Problem 38 - Calculus

Question

Give an example where \(u(x) / v(x)\) is increasing but \(d u / d x=\) \(d v / d x=1\)

Step-by-Step Solution

Answer

Functions \(u(x) = x\) and \(v(x) = x + 1\) satisfy the conditions.

Step 1: Understand the Question

The task is to find functions \(u(x)\) and \(v(x)\) such that their derivatives are both 1, and the ratio \(u(x)/v(x)\) is increasing.

Step 2: Choose Simple Functions

Let's choose linear functions for simplicity. Set \(u(x) = x + c_1\) and \(v(x) = x + c_2\), where the derivatives \(\frac{d}{dx}(u(x)) = \frac{d}{dx}(v(x)) = 1\).

Step 3: Find Conditions for Increasing Ratio

The ratio \(\frac{u(x)}{v(x)}\) will be increasing if its derivative is positive. Compute the derivative using the quotient rule:\[\left(\frac{u}{v}\right)' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\].

Step 4: Apply the Derivative Rule

Apply \(u'(x) = 1\) and \(v'(x) = 1\) in the quotient derivative formula:\[\left(\frac{u}{v}\right)' = \frac{(x + c_2) - (x + c_1)}{(x + c_2)^2} = \frac{c_2 - c_1}{(x + c_2)^2}\].

Step 5: Ensure Positive Derivative

For the ratio to be increasing, set \(c_2 - c_1 > 0\). Choose, for example, \(c_1 = 0\) and \(c_2 = 1\), which makes \(c_2 - c_1 = 1 > 0\).

Step 6: Validate the Functions

With \(u(x) = x\) and \(v(x) = x + 1\), both derivatives are 1, and the ratio \(\frac{u(x)}{v(x)} = \frac{x}{x+1}\) has a positive derivative by the condition we derived.

Key Concepts

Derivative

The concept of the derivative is central to calculus and it's all about rates of change. Consider a function, say, the position of a car as a function of time. The derivative of this function gives the car's speed at any given moment. In general, the derivative of a function \(f(x)\) represents how sensitive \(f(x)\) is to changes in \(x\). In simpler terms, it tells you how much \(f(x)\) will increase or decrease if \(x\) is slightly increased.

When computing derivatives, different rules apply depending on the type of function you're dealing with:

• For a constant function, the derivative is zero because constants don't change.
• The derivative of \(x^n\) is \(nx^{n-1}\).
• If \( f(x) = ax + b \), the derivative is \( a \) because linear functions change at a constant rate.

In our problem, the functions \(u(x) = x + c_1\) and \(v(x) = x + c_2\) both have derivatives equal to 1 because they are linear, with a slope of 1.

Quotient Rule

The quotient rule is a technique used in calculus for finding the derivative of a quotient of two functions. It is particularly useful when you have a function that is the division of two other functions, like \(\frac{u(x)}{v(x)}\).

The formula for the quotient rule is:\[\left(\frac{u}{v}\right)' = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}\]This formula results from applying the product and chain rules to derivatives. Notice that you'll need the derivatives of \(u(x)\) and \(v(x)\), as well as the original functions themselves.

• In the exercise, by setting \(u'(x) = 1\) and \(v'(x) = 1\), the derivative becomes \(\frac{c_2 - c_1}{(x + c_2)^2}\).
• The derivative is positive when \(c_2 - c_1 > 0\), which is why the choice of constants is crucial here.

Therefore, by ensuring \(c_2 > c_1\), you secure an increasing function as the derivative of the quotient remains positive.

Increasing Function

An increasing function is one where, as you move from left to right across its graph, the function continues to rise. In formal terms, a function \(f(x)\) is increasing on an interval if, for any \(x_1 < x_2\) in that interval, \(f(x_1) < f(x_2)\).

For our exercise, the ratio \(\frac{u(x)}{v(x)}\) should be increasing, which means its derivative must be greater than zero for all \(x\).

• The derivative \(\frac{c_2 - c_1}{(x + c_2)^2}\) is positive if \(c_2 - c_1 > 0\).
• This ensures that any small increase in \(x\) leads to an increase in the ratio \(\frac{u(x)}{v(x)}\).

By selecting \(c_1 = 0\) and \(c_2 = 1\), we ensured that \(c_2 - c_1 = 1 > 0\). This makes the ratio increase for all \(x\), as desired.