Problem 8 - Technical Mathematics with Calculus

Question

If \(y\) is directly proportional to the five-halves power of \(x,\) and \(y\) has the value 55.3 when \(x\) is 17.3 (a) Find the constant of proportionality. (b) Write the equation \(y=f(x)\) (c) Find \(y\) when \(x=27.4\) (d) Find \(x\) when \(y=83.6\)

Step-by-Step Solution

Answer

The constant of proportionality 'k' is approximately '0.766'. The equation of proportionality is 'y ≈ 0.766 * x^(5/2)'. The value of 'y' when 'x=27.4' is approximately '124.252'. The value of 'x' when 'y=83.6' is approximately '22.084'.

Step 1: Understand direct proportionality with a power

Recognize that if 'y' is directly proportional to the 'five-halves' power of 'x', we can write this relationship as 'y = k * x^(5/2)', where 'k' is the constant of proportionality and 'x^(5/2)' represents 'x' raised to the power of five-halves.

Step 2: Calculate the constant of proportionality

Using the given values, when 'x' is 17.3, 'y' is 55.3, substitute these values into the equation 'y = k * x^(5/2)' to find 'k'. The equation becomes '55.3 = k * 17.3^(5/2)'. Solve for 'k' by dividing both sides of the equation by '17.3^(5/2)'.

Step 3: Solve for the constant 'k'

Calculate '17.3^(5/2)' and then divide '55.3' by this number to find the constant 'k'.

Step 4: Write the equation 'y=f(x)'

After finding the constant of proportionality 'k', substitute it back into the equation 'y = k * x^(5/2)' to get the function 'y = f(x)'.

Step 5: Find 'y' when 'x=27.4'

With the equation 'y = k * x^(5/2)', substitute 'x = 27.4' and calculate the value for 'y' using the previously found constant 'k'.

Step 6: Find 'x' when 'y=83.6'

To find 'x' when 'y = 83.6', substitute '83.6' for 'y' in the equation 'y = k * x^(5/2)' and solve for 'x'. You will need to isolate 'x' by dividing both sides by 'k' and then taking the 'two-fifths' power (the reciprocal of five-halves) of both sides.

Key Concepts

Constant of Proportionality

When two quantities are directly proportional, there's a special number known as the constant of proportionality. This unique value determines how much one variable will change in relation to the other. Imagine it as the secret ingredient that keeps the relationship consistent. For example, if we have the equation \(y = k \times x^{\frac{5}{2}}\), where y varies directly as the five-halves power of x, k here is the elusive constant of proportionality. To pinpoint the value of k, we can use known values of x and y from a specific instance.

As shown in the step-by-step solution, by substituting \(y = 55.3\) and \(x = 17.3\), we can calculate k. This constant is paramount because once we unravel it, we can forecast y for any value of x, and reciprocally, determine x for any given y in the power relationship described. As a tip, always ensure the equation is correctly set up with known variables before solving for k.

Power Functions

Power functions represent relationships where one variable is to the power of another, and they are ubiquitous in mathematics. When we refer to a five-halves power function, we're talking about an equation like \(y = x^{\frac{5}{2}}\), where the exponent is \(\frac{5}{2}\). This particular kind of power function signifies a direct proportionality with an additional twist—the influence of x on y is not linear but is instead amplified by the exponent.

In the context of our problem, this non-linear scaling factor means that as x grows, y increases at a rate that is the square root of x cubed, or in simpler terms, more than linearly but less than cubically. Familiarity with power functions aids in understanding the behavior of variables and in anticipating how significant changes in x will affect y. It’s also crucial in identifying how to manipulate equations of this sort to solve for unknown values—by using roots and powers inversely.

Solving Equations

Solving equations is at the heart of algebra. Equations are like puzzles that challenge us to find the missing piece—or, in mathematical terms, the unknown variable. The problem serves as a classic example: we’ve been given a complex equation with a power function and asked to solve for y and x in various scenarios. To tackle this, we use algebraic techniques such as substitution, isolating the variable, and applying inverse operations like taking roots.

Critical steps include ensuring that we substitute known values accurately and perform operations on both sides of the equation to maintain equality. When searching for the elusive x in \(y = k \times x^{\frac{5}{2}}\), remember to exercise patience—it's necessary to divide by k and then solve for x by inverting the exponent, which means taking the reciprocal power of \(\frac{5}{2}\). If you follow these processes carefully, you'll not only solve the equation at hand but also solidify a core skill that's fundamental in math and beyond.