Math Problem Statement

Use differentials to approximate the value of the cube root of 63. Round your answer to four decimal places, if necessary.

Solution

To approximate the value of 633\sqrt[3]{63} using differentials, we follow these steps:

Step 1: Define the function

Let f(x)=x3=x1/3f(x) = \sqrt[3]{x} = x^{1/3}, so we want to approximate f(63)f(63).

Step 2: Choose a point close to 63

We'll choose x0=64x_0 = 64 because 643=4\sqrt[3]{64} = 4, which is an easy value to compute and close to 63.

Step 3: Find the derivative

We need the derivative of f(x)=x1/3f(x) = x^{1/3}. Using the power rule: f(x)=13x2/3=13x23f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3\sqrt[3]{x^2}} At x0=64x_0 = 64, this becomes: f(64)=136423=13(64)2/3=13×16=148f'(64) = \frac{1}{3\sqrt[3]{64^2}} = \frac{1}{3(64)^{2/3}} = \frac{1}{3 \times 16} = \frac{1}{48}

Step 4: Use the linear approximation formula

The differential approximation formula is: f(x)f(x0)+f(x0)(xx0)f(x) \approx f(x_0) + f'(x_0)(x - x_0) Substituting x0=64x_0 = 64, f(64)=4f(64) = 4, and f(64)=148f'(64) = \frac{1}{48}, we get: f(63)4+148(6364)f(63) \approx 4 + \frac{1}{48}(63 - 64) f(63)4+148(1)f(63) \approx 4 + \frac{1}{48}(-1) f(63)4148=40.0208=3.9792f(63) \approx 4 - \frac{1}{48} = 4 - 0.0208 = 3.9792

Conclusion:

Using differentials, the approximate value of 633\sqrt[3]{63} is 3.97923.9792.

Would you like further details or have any questions?

Related Questions:

  1. What is the purpose of differentials in calculus approximations?
  2. How can we apply differentials to approximate values for higher powers, such as x4\sqrt[4]{x}?
  3. Why is x0=64x_0 = 64 a better choice than other numbers close to 63 for this problem?
  4. What happens if we try to approximate 623\sqrt[3]{62} using differentials?
  5. How does the error in a differential approximation compare to other approximation methods?

Tip:

When using differentials for approximation, always choose a point x0x_0 near the target value where the function is easy to evaluate exactly.

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Math Problem Analysis

Mathematical Concepts

Differentials
Approximations
Derivatives

Formulas

f(x) = x^{1/3}
f'(x) = (1/3) * x^{-2/3}
f(x) ≈ f(x_0) + f'(x_0) * (x - x_0)

Theorems

Linear Approximation Theorem

Suitable Grade Level

Grades 11-12 (or early university calculus)