use differentials to approximate the value of the squareroot of 4.3
Let
f(x)=sqrt(x)=x^(1/2)
f'(x)=1/(2sqrt(x))
and
f(x0+Δx)=f(x0)+f'(x0)*Δx (approx.)
or
Δx=(f(x0+Δx)-f(x0))/f'(x0) approx.
Knowing that 2^2=4, and 2.1^2=4.41
Try x0=2
Δx=(4.3-2^2)/(1/(2*2)
=0.3/(4)
=0.075
or
Approximately, x+Δx=2+.075=2.075
(check: 2.075^2=4.305625)
Try x0=2.1, x0^2=4.41...
to get a still better approximation.
To approximate the value of the square root of 4.3 using differentials, we can start by considering a small change (d) in the value of 4.3 and finding the corresponding change in the square root.
Let's define the function f(x) as the square root of x. In this case, we want to evaluate f(4.3).
We can find the derivative of f(x) using calculus, which is given by:
f'(x) = 1 / (2 * sqrt(x))
Now, let's calculate the differential of f(x) using the formula:
df = f'(x) * dx
Since we are interested in the change of f(x) at x=4.3, we can rewrite the equation as:
df = f'(4.3) * dx
Substituting x = 4.3 and dx = (4.3 - 4), we get:
df = f'(4.3) * (4.3 - 4)
Now, let's evaluate f'(4.3) by substituting x = 4.3 into the derivative function:
f'(4.3) = 1 / (2 * sqrt(4.3))
df = (1 / (2 * sqrt(4.3))) * (4.3 - 4)
Simplifying this equation will give us the approximate change in the square root of 4.3 for a small change in the value of 4.3.
After finding df, we can estimate the value of the square root of 4.3 by adding the change to the initial value. The approximation would be:
sqrt(4.3) ≈ sqrt(4) + df
Note: While differentials provide a good approximation for small changes, it might not be as accurate as other methods like using calculators or numerical methods.