Question

Write an expression to evaluate the relative error in g (given that g= 2h/t^2) in terms of h,Δh,t,Δt

Answer 1

An expression which can be used to evaluate the relative error in g (in terms of h, Δh, t, and Δt) is: $\delta = \frac{t^2(h\;+\Delta \;h) }{h(t\;+\;\Delta t)^2} - 1$

What is relative error?

Relative error can be defined as a measure of the ratio of an absolute (real) value of a measurement to an expected (theoretical) value. Also, it's independent of the magnitude of its values.

How to evaluate the relative error in g?

In order to write this expression, we would divide the absolute (real) value by the expected (theoretical) value as follows:

$\delta = \frac{g(h\;+\;\Delta h,t \;+ \;\Delta t) - \;g(h,t)}{g(h,t)} \\\\\delta = \frac{g(h\;+\;\Delta h,t \;+ \;\Delta t) }{g(h,t)} - 1\\\\\delta = \frac{t^2(h\;+\Delta \;h) }{h(t\;+\;\Delta t)^2} - 1$

Note: g = 2h/t²

Read more on relative error here: brainly.com/question/13370015

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