Use differentials (or equivalently, a linear approximation) to approximate ln(1.02) as follows: Let f(x)=ln(x) and find the equation of the tangent f(x) line to at a "nice" point near 1.02 . Then use this to approximate ln(1.02) .
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To approximate ln(1.02) using differentials, we first define the function f(x) = ln(x). Now, we need to find the equation of the tangent line to f(x) at a "nice" point near 1.02.
To do this, let's calculate the derivative of f(x) with respect to x. The derivative of ln(x) is 1/x.
Next, we need to find the slope of the tangent line, which is the value of the derivative at the point (1.02, ln(1.02)). So, substitute x = 1.02 into the derivative: f'(1.02) = 1/1.02.
The equation of the tangent line is given by: y - ln(1.02) = (1/1.02)(x - 1.02).
Now, we will use this equation to approximate ln(1.02). We substitute x = 1.02 into the equation to find the corresponding y-value, which will be our approximation.
ln(1.02) ≈ ln(1.02) - ln(1.02) = (1/1.02)(1.02 - 1.02) ≈ 0.
Therefore, the approximation of ln(1.02) using differentials is approximately 0.
To approximate ln(1.02) using differentials or linear approximation, we start by defining the function f(x) = ln(x).
Now, let's find the equation of the tangent line to the graph of f(x) at a "nice" point near 1.02. To do this, we need to find the slope of the tangent line.
The slope of the tangent line can be found using the derivative of f(x). In this case, f'(x) = 1/x since the derivative of ln(x) is 1/x.
Now, let's find the slope of the tangent line at x = 1.02:
f'(x) = 1/x
f'(1.02) = 1/1.02
So, the slope of the tangent line at x = 1.02 is approximately 1/1.02.
To find the equation of the tangent line, we also need a point that lies on the line. We'll use the point (1, ln(1)) since it's a "nice" point that lies on the graph of f(x).
Using the point-slope form of a linear equation, the equation of the tangent line is:
y - ln(1) = (1/1.02)(x - 1)
Simplifying the equation, we get:
y = ln(1) + (1/1.02)(x - 1)
Since ln(1) = 0, the equation becomes:
y = (1/1.02)(x - 1)
Now, let's use this tangent line equation to approximate ln(1.02).
Substituting x = 1.02 into the equation, we have:
y ≈ (1/1.02)(1.02 - 1)
Simplifying, we find:
y ≈ (1/1.02)(0.02)
Calculating this expression, we find:
y ≈ 0.0196
Therefore, ln(1.02) is approximately equal to 0.0196, using differentials or linear approximation.