Angle to a particle (part 2) The figure in Exercise 81 shows t...

Question

Angle to a particle (part 2) The figure in Exercise 81 shows the particle traveling away from the sensor, which may have influenced your solution (we expect you used the inverse sine function). Suppose instead that the particle approaches the sensor (see figure). How would this change the solution? Explain the differences in the two answers.

Accepted Answer

Welcome back, everyone. A street light of height H is fixed at point A on the ground. A person walks away from the street light along a straight path at a constant speed. Let B be the point where the person is currently standing, and let C be the distance between points A and B on the ground. Define theta as the angle between the street light and the line connecting the top of the street light to point B. Find the theta divided by DC. So essentially we want to express our angle theta and we can do that using the tangent function, right? Let's recall that if we take tangent of theta, this is simply the ratio between the opposite side and the adjacent side. So we are essentially taking our side C within the right triangle. Notice that we are basically treating this as a right triangle. And we're dividing C by H. Where each Is a constant. So now how do we express theta? Well, we are taking the inverse of 10. So theta is basically the inverse of 10 of C divided by H. and now we want to differentiate the theta. Divided by the C, so we're differentiating theta relative to C. This means that we're taking the derivative of the inverse of 1 of C divided by H. Now let's recall that the derivative of the inverse of 10 is 1 divided by 1 plus the independent variable squared. So in this case our variable is c divided by H. We have to square it and now because it is an inner function, not just in dependent variable, we have a ratio between C and H. We have to apply the chain rule and multiply by the derivative of C. Divided by h, so we have applied the chain rules successfully, and we have to understand that the derivative of C is 1. So this term simply becomes 1 divided by H because 1 divided by H is a constant and we're multiplying it by the derivative of c. So we can say that. The derivative of theta with respect to C is equal to 1. Divided by 1 + C divided by h2d multiplied by 1. Divided by H. And now we just want to simplify. This gives us one divided by. H multiplied by 1 + C divided by h squared. Now let's distribute the terms. So in the numerator we have one. In the the nominator we got H. Multiplied by One, so this gives usage. And then we're multiplying H by C2 divided by H2. So this gives us H C squared. Divided by H squared and then we can cancel out h on top and H squared on the bottom. Let's find the common denominator. We get one divided by. The common denominator would be H. This gives us H2 + C2 and now we're taking the reciprocal of that bottom fraction, which gives us H. Divided by H2 + C2 which essentially is our final answer to this problem. Thank you for watching.

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