Problem 39 - Calculus of a Single Variable

Question

Approximating Function Values In Exercises \(37-40,\) use differentials to approximate the value of the expression. Compare your answer with that of a calculator. $$ \sqrt[4]{624} $$

Step-by-Step Solution

Answer

The differential approximation of \(\sqrt[4]{624}\) is approximately 4.99 which is very close to the actual calculator value of approximately 4.987.

Step 1: Convert the expression into an easier workable format

The given function can be expressed in terms of exponents as 624^(1/4). This will serve as the base function for finding the differentials.

Step 2: Calculate the linear approximation

The formula for linear approximation is \(L(x) = f(a) + f'(a) *(x - a)\). For this exercise, a close exact square is 625 and it is very close to 624. So, choose 625 for 'a' such that the base function becomes f(x) = x^(1/4) and 'a' becomes 625.

Step 3: Calculate f(a) and f'(a)

For a = 625 and f(x) = x^(1/4), calculate f(a) and f'(a). f(a) = 625^(1/4) = 5 and f'(x) = 1/4*x^(-3/4) so f'(a) = 1/4*625^(-3/4) = 1/100.

Step 4: Substitute x, a, f(a) and f'(a) into the linear approximation formula

Substitute x = 624, a = 625, f(a) = 5, f'(a) = 1/100 into the linear approximation formula \(L(x) = f(a) + f'(a) *(x - a)\) and find L(624) = L(x) = 5 - 1/100 * (625 - 624) = 5 - 1/100 = 4.99.

Step 5: Compare the result to that of a calculator

Calculate \(\sqrt[4]{624}\) using a calculator and it returns approximately 4.987... which is close to the linear approximation value 4.99

Key Concepts

Linear Approximation

When it comes to understanding how functions behave, linear approximation is an incredibly valuable tool. It allows us to estimate the value of a function at a point close to a known value without complex calculations. The core idea behind it is to use the tangent line at a known point on the function, which serves as a good approximation for the function nearby.

For instance, if we want to approximate the expression \( \sqrt[4]{624} \) without a calculator, linear approximation comes to the rescue. The process begins by identifying a value close to 624 that is convenient for computation, like 625, because we can easily compute \( \sqrt[4]{625} \) by hand. We then construct the function \( f(x) = x^{1/4} \) and find its value and derivative at the point 625. By applying the linear approximation formula \( L(x) = f(a) + f'(a) \cdot (x - a) \) where \( a \) is the known value, we obtain an approximated result. This method yields \( 4.99 \) for \( \sqrt[4]{624} \) which is impressively close to the actual calculator value.

Differentials in Calculus

Differentials in calculus play an essential role in approximation methods, including linear approximation. A differential, denoted as \( dy \) or \( df \) for function \( f \) is an infinitesimally small change in the function's value corresponding to a small change \( dx \) in its input value. It allows us to model how tiny variations in input affect the output, forming the basis for understanding rates of change.

In the context of the approximation problem \( \sqrt[4]{624} \) we use the differential to compute the change in \( f(x) \) as \( x \) changes from 625 to 624. To calculate this small change, we find \( f'(x) \) or \( df/dx \) when \( x = 625 \) and apply it to the change in \( x \) (which is \( -1 \) in our case). The product of this and \( dx \) gives us the differential, which we use to adjust \( f(625) \) to approximate \( f(624) \) effectively.

Function Exponentiation

Function exponentiation is the process of raising a function to a power, which plays a pivotal role in both theoretical and applied mathematics. For instance, the function \( f(x) = x^{1/4} \) involves the fourth root, which is the same as raising \( x \) to the power of \( 1/4 \).

The exponent \( 1/4 \) signifies that we are looking for the fourth root, a concept frequently used in various domains, such as geometry and physics. When we differentiate this function as seen in the exercise, \( f'(x) \), we apply the power rule of differentiation. This rule tells us how to deal effectively with exponentiation when finding the derivative. In the context of approximating \( \sqrt[4]{624} \) knowing how to work with function exponentiation and its derivatives enables us to estimate values with precision and facilitates easier computation without the necessity of a calculator.