Journey to the Taco Cart INQUIRY PROBLEM Lets

Journey to the Taco Cart INQUIRY PROBLEM – Let’s “taco bout” it. Franco Prieto, Tanvir Khaira, Tarinder Kullar

Background Ben and “I” were walking on the sand when they spot a taco cart in the distance. Being the mathematical curious people they are, they decided to have a race to see who could reach the taco cart first. Ben decided to take the straight line distance to the taco cart across the sand. On the other hand, “I” figured it’s much faster to walk on the sidewalk so he decided to cut through the sand towards the sidewalk and complete his journey on the sidewalk. So… WHO MADE IT THERE FIRST?

PART ONE Who will reach the cart first?

Our Prediction We predict that Ben will reach the taco cart first because by the time “I” reaches the sidewalk, Ben will have already covered a large portion of the sand meanwhile “I” still has to cover a large distance on the sidewalk in the same amount of time compared to Ben to catch up to him, which is why we predict Ben to reach the taco cart first.

Other Factors Involved Other information that is required to make a more definitive answer include: • At least two of the following distances: • The distance from Ben and I to the sidewalk. • The distance from Ben and I to the cart as Ben and I move parallel to the cart until the are across from the cart. • The speed at which Ben and I move on the sand. • The speed at which Ben and I move on the sidewalk. • Whether or not either Ben or I make any stops or change their speed.

WE WERE GIVEN… Dimensio ns and Speed on Road: 5 ft/s Speed on Sand: 2 ft/s

What does this change? This changes our prediction drastically. Now that we have some sort of perspective to compare the situation with, it appears that either “I” and Ben could reach the cart first.

With this extra information given we can determine that it was the person “I” who reached the taco cart first. Because it took Ben 5 min 25 sec to reach the cart and “I” only took 4 min 35 sec, “I” reached the taco cart first.

PART TWO What is the exact position of the cart on the road such that both Ben and I will reach it at the exact same time?

Model We know the vertical distance up to the side walk will remain 325. 6 ft because we are only moving the cart up or down the sidewalk, not up or down the beach. X X = distance “I” walks on the sidewalk 325. 6 Ben’s distance was determined using Pythagoras and is in terms of X Knowing this, we can create an expression for time for both “I” and Ben. TBen = TI = By creating two expressions of time in terms of X, we made a system of equations.

Calculations (Algebraically) Algebraically: From the previous slide, we need to solve for x in order to determine the distances they both walk as well as the time it took for them to complete it. Since the time it took for them to walk is equal, we can create a systems of equations where TBen = TI. After plugging in the values we can solve for x. 1. Multiply both sides by 10 to get rid of the denominators. 2. Square both sides to get rid of the root. 3. Expand combine like terms on one side. 4. Lastly, solve for what values of x will reach zero using the zero product principle. X = 0 ft, 310. 1 ft The cart must be either 0 ft or 310. 1 ft from the vertex of the triangle for Ben and I to be able to reach the cart at the same time.

Graphically we use the same equation and find where Y 1 and Y 2 intersect. With distance as your x-axis and time as your y-axis. X = 0 ft , 310. 1 ft The cart must be either 0 ft or 310. 1 ft from the vertex of the triangle for Ben and I to be able to reach the cart at the same time.

c i a r b e g l A

Graphically

To find the time it will take for Ben and I to reach the taco cart you can plug in the distances ‘x’ into either time function because they both should provide equal time values. To find the distance “I” travelled you take the sum of the distances “I” travelled on the sand the distance ‘x’ we just found that “I” must walk on the sidewalk. To find the distance Ben travelled use Pythagorean theorem. I : 325. 6 ft + 310. 1 ft = 635. 7 ft The time taken by Ben to move 449. 2 ft and I to move 635. 7 ft will be 3 min 45 sec if the taco cart is moved 310. 1 ft from the right angle corner of the triangle.

PART THREE What is the optimum path for Ben and “I” to take that results in the fastest time?

Optimum path to the cart that results in the fastest time There a number of paths that can be taken to reach the taco cart by Ben and I. However, there is one path in particular that will yield the shortest time. For instance, on the picture to the right each of the coloured lines represent possible paths that Ben and I can take to reach the taco cart. The optimum path can be determined numerically, graphically or algebraically.

