A man of height 1.7 meters walks away from a 5-meter lamppost at a speed of 1.2m/s Find the rate at which his shadow is increasing in length. (Use decimal notation. Give your answer to three decimal places.) be completel

To solve this problem, we can use related rates in calculus. Let's denote the following variables:<br /><br />- \( h = 1.7 \) meters (height of the man)<br />- \( H = 5 \) meters (height of the lamppost)<br />- \( x \) meters (distance of the man from the lamppost)<br />- \( s \) meters (length of the shadow)<br /><br />We need to find the rate at which the length of the shadow is increasing, denoted as \( \frac{ds}{dt} \).<br /><br />First, let's set up the relationship between the distances and heights using similar triangles. The triangle formed by the lamppost and its shadow is similar to the triangle formed by the man and his shadow.<br /><br />Using similar triangles, we have:<br />\[<br />\frac{H}{x + s} = \frac{h}{s}<br />\]<br /><br />Substituting the given values for \( H \) and \( h \):<br />\[<br />\frac{5}{x + s} = \frac{1.7}{s}<br />\]<br /><br />Cross-multiplying to solve for \( s \):<br />\[<br />5s = 1.7(x + s)<br />\]<br /><br />Expanding and simplifying:<br />\[<br />5s = 1.7x + 1.7s<br />\]<br />\[<br />5s - 1.7s = 1.7x<br />\]<br />\[<br />3.3s = 1.7x<br />\]<br />\[<br />s = \frac{1.7x}{3.3}<br />\]<br /><br />Now, differentiate both sides with respect to time \( t \):<br />\[<br />\frac{ds}{dt} = \frac{1.7}{3.3} \cdot \frac{dx}{dt}<br />\]<br /><br />Given that the man walks away from the lamppost at a speed of \( \frac{dx}{dt} = 1.2 \) m/s, we substitute this value into the equation:<br />\[<br />\frac{ds}{dt} = \frac{1.7}{3.3} \cdot 1.2<br />\]<br /><br />Calculating the numerical value:<br />\[<br />\frac{ds}{dt} = \frac{1.7 \times 1.2}{3.3}<br />\]<br />\[<br />\frac{ds}{dt} = \frac{2.04}{3.3}<br />\]<br />\[<br />\frac{ds}{dt} \approx 0.618 \text{ m/s}<br />\]<br /><br />Therefore, the rate at which the man's shadow is increasing in length is approximately \( 0.618 \) meters per second.