CONTEXTUAL APPLICATIONS OF DERIVATIVES

Derivatives in Context

The derivative can help us find the
    Note:
    \(\frac{dy}{dx}\) is the same as \(\frac{d}{dx}(f(x))\) where \(y = f(x)\).
    So, \(\frac{dy}{dx}\) is essentially the same as \(f'(x)\) and is used interchangeably.
  1. instantaneous velocity
  2. rate of change
  3. average speed
  4. slope of a curve at a particular point

To be specific, the derivative (often notated as \(\frac{dy}{dx}\)) denotes the value produced when an infinitesimally small change in y is divided by an infinitesimally small change in x.

Through this notation, we can think of derivatives as the rate of change over a very very small interval. On the other hand, the notation \(f'(x)\) can be thought of as the derivative without thinking of it as a slope (rate of change).

Position, Velocity, Acceleration

Suppose \(x(t)\) is a function that gives you the position of a particle with respect to time.

Then the derivative of \(x(t)\) with respect to time is the velocity (or \(v(t)\)).
Then the second derivative of \(x(t)\) with respect to time is the acceleration (or \(v(t)\)).

Position: \(x(t)\)
Velocity: \(v(t) = x'(t)\) or \(\frac{dx}{dt}\)
Acceleration = \(a(t) = v'(t)\) or \(\frac{dv}{dt} = x''(t)\) or \(\frac{d^2x}{dt^2}\)

Example: The position of a car at time t is given by the equation \(x(t) = t^4 - t^3 + 19\). Find the position, velocity and acceleration at t = 2 seconds.

\(x(t) = t^4 - t^3 + 19\)
\(v(t) = \frac{dx}{dt} = 4t^3 - 3t^2\)
\(a(t) = \frac{dv}{dt} = 12t^2 - 6t\)

At t = 2 seconds,

Can \(x\) be negative?
Yes, it can!!! This is because x is the position
of the particle in some space, not the distance.
If it were distance, then it cannot be negative.

\(x(2) = 2^4 - 2^3 + 19 = 27\)
\(v(2) = 4(2)^3 - 3(2)^2 = 20\)
\(a(2) = 12(2)^2 - 6(2) = 36\)

Note:
\(v(2)\) can also be denoted as \(\frac{dx}{dt}|_{t=2}\)

So, What does this mean?

  • Since x is positive, we can assume that, relative to some object, the car is in the supposed positive direction (we get to choose what we define as positive direction).
  • Since velocity AND position are positive, the car will continue to travel in the positive direction at t = 2 seconds.
  • Since both velocity AND acceleration are positive, the car is speeding up at t = 2 seconds.

Try it yourself:

  1. The position of a bee is given by the function \(x(t) = \sin(2t)\). Find the velocity and acceleration of the bee at t = 5 seconds. Is the bee speeding up or slowing down?

BUT GUESS WHAT, DERIVATIVES ARE ALSO USEFUL IN CONTEXTS OUTSIDE OF MOTION!

Example: At a given moment t, a family of bacteria has a population of \(B = 17t - 3^t\). After how many hours will the basteria stop reproducing?

\(B = 17t - 3^t\)
\(\frac{dB}{dt} = 17 - (\ln(3))3^t\)

When \(\frac{dB}{dt} = 0\), the family of bacterias will stop reproducing.

\(0 = 17 - 3^t\ln(3)\)
\(17 = 3^t\ln(3)\)

Note:
Always show three decimal places on you final answer!!
Round you answer ONLY on the final step!!

\(3^t = \frac{17}{\ln(3)}\)
\(t = \log_3(\frac{17}{\ln(3)})\) = 2.493 hours

Related Rates

What does related rates mean?

Personally, the definition is not important.
Conceptual understanding, though, is ABSOLUTELY ESSENTIAL.

Just as the name sounds, related rates is the rate of change of A with respect to something that is somehow related to the rate of change of B.
(Chain rule + Implicit Differentiation is everyone's best friend in related rates).

Example 1: A spherical snowball is melting. Its radius decreases at a constant rate of 2cm/minute from an initial value of 70cm. How fast is the volume of the snowball decreasing in half an hour.

The question tells us that \(\frac{dr}{dt} = -2\)cm/min and that \(r_i = 70\)cm.
"Volume of the snowball decreasing" is the same thing as \(\frac{dV}{dt}\), which we need to find.