Numerically: Numerically we can test a series of paths, displayed in a table of values, to try and find the path that gives us the fastest route to the taco cart. By moving the vertex as shown in these two diagrams we can create new paths each time. We can then subtract the distance we moved from the vertex too find the length walked by Ben and I on the sidewalk. Next using Pythagoras we can determine the length walked on the sand because we know the distance from the vertex and we know the side length of the triangle is a constant (325. 6 ft).

dv represents the distance from the vertex in ft. dx represents the distance travelled on the sidewalk by Ben and I in ft. ds represents the distance travelled on the sand by Ben and I in ft. t represents the total time taken during that trial path in seconds. Because we know the time stops decreasing and starts increasing over the course of the third and fifth trial paths we know that the minimum amount of time taken was reached in that interval.

We can now create another subinterval to narrow down the correct path even further using the same kind of calculations. dv 100 120 140 160 180 200 dx 462. 6 442. 6 422. 6 402. 6 380. 6 362. 6 ds 358. 5 347. 0 354. 4 362. 8 372. 0 382. 1 t 261. 8 262. 0 261. 7 261. 9 262. 1 263. 3 Minimum You can continue making more subintervals to narrow down the distance from the vertex even more but we are going to end here and estimate our distance from the vertex as about 140 ft and our total time taken is about 261. 7 seconds.

Graphically: To determine the quickest path possible graphically we must first find an algebraic expression for the time taken during each path with respect to the distance from the vertex. And then find its minimum which will tell us how far from the vertex should Ben and I move toward a point on the sidewalk for which they can then after walk along the sidewalk. - c is the same as ds from the previous table which represented the distance travelled on the sand by Ben and I in ft. - d is the same as dx which represents the distance travelled on the sidewalk by Ben and I in ft. - x is the same as dv which represents the distance from the vertex in ft. - t represents the time taken in seconds just as it did before. - The expressions for the variables c and d were gotten the same way as described before. - The function t(x) is gotten simply by taking the distance travelled on sand on the sidewalk by the corresponding velocities of Ben and I on the sand sidewalk. Which are then added up to get the total time.

The distance from the vertex is found to be 142. 1 ft which gives for a total time of 261. 7 seconds.

Graphically it was determined the path in which the distance x that represents the distance from the vertex would be 142. 1 ft which would then yield a total time of 261. 7 seconds is the quickest path. Numerically is was estimated that about a distance of 140 ft would yield the fastest time which takes about 261. 7 seconds. The answer determined graphically however is more precise as these values for distance from the vertex and time are exact where as even though the estimated time is the same for the numeric method, the distance is different. However, the numeric method involved more rounding of numbers so the distance may be just slightly inaccurate.

Algebraically: To solve this problem algebraically it will require the use of calculus. By finding the derivative of our t(x) function we can then find critical values for which there might be a maximum or a minimum. In this case we are looking for a minimum. To be more specific the minimum time required to reach the taco cart. Step One is to find the derivative of the t(x) function using the Quotient Rule for differentiating

Next you want to let the derivative of your t(x) function to equal zero because when your derivative equals zero that indicates that the slope at that location x for the function t(x) is equal to zero therefore being a maximum or minimum. We can then get the time taken by plugging this value into our original t(x) function. The point from the vertex Ben and I must head toward before walking along the sidewalk to reach the taco cart is 142. 1 ft. This will take a total time of 261. 7 seconds.

Optimum Path to reach the Taco Cart the quickest 142. 1 ft 420. 5 ft Ben and I 355 325. 6 ft . 3 ft Taco Cart

PART FOUR Reflections

What did we enjoy the most? It was enjoyable because it’s always nice to apply lessons learnt at school into a real life situation. It’s a nice change of pace from textbook questions, that’s for sure. It also gave us a chance to use some of our basic math skills like using Pythagoras in conjunction with our newer math skills like differentiation. The first two parts are better suited for a Math 9 or 10 class and part 3 required calculus to do efficiently and accurately. The worst part was solving problems numerically. The calculations were very simple but the entire idea of making a table was very tedious.

What skills were developed and how can it be applied to our own benefit? Some skills and strategies this problem helped us develop was our problem interpreting skills, because we used the information provided to create our own functions to use and solve the problem with. We were able to further develop our differentiation skills and practice our optimization applications. These skills will help us in learning by practicing our ability to use different methods in finding the same answer. Not only that but optimization is an important ability to know how to use in real life. For instance: optimizing profit using the least amount of material costs for a manufacturing business. This assignment has also introduced us to the very useful Wolfram and Desmos utilities.

What did we like about Desmos & Wolfram? We found the graphing capabilities on Wolfram Alpha and Desmos to be most useful because it saves us a lot of time in not having to graph and it gives us the points we need. Similarly, we can save a lot of time because Wolfram also can be used to find the derivative of a function or check our answer.

Enjoy your Tacos!