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21. Applications of Differentiation

Dated: 30-10-2024

We come across a lot of problems in world related to related rates.
We want to know how one quantity changes with respect to another quantity.

Example

Imagine a 5 ft ladder leaning against a wall, with a base of 4 ft, sliding off at 2 ft / sec speed.
We are interested to know at which speed, the top of the ladder is falling down at that instant.
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Solution

Let us first assign some variables.
Let us say, \(h = 5\) (hypotenus), \(x\) being the base and \(y\) being the height.
We know that ladder, being a concrete object, will not change its length as it falls down.
Therefore, we can create a function1 using the Pythagorus Theorem as such:

\[y^2 + x^2 = 5^2\]

Before touching this equation, let us look at more data.
We know that the base of the ladder at the captured instant is moving at the rate of 2 ft / sec.
Since displacement or increase in base is a function1 of time, we can write the rate of change as such:

\[\frac{dx}{dt} = \frac{2 sec}{ft}\]

And we are asked to find same but for the vertical direction, which would be:

\[\frac{dy}{dt}\]

Using the implicit differentiation2 with respect to t on the original equation,

\[2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\]

After simplification, we get

\[\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\]

We already know that x = 4, to know the value of y at that instance, we use our equation which we discussed in the beginning.

\[y^2 + x^2 = 5^2\]
\[y = \sqrt{25 - 16}\]
\[y = \sqrt{9}\]
\[y = 9\]

Now, we can plug these value inside our equation which we got by differentiation.

\[\frac{dy}{dt}|_{x=4}=-\frac{4}{3}(2)\]
\[= - \frac{8 ft}{3 sec}\]

Increasing Functions

In the context of talking over a specific interval,3 a function1 is called increasing if its y value increases as x value increases over that interval.3

Decreasing Functions

Similarly, a function1 is called decreasing if its y value decreases as the x value increases over the interval.3

Constant Functions

If the y value of a function1 remains the same over the interval3 then it is called constant function.

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Theorems

  1. if \(f^{\prime}(x) > 0\) for all values of \(x\) in \((a, b)\) then \(f(x)\) is an increaing function.
  2. if \(f^{\prime}(x) < 0\) for all values of \(x\) in \((a, b)\) then \(f(x)\) is an decreasing function.
  3. if \(f^{\prime}(x) = 0\) for all values of \(x\) in \((a, b)\) then \(f(x)\) is an constant function.

Example

Find the intervals3 where the graph of \(f(x) = x^2 - 4x + 3\) is increasing and decreasing.

Solution

First, we differentiate4 the function.1

\[f^{\prime}(x) = 2x - 2 = 2(x - 2)\]
Decreasing Interval
\[f'(x) < 0 \Rightarrow 2(x-2) < 0 \Rightarrow -\infty < x < 2\]
\[(-\infty, 2]\]
Increasing Interval
\[f'(x) > 0 \Rightarrow 2(x-2) > 0 \Rightarrow 2 < x < +\infty\]
\[[2, \infty)\]

Concavity of Functions

Let \(f(x)\) be differentiable on an interval3 then.

  1. \(f(x)\) is concave up if it is increasing on the interval3
  2. \(f(x)\) is concave down if it is decreasing on the interval3

Theorems

  1. If \(f^{\prime \prime}(x) > 0\) on an open interval3 \((a, b)\) then \(f(x)\) is concave up on that interval.3
  2. If \(f^{\prime \prime}(x) < 0\) on an open interval3 \((a, b)\) then \(f(x)\) is concave down on that interval.3

Example

\[f(x) = x^2 - 4x + 4\]
\[f^{\prime}(x) = 2x - 4\]
\[f^{\prime \prime}(x) = 2\]

Since \(f^{\prime \prime}(x) > 0\) for all \(x\), the graph is concave up at interval \((-\infty, +\infty)\).

References

Read more about notations and symbols.

  1. Read more about functions. ↩↩↩↩↩↩

  2. Read more about implicit differentiation. ↩

  3. Read more about intervals. ↩↩↩↩↩↩↩↩↩↩↩↩

  4. Read more about differentiation. ↩