We have \(\frac{dr}{dt}\), but we need \(\frac{dV}{dt}\).

SO we asks ourselves,

"What is the relationship between the radius (r) and volume (V) of the sphere?"

(cause spherical snow ball = sphere)

OH! \(V = \frac{4}{3}\pi r^3\)

Genius!!! But we need \(\frac{dV}{dt}\)... hmm...

OHHHH! We take the derivative of V in terms of t. We get,

\(\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}\)

But we are not done yet. We need to find out what the volume is in 30 minutes.
At the start r = 70 cm and it decreases at 2 cm per minute

So, \(r(t) = 70 - 2t\)
Then in 30mins, \(r(30) = 70 - 2(30) = 10\) cm
So \(\frac{dV}{dt}|_{t=30} = 4\pi(r(30))^2\frac{dr}{dt}|_{t=30} = 4\pi(10)^2(-2) = -800\pi\) cm3/min

Example 2: A 26 foot ladder stands against a high wall. If the foot of the ladder is sliding away at 15 ft/sec, how fast is the top of the ladder sliding down the wall when the top of the ladder in 10 ft from the ground?

"What is the relationship between x and y?""
Oh!! The pythagoren Theorem!!!
So, \(x^2 + y^2 = 26^2\)

As always, take the derivative in terms of time (here you take the derivative implicitly)

\(2x\cdot\frac{dx}{dt} + 2y\cdot\frac{dy}{dt} = 0\)

We know that \(\frac{dx}{dt} = 15\) ft/sec. We want \(\frac{dy}{dt}\) when \(y = 10\) ft.
We'll also need to know the value of \(x\) when \(y = 10\). To do this we'll use the pythagorean theorem!

\(x^2 + y^2 = 26^2\)
\(x^2 + 10^2 = 26^2\)
So, x = 24

Now, we can plug all the values into \(2x\cdot\frac{dx}{dt} + 2y\cdot\frac{dy}{dt} = 0\)

Why is \(\frac{dy}{dt}\) negative?
This is because y decreases in value as the time increase.

\(2(24)(15) + 2(10)\frac{dy}{dt} = 0\)
\(\frac{dy}{dt} = \frac{-2(24)(15)}{2(10)} = -36\) ft/sec

Try it out:

  1. A balloon is expanding at \(20\pi\) in3/sec. How fast is the SA of the balloon expanding when the balloon has a radius of 4 inches.

L'Hopital's Rule

L'Hopital's Rule (pronounced: lo-pi-tals) is another tool that can be used to solve certain special limits.
So, When can you use L'Hopital's rule?? When a limit is in indeterminate form.

If \(\frac{f(c)}{g(c)}\) is in indeterminate form, \(f'(c)\) & \(g'(c)\) exists, and \(g'(c) \neq 0\)
Then, \(\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}\)

Indeterminate Form (def.):
Anything that can't be defined by
maths is called an indeterminate form.
Example: \(\frac{0}{0}, \frac{\infty}{\infty}, \frac{-\infty}{-\infty}\), etc

Example: What does \(\lim_{x \to 0} \frac{\sin x}{x}\) evaluate to?

By direct substitution, we get: \(\frac{\sin 0}{0} = \frac{0}{0}\)
... this is in indeterminate form. SO we use L'Hopital's Rule!

Solution: \(\lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1\)

Example: What does \(\lim_{x \to \frac{\pi}{2}} \frac{\sec x}{1 + \sec x}\) evaluate to?

\(\lim_{x \to \frac{\pi}{2}} \frac{\sec x}{1 + \sec x} = \frac{\sec \frac{\pi}{2}}{1 + \sec \frac{\pi}{2}} = \frac{\frac{1}{0}}{1 + \frac{1}{0}} = \frac{\infty}{\infty} =\) indeterminate form! Therefore, L'Hopital's Rule.

Solution: \(\lim_{x \to \frac{\pi}{2}} \frac{\sec x}{1 + \sec x} = \lim_{x \to \frac{\pi}{2}} \frac{\sec x \tan x}{\sec x \tan x} = \lim_{x \to \frac{\pi}{2}} 1 = 1\)

Try it out:

  1. \(\lim_{x \to \infty} \frac{x^5}{e^{5x}}\)
  2. \(\lim_{x \to 0} \frac{e^{3x} - e^x}{x}\)
  3. \(\lim_{x \to 0} \frac{e^{3x} - e^x}{e^x}\)