Use differentials to approximate the cube root of 66.

Answer

To approximate the cube root of 66 using differentials, we can follow these steps:

### Step 1: Identify a nearby perfect cube
We need to find a perfect cube that is close to 66. The perfect cube near 66 is 64, which is \(4^3\).

### Step 2: Define the function and the error
Let \(y = \sqrt[3]{x}\). We can express the input \(x\) as \(x = 64 + \Delta x\), where \(\Delta x = 66 - 64 = 2\).

### Step 3: Calculate \(y\) at the nearby perfect cube
Now, we calculate \(y\) at \(x = 64\):
\[ y = \sqrt[3]{64} = 4 \]

### Step 4: Find the derivative \(dy/dx\)
Next, we need to find the derivative of \(y\) with respect to \(x\):
\[ y = x^{1/3} \]
Using the power rule, we find:
\[ \frac{dy}{dx} = \frac{1}{3} x^{-2/3} = \frac{1}{3 \sqrt[3]{x^2}} \]

### Step 5: Evaluate the derivative at \(x = 64\)
Now, we evaluate \(\frac{dy}{dx}\) at \(x = 64\):
\[ \frac{dy}{dx} \bigg|_{x=64} = \frac{1}{3 \cdot (64)^{2/3}} = \frac{1}{3 \cdot 16} = \frac{1}{48} \]

### Step 6: Calculate \(\Delta y\)
Using the formula \(\Delta y = \frac{dy}{dx} \cdot \Delta x\), we can find \(\Delta y\):
\[ \Delta y = \frac{1}{48} \cdot 2 = \frac{2}{48} = \frac{1}{24} \]

### Step 7: Approximate the cube root of 66
Finally, we can approximate the cube root of 66:
\[ \sqrt[3]{66} \approx y + \Delta y = 4 + \frac{1}{24} \]
Calculating this gives:
\[ 4 + \frac{1}{24} = 4 + 0.04167 \approx 4.04167 \]

### Final Answer
Thus, the approximate value of the cube root of 66 is:
\[ \sqrt[3]{66} \approx 4.04167 \]