Some bats determine their distance to an object by detecting the difference in intensity between cchoes.(a) If intensity falls off at a rate that is inversely proportional to the distance squared, show that the echo intensity is inversely proportional to the fourth power of distance. (b) The bat was originally \(0.60 \mathrm{m}\) from one object and \(1.10 \mathrm{m}\) from another. After flying closer, it is now \(0.50 \mathrm{m}\) from the first and at \(1.00 \mathrm{m}\) from the second object. What is the percentage increase in the intensity of the ccho from each object?

Let's denote intensity as \(I\) and distance as \(d\). We know that intensity falls off at a rate that is inversely proportional to \(d^2\). This means that we can write this relationship as, \(I = k_1/d^2\), where \(k_1\) is the constant of proportionality. When considering the echoes, we have two distances at play, the distance from the bat to the object and the distance back from the object to the bat. Essentially, the echo intensity is affected by the distance twice. So, we model this echo intensity relationship as: \(EI = k_2(\frac{1}{d^2})^2\) \(EI = k_2(\frac{1}{d^4})\) Comparing with the intensity relationship, we can see that the echo intensity, \(EI\), is inversely proportional to the fourth power of distance, \(d^4\). This proves part (a).

We need to calculate the percentage increase in the echo intensity for both objects when the bat moves closer. Let's denote the initial distances for object 1 and object 2 as \(d_{1i}\) and \(d_{2i}\). Similarly, the new distances after the bat moves closer will be \(d_{1f}\) and \(d_{2f}\). We are given these values as: \(d_{1i} = 0.60\,\text{m},\enspace d_{2i} = 1.10\,\text{m}\) and \(d_{1f} = 0.50\,\text{m},\enspace d_{2f} = 1.00\,\text{m}\) We know that echo intensity is inversely proportional to \(d^4\), so we can write the echo intensities for both objects initially and after the bat moves closer as: \(EI_{1i} = k_3/d_{1i}^4,\enspace EI_{2i} = k_3/d_{2i}^4\) and \(EI_{1f} = k_3/d_{1f}^4,\enspace EI_{2f} = k_3/d_{2f}^4\) We want to find the percentage increase in intensity for both objects. The formula for percentage increase is: \(\text{Percentage Increase} = \frac{\text{Final Value - Initial Value}}{\text{Initial Value}} \times 100\%\) For object 1: \(\text{Percentage Increase}_{1} = \frac{EI_{1f} - EI_{1i}}{EI_{1i}} \times 100\% = \frac{k_3((1/d_{1f}^4)-(1/d_{1i}^4))}{k_3(1/d_{1i}^4)} \times 100\% = \frac{(1/d_{1f}^4)-(1/d_{1i}^4)}{(1/d_{1i}^4)} \times 100\% \) Insert the values of the distances: \(\text{Percentage Increase}_{1} = \frac{(1/0.5^4)-(1/0.6^4)}{(1/0.6^4)} \times 100\% \approx 166.29\%\) Similarly, for object 2: \(\text{Percentage Increase}_{2} = \frac{EI_{2f} - EI_{2i}}{EI_{2i}} \times 100\% = \frac{k_3((1/d_{2f}^4)-(1/d_{2i}^4))}{k_3(1/d_{2i}^4)} \times 100\% = \frac{(1/d_{2f}^4)-(1/d_{2i}^4)}{(1/d_{2i}^4)} \times 100\% \) Insert the values of the distances: \(\text{Percentage Increase}_{2} = \frac{(1/1^4)-(1/1.1^4)}{(1/1.1^4)} \times 100\% \approx 47.73\%\) Thus, the percentage increase in intensity for object 1 is approximately 166.29% and for object 2 is approximately 47.73%.