The centripetal acceleration of a particle moving in a circle is a = v2r , where v is the velocity and r is the radius of the circle. Approximate the maximum percent error in measuring the acceleration due to errors of 1% in v and 5% in r.
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Use differential (propagation of relative errors): da/a = 2(dv/v) − (dr/r). Show more…
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The centripetal acceleration of a particle moving in a circle is $a=v^{2} / r$, where $v$ is the velocity and $r$ is the radius of the circle. Approximate the maximum percent error in measuring the acceleration due to errors of $3 \%$ in $v$ and $2 \%$ in $r$.
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The centripetal acceleration of a particle moving in a circle is $a=v^{2} / r,$ where $v$ is the velocity and $r$ is the radius of the circle. Approximate the maximum percent error in measuring the acceleration due to errors of $3 \%$ in $v$ and $2 \%$ in $r$.
The centripetal acceleration of a particle moving in a circle is given by $a(r, v)=\frac{v^{2}}{r},$ where $v$ is the velocity and $r$ is the radius of the circle. Approximate the maximum percent error in measuring the acceleration resulting from errors of 3% in v and 2% in r. (Recall that the percentage error is the ratio of the amount of error over the original amount. So, in this case, the percentage error in $a$ is given by $\frac{d a}{a} . )$
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