What are special triangles? How do they apply to the unit circle?
Trig functions take an angle as an "argument" and are equal to the ratio of one side of the triangle to another side of the triangle, this ratio is dependent on the function being used. The unit circle is a circle with radius one. When talking about the unit circle, cosine represents x and sine represents y, and we know that the radius is one. Special triangles that have a hypotenuse of one, meaning they fit on the unit circle (radius = 1). If we use the Pythagorean Theorem or the ratio of the sides, i.e., trig functions, we can find the sides of these triangles. This ratio of x, y, and the hypotenuse gives the value of the trig functions with the interior angle of a special triangle.
What does it mean to measure the area under a curve?
The area under the curve is defined as an integral in calculus. Given a curve, f(x), we can start to look at the area but breaking the domain into small parts and summing the area of the blocks created by breaking up the domain (I would draw a picture and start by using bigger blocks where there is clearly an error in the estimation). As you can see in the picture, there is area under-accounted and over-accounted, so when we add up the area of the blocks our estimation will be inaccurate. What if we drew smaller blocks? I would then draw blocks approx. half the size of the original. From the picture we can tell that the estimation will be closer. I would then repeat this process again showing that as we make the blocks smaller and smaller the estimation will get closer and closer to the actual value of the area under the curve. What if we made these blocks smaller and smaller? Our estimation would be getting closer to the area under the curve (this should be visible in the picture because more area under the curve would be accounted for and there would be less over-estimation). This is the definition of an integral, the sum of infinitely small sections that give us a precise estimation of the area of under the curve. The sum of these blocks are called Reimann Sums. The pictures I would draw would be similar to the picture found at this link: created.https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Riemann_sum_convergence.png/1200px-Riemann_sum_convergence.png
At 9am car A began a journey from a point traveling at 40 mpg. At 10am another car, car B, started traveling from the same point as car A traveling in the same direction at 60 mph. At what time will car B pass car A?
First, I would have the student go through the problem and pull out all relevant information, assign variables where needed, and determine what the equation is asking. I would assign ra =40 mph, rb =60 mph as given. Next, we would assign the time that car A left (ta) to be t and I would explain that since car B is leaving one hour later, we have to assign tb = (t-1) since it is the time relative to car A. I would then start explaining the problem they are being asked by drawing a position-time graph displaying both cars and their point of intersection. I would emphasize that the slope of car B is steeper than car A since it is covering more distance in a shorter amount of time. As shown in the picture, we can then see that we are looking for the time that the cars have travelled the same distance. I would then introduce the equation distance = rate x time and explain to the student using the picture that the cars have traveled the same distance on the point of intersection on the graph. I would then relate the graph to the equation, since we are looking for the distance that car A and car B are the same we can set these values equal to each other, thus giving us the equation rata=rbtb. We can then plug in the values and solve for the time. We then must answer the question being asked. The time (t) found in the equation is the time it takes for the cars to travel the same distance, we must plug this value back in to find the how long it takes for the cars to meet.