The grade 12 class picnic
The grade 12 class decided that they should go together for a picnic. Wendy, the class rep, had made this decision by talking to a few of her friends. This may be the last chance for them to get together like this because after the school year, everyone might go to different directions – colleges, universities, work or internships – who knew. A notice was placed on the school information board were anyone could put a notice. The site and the date for the picnic were also given. The picnic would be on a weekend as it was not a picnic arranged by the school authorities. Wendy and her friends thought that the class students were mature enough to make their own decisions. All the expenses would have to be borne by students. They would also have to be responsible for their own rides. Wendy also asked the students to contact her if they wanted to go.
There were 150 students in the class but only 25 wanted to go. Others thought that it was too close to the final exams or that they had other responsibilities. Johnny had asked her mom to borrow her car. He also offered his girfriend Sara a ride. He would also take two other friends with him. At the picnic, there was a potluck lunch. Everyone had carried a small item or two for many people to share. They also had lots of fun and games after which there was the free time. It was decided that everyone could go around and come back in an hour. After coming back, they could have some pop or water before saying bye to everyone.
Stroll along the lake near picnic grounds
Johnny and Sara took a stroll along the lake next to the picnic grounds. It was a huge lake with a stony beach. There was a small path around it for walking. It was protected by large trees that provided shade. The love birds were walking holding hands which were swinging back and forth together. While they were enjoying, they saw a large information board which read:
“CAUTION: Please, do not picnic very close to the lake. Water can rise suddenly. In 2007, there was a sudden rain lasting several hours during which the level of water in the lake rose fast at the rate of 1 centimeter per minute. Aerial photographs showed that the area of the lake also increased at the rate of 1 square meter per minute. The water reached close to the path. Many people had to clear fast because they were sitting and enjoying too close to the lake.”
Johnny and Sara decided to walk back to the picnic grounds. They said bye to everyone and went back.
Johnny’s conundrum
At home, Johnny was sitting in a deep thought.
Sara: What’s the matter Johnny? You look lost.
Johnny: I am still thinking about what we read in the notice at the lake. I am trying to figure out how much water had been adding to the lake at the peak of the rainfall.
Sara: Wasn’t the rise rate of 1 centimeter per minute and increase in the lake area at at the rate of 1 square meter per minute ?
Johnny: Yes, but that does not tell me how much water. Should I just multiply 1 centimeter per minute by 10,000 square centimeters (1 square meter) and say it was 10,000 cubic centimeters per minute ?
Sara: That’s not the proper way. I think you have to use the product rule.
Johnny: What’s that?
Product formula for differentiation
Sara: These are both functions of time. Let’s say the rise in the water level is f1(t) and increase in the area is f2(t). The rate of change in f1(t) will be f ‘1(t) and the change in f2(t) will be f ‘2(t). If the change in the amount of water is fw(t), then according to the product formula:
f ‘w(t) = f1(t) . f ‘2(t) + f2(t) . f ‘1(t).
Johnny: Why bother with this complicated stuff? What’s wrong with the way I did it ?
Sara: There is a lot of information which you might miss. Let’s say, that when the rain pour started, the lake had a smaller area and its water level rose by 1 centimeter per minute and also later on during that period the rate of rise was still 1 centimeter per minute when the lake had become bigger. Then the f ‘w(t) would be much greater when the lake was wider than in the beginning. This product formula forces you to think of such a possibility.
Johnny: Give me some more example where the product formula would have to be applied.
Sara browsed the Internet for a minute and then said, here is one:
“Jack collects wild berries from the Tinka valley and then sells them in the city to make his living. He is worried. Many housing developments keep coming in the valley every year so that the area A for the growth of the berry decreases at the rate of dA/dt square meters per year. At the same time, the pollution in the valley increases every year so that the growth G per square kilometer of the berries in the wild decreases at rate of dG/dt per year. What is the net decrease in the berries available for collection from the valley?”
Johnny: Are there more examples?
Applications of the product formula
Sara: Lots of them. They relate to economics, businesses, engineering, medicine and ecology.
Johnny: That’s interesting. I know that the notice at the lake gave us the values for the rates of increase in the water level and area of the lake, and we figured out the rate of the increase in the amount of water from the lake by using
f ‘w(t) = f1(t) . f ‘2(t) + f2(t) . f ‘1(t).
But if they gave us the rates for the increase in the amount of water f ‘w(t) and rise in the level f ‘1(t), could we also have figured out the change in the rate of increase in the area of the lake f ‘2(t).
Sara: Yes, our book says that then we can use the quotient rule which would be
f ‘2(t) = (fw(t)/ f1(t))’ = (f1(t). f ‘w(t) – fw(t).f ‘1(t)/(f1t)2
Actually the book wrote the two general formulae as
Product rule: d(uv)/dx = vdu/dx +udv/dx
Quotient rule: d(u/v)/dx = (v(du/dx) – u(dv/dx))/v2
Johnny: I pity the guys who didn’t go to the picnic because it was too close to the final exams. Actually, it helped me. I might even ace my Calculus. Thanks to you and the picnic that brought these issues.
Challenge
Johnny’s grades in class 8 were really poor. Then he met Sara. Sara was not only smart and had good grades but also nice to Johnny. She loved him and also gave him tips on improving his grades. Johnny also started to study more. His math teacher told him that his grades had been improving over the years with time with the function G(t) = (t-.006t3)(1+0.5t) where t is the time in number of semesters. Johnny has now been with Sara for 6 semesters. Can you tell him whether his grades are expected to improve after this semester if this trend continues?
Solution: Johnny’s grades follow the trend G(t) = (t-.006t3)(1+0.5t)
What he needs to find out if whether dG/dt is positive at t =6.
You can either simplify the function first and differentiate or use the product rule. Let us use the product rule
G(t) = (t-.006t3)(1+0.5t)
dG/dt = (1-.018t2)(1+0.5t) + (t-.006t3) x 0.5
dG/dt = At t = 6, dG/dt = (1-.018×62)(1+0.5×6) + 0.5(6-.006x 63) = 1.408 + 2.352 = 3.76
Thus dG/dt is positive at t = 6 and Johnny can smile since his grades will keep improving.
Math Stories For You 