Respuesta :
Using the Newton method, the next estimate of the height after one iteration is: h = 0.1474m.
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The Newton's method to find a solution [tex]x[/tex] of a function [tex]f(x)[/tex] is given by:
[tex]x_{n+1} = x_n - \frac{f(x_n)}{f^{\prime}(x_n)}[/tex]
In which n+1 is the next iteration.
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In the context of this problem, the height is:
[tex]h_{n+1} = h_n - \frac{V(h_n)}{V^{\prime}(h_n)}[/tex]
Thus, for the first iteration:
[tex]h_1 = h_0 - \frac{V(h_0)}{V^{\prime}(h_0)}[/tex]
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The function, and its derivative, are:
[tex]V(h) = \frac{3r \pi h^2 - \pi h^3}{3}[/tex]
The radius is [tex]r = 3[/tex], thus:
[tex]V(h) = \frac{9\pi h^2 - \pi h^3}{3}[/tex]
[tex]V^{\prime}(h) = \frac{18\pi h - 3\pi h^2}{3}[/tex]
Simplifying by 3:
[tex]V^{\prime}(h) = \pi(6h - h^2)[/tex]
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The initial estimate is [tex]h_0 = 0.3[/tex], thus:
[tex]V(0.3) = \frac{9\pi (0.3)^2 - \pi (0.3)^3}{3} = 0.82[/tex]
[tex]V^{\prime}(0.3) = \pi[6(0.3) - (0.3)^2] = 5.3721[/tex]
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The estimate, after 1 iteration, is:
[tex]h_1 = h_0 - \frac{V(h_0)}{V^{\prime}(h_0)}[/tex]
[tex]h_1 = 0.3 - \frac{0.82}{5.3721}[/tex]
[tex]h_1 = 0.1474[/tex]
The estimate of the height after one iteration is of h = 0.1474m.
A similar problem is given at https://brainly.com/question/17031314
