small and thus will find instantaneous velocity: `v=lim_(h->0)(s(t+h)-s(t))/h`. But this is again derivative of function `s(t)` with respect to time.

So, `s'(t)=(g/2 t^2)'=g/2 *2t=g t=9.81t` and, for example, instantaneous velocity at time `t=3` is `s'(3)=9.81*3=29.43 m/s`.

In general, suppose an object moves along a straight line according to an equation of motion `s=f(t)`, where `s` is the displacement (directed distance) of the object from the origin at time `t` . The function `f` that describes the motion is called position function.

Fact. Velocity is derivative of position function: `v=s'`.

Example 2. Suppose that a distance the car travelled at any time `t` is given as `s(t)=t^2+t+5` metres. This means, that for example after 2 seconds car travelled `2^2+2+5=11` metres. Find velocity of car after 4 seconds.

Since velocity of the car is derivative of position function then `v(t)=s'(t)`: `v(t)=(t^2+t+5)'=(t^2)+(t)'+(5)'=2t+1`.

Thus, velocity of car after 4 seconds is `v(4)=2*4+1=9 m/s`.

Note how close are tangent and velocity problems. In both cases we need to find some quantity at point. To do this we take interval and then shorten it to limiting value (with tangent problems it is length of line P and Q, in velocity problem - time interval).

In fact we can extend this idea to any problem.

Suppose that we are given a function `y=f(x)`. On interval `[x_1,x_2]` change in `x` is `Delta x=x_2-x_1` and corresponding change in `y` is `Delta y=y_2-y_1`.

Then rate of change of `y` with respect to `x` is `(Delta y)/(Delta x)=(y_2-y_1)/(x_2-x_1)=(f(x_2)-f(x_1))/(x_2-x_1)`.

Instantaneous rate of change is `lim_(Delta x->0)(Delta y)/(Delta x)=lim_(Delta x->0)(f(x_2)-f(x_1))/(x_2-x_1)`.

But we recognize in this limit derivative. Indeed, if we set `x_2-x_1=h` then `h->0` as `Delta x->0` and `x_2=x_1+h`.

Now, instantaneous rate of change can be rewritten as `lim_(h->0)(f(x_1+h)-f(x_1))/h` which is derivative at point `x_1`.

Fact. The derivative `f'(a)` is instantaneous rate of change of `y=f(x)` with respect to `x` when `x=a`.

For example, suppose that `N(t)` is population of bacteria at any time `t`. Then `N'(t)` is rate of change of population at any given time `t`, i.e. how fast population is changing at any time `t`.

Example 3. Suppose that population of bacteria at any time `t` (`t` is measured in hours) is `N(t)=100e^t+2t^2`. How fast population is increasing after two hours.

We need to find `f'(2)`. Since `N'(t)=100e^t+4t` then `N'(2)=100*e^2+4*2~~747` bacteria per hour.

So, after two hours population of bacteria grows at the rate of 747 bacteria per hour.