A car travels down a highway at 35m/s An observer stands 200 m from the highway. (a) How fast is the distance from the observer to the car increasing when the car passes in front of the observer? (Use decimal notation.Give your answer to three decimal places.) (d)/(dx)approx square cos Enter numeric value (b) How fast is the distance increasing 50 s later? (Use decimal notation. Give your answer to three decimal places.) (d^4)/(dt)approx square

To solve this problem, we need to use the concept of related rates in calculus. We have a right triangle where the car's position on the highway is one leg, the distance from the observer to the highway is the other leg, and the hypotenuse is the distance from the observer to the car.<br /><br />Let's define:<br />- \( x(t) \): the horizontal distance from the observer to the car at time \( t \).<br />- \( h(t) \): the hypotenuse, or the distance from the observer to the car at time \( t \).<br />- The observer is 200 m away from the highway, so this is a constant vertical distance.<br /><br />Given:<br />- The car travels at a speed of \( \frac{dx}{dt} = 35 \, \text{m/s} \).<br /><br />(a) When the car passes directly in front of the observer, \( x = 0 \). At this moment, the hypotenuse \( h \) is simply the vertical distance, which is 200 m.<br /><br />Using the Pythagorean theorem:<br />\[ h^2 = x^2 + 200^2 \]<br /><br />Differentiating both sides with respect to time \( t \):<br />\[ 2h \frac{dh}{dt} = 2x \frac{dx}{dt} \]<br /><br />At the moment when the car is directly in front of the observer (\( x = 0 \)):<br />\[ h = 200 \]<br />\[ 2 \times 200 \times \frac{dh}{dt} = 2 \times 0 \times 35 \]<br />\[ 400 \frac{dh}{dt} = 0 \]<br />\[ \frac{dh}{dt} = 0 \]<br /><br />So, when the car passes directly in front of the observer, the rate at which the distance from the observer to the car is increasing is \( \boxed{0.000} \, \text{m/s} \).<br /><br />(b) 50 seconds later, the car has traveled:<br />\[ x = 35 \times 50 = 1750 \, \text{m} \]<br /><br />Now, calculate the new hypotenuse \( h \):<br />\[ h^2 = 1750^2 + 200^2 \]<br />\[ h^2 = 3062500 + 40000 \]<br />\[ h^2 = 3102500 \]<br />\[ h = \sqrt{3102500} \approx 1761.397 \, \text{m} \]<br /><br />Using the differentiated equation again:<br />\[ 2h \frac{dh}{dt} = 2x \frac{dx}{dt} \]<br />\[ 2 \times 1761.397 \times \frac{dh}{dt} = 2 \times 1750 \times 35 \]<br />\[ 3522.794 \frac{dh}{dt} = 122500 \]<br />\[ \frac{dh}{dt} = \frac{122500}{3522.794} \approx 34.769 \]<br /><br />Therefore, 50 seconds later, the rate at which the distance from the observer to the car is increasing is approximately \( \boxed{34.769} \, \text{m/s} \).