Find the approximate values of the following using differentials
√(25.01)
3√(125.1)
√(36.01)
To find the approximate values using differentials, we can use the formula:
dy ≈ f'(x) * dx
1) √(25.01):
Let f(x) = √x
f'(x) = 1 / (2√x)
Let x = 25 and dx = 0.01.
dy ≈ f'(x) * dx
dy ≈ (1 / (2√25)) * 0.01
dy ≈ (1/10) * 0.01
dy ≈ 0.001
Approximate value: √(25.01) ≈ 5 + 0.001 = 5.001
2) 3√(125.1):
Let f(x) = 3√x
f'(x) = 1 / (3x^(2/3))
Let x = 125 and dx = 0.1.
dy ≈ f'(x) * dx
dy ≈ (1 / (3 * (125^(2/3)))) * 0.1
dy ≈ (1 / (3 * (125^(2/3)))) * 0.1
dy ≈ 0.004
Approximate value: 3√(125.1) ≈ 5 + 0.004 = 5.004
3) √(36.01):
Let f(x) = √x
f'(x) = 1 / (2√x)
Let x = 36 and dx = 0.01.
dy ≈ f'(x) * dx
dy ≈ (1 / (2√36)) * 0.01
dy ≈ (1/12) * 0.01
dy ≈ 0.000833
Approximate value: √(36.01) ≈ 6 + 0.000833 = 6.000833
To find the approximate values of these expressions using differentials, we can use the formula:
Δy ≈ dy ≈ f'(x)⋅Δx
Where Δy is the change in y, dy is the differential of y, f'(x) is the derivative of the function, and Δx is the change in x.
Let's calculate the approximate values step-by-step.
1. Approximating √(25.01):
Let's start by finding the derivative of √x with respect to x:
f(x) = √x
f'(x) = 1 / (2√x)
Now, substitute x = 25.01 into the derivative:
f'(25.01) = 1 / (2√25.01) ≈ 1 / (2 * 5) = 1/10 = 0.1
Assuming Δx is small, let's take Δx = 0.01. Now, we can find dy:
dy ≈ f'(x)⋅Δx ≈ 0.1 * 0.01 = 0.001
Therefore, the approximate value of √(25.01) is:
√(25.01) ≈ √25 + 0.001 ≈ 5 + 0.001 ≈ 5.001
2. Approximating 3√(125.1):
Let's find the derivative of 3√x with respect to x:
f(x) = 3√x
f'(x) = 3 / (3√x^2)
Now, substitute x = 125.1 into the derivative:
f'(125.1) = 3 / (3√(125.1)^2) ≈ 3 / (3 * 11.18) ≈ 3 / 33.54 ≈ 0.089
Assuming Δx is small, let's take Δx = 0.1. Now, we can find dy:
dy ≈ f'(x)⋅Δx ≈ 0.089 * 0.1 = 0.0089
Therefore, the approximate value of 3√(125.1) is:
3√(125.1) ≈ 3√125 + 0.0089 ≈ 5 + 0.0089 ≈ 5.0089
3. Approximating √(36.01):
Let's find the derivative of √x with respect to x:
f(x) = √x
f'(x) = 1 / (2√x)
Now, substitute x = 36.01 into the derivative:
f'(36.01) = 1 / (2√36.01) ≈ 1 / (2 * 6) = 1/12 ≈ 0.083
Assuming Δx is small, let's take Δx = 0.01. Now, we can find dy:
dy ≈ f'(x)⋅Δx ≈ 0.083 * 0.01 = 0.00083
Therefore, the approximate value of √(36.01) is:
√(36.01) ≈ √36 + 0.00083 ≈ 6 + 0.00083 ≈ 6.00083