# Thread: determine approximate percentage error using binomial expansion

1. ## determine approximate percentage error using binomial expansion

The shear stress τ in a shaft of diameter
D under a torque T is given by:
τ = kT
πD^3 .
Determine the approximate percentage error
in calculating τ if T is measured 3% too
small and D 1.5% too large.

this is a question from J bird engineering mathematics book. it gives the answer but i need to know how to do the method.

any help much appreciated

2. ## Approximate increases using binomial expansion

Hello decoy808
Originally Posted by decoy808
The shear stress τ in a shaft of diameter
D under a torque T is given by:
τ = kT
πD^3 .
Determine the approximate percentage error
in calculating τ if T is measured 3% too
small and D 1.5% too large.

this is a question from J bird engineering mathematics book. it gives the answer but i need to know how to do the method.

any help much appreciated
If $\displaystyle y = kx^n$, and $\displaystyle x$ and $\displaystyle y$ increase by small amounts $\displaystyle \delta x$ and $\displaystyle \delta y$, then

$\displaystyle y+ \delta y = k(x+\delta x)^n$

$\displaystyle \Rightarrow \frac{y+ \delta y}{y }= \frac{k(x+ \delta x)^n}{kx^n}$

$\displaystyle \Rightarrow 1 + \frac{\delta y}{y}= \left(1 + \frac{\delta x}{x}\right)^n$

$\displaystyle \approx 1 + n\frac{\delta x}{x}$, ignoring higher powers of $\displaystyle \frac{\delta x}{x}$, which we may do for small increases in $\displaystyle x$

$\displaystyle \Rightarrow$ the fractional increase in $\displaystyle y \approx n \times$ the fractional increase in $\displaystyle x$.

It is very easy show that this may be extended to a formula for $\displaystyle y$ which involves powers of any number of variables.

So, using the same method you may prove that if $\displaystyle \tau = \frac{k}{\pi}T\cdot D^{-3}$, then

$\displaystyle \frac{\delta \tau}{\tau} \approx \frac{\delta T}{T} -3 \frac{\delta D}{D}$

With the percentage errors you have been given

$\displaystyle \frac{\delta T}{T} = -0.03$ and $\displaystyle \frac{\delta D}{D} = 0.015$

So $\displaystyle \frac{\delta \tau}{\tau} \approx -0.03 -0.045 = -0.075 = -7.5$%

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