Two buildings of height 20 feet and 40 feet, respectively,...

Two buildings of height 20 feet and 40 feet, respectively, are 60 feet apart. Suppose that the intensity of light at a point between the buildings is proportional to the angle $\theta$ in the figure. (a) If a person is moving from right to left at $4 \mathrm{ft} / \mathrm{s},$ at what rate is $\theta$ changing when the person is exactly halfway between the two buildings? (b) Find the location at which the angle $\theta$ is maximum.

Two buildings of height 20 feet and 40 feet, respectively, are 60 feet apart. Suppose that the intensity of light at a point between the buildings is proportional to the angle $\theta$ in the figure.
(a) If a person is moving from right to left at $4 \mathrm{ft} / \mathrm{s},$ at what rate is $\theta$ changing when the person is exactly halfway between the two buildings?
(b) Find the location at which the angle $\theta$ is maximum.

Key Concepts

Related Rates

Related rates problems involve determining the rate at which one quantity changes with respect to time given the rate of change of another related quantity. This typically requires forming an equation that connects the various quantities, differentiating that equation with respect to time, and then solving for the unknown rate.

Implicit Differentiation

Implicit differentiation is used when differentiating equations in which the variables are interdependent and not easily separable. This technique is essential in related rates problems where multiple quantities are related, allowing one to differentiate each side of the equation with respect to time without having to solve for one variable in terms of the other.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as the arctan function, help express an angle in terms of a ratio of sides in a right triangle. They play a crucial role in problems where geometric angles must be related directly to algebraic expressions involving other quantities, thus enabling the link between spatial configurations and their corresponding angles.

Optimization in Calculus

Optimization in calculus involves finding the maximum or minimum values of a function under given constraints. This is typically achieved by taking the derivative of the function with respect to the variable of interest, setting it equal to zero, and solving the resulting equation to find potential extrema, which are then evaluated to determine the optimal value.

Trigonometric Relationships in Geometry

Trigonometric relationships are foundational in connecting geometric figures with algebraic expressions. By using functions like sine, cosine, and tangent, one can relate angles to side lengths, thereby translating physical, spatial problems into mathematical equations that are more easily analyzed and solved using calculus.

Transcript

00:01 And this problem, we are dealing with an application of differentiation, which is related rates.

00:07 And what this is essentially doing is we're taking differentiation one step further.

00:12 And now we can apply it to real world problems or applied problems, which is what the question that we're given is.

00:20 So first, what i did is let's draw a picture of what's going on.

00:25 I think with related rates problems, that's the first thing that you should always do.

00:28 So essentially what we have is we have these two heights here, 140 and 120.

00:35 We know they're 60 feet apart.

00:38 And we have this beam of light that's being essentially shine from each side and it meets at an angle theta.

00:48 And the first thing that we want to know is the rate of which theta is moving.

00:53 So to do that is we're going to need some trigonometric relationships from this.

00:59 This picture.

01:00 What you can see is we have two triangles here, which is going to make our derivative much easier.

01:07 So the first thing that we can do is we can say theta equals pi minus the inverse tangent of 40 over 60 minus x minus the tangent inverse of 20 over x.

01:19 This is simply coming from the triangle relationships of these two triangles.

01:24 So now what we can do is we can derive that.

01:29 So d theta dx is going to be equal to negative 40x times 1 over 60 minus x squared, divided by 1 plus 40 over 60 minus x squared, plus 20 over x squared over 1 plus 20 over x squared.

01:46 Now in this derivative, you probably notice we had to use some differentiation rules because we had the quotients inside of trigonometric.

01:55 Functions...

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