Related Rates and calculus problems for real life situaions
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Related Rates Problems
1. A tiger escapes from a truck, right in front of the Empire State Building. I start running west along 34th Street at 2.5 m/s, while my friend takes off north on Fifth Avenue at 3 m/s.
Draw a diagram of this situation.
How fast is the distance between my friend and I growing after 12 seconds?
After 12.8 minutes, I've hit the river; I stop to catch my breath. My friend is still running up Fifth Avenue. At what rate is the angle between my friend and I increasing at 13 minutes? How about at 15 minutes? (NB: Convert minutes to seconds before computing)
2. Huckleberry Finn is drifting down the Mississippi on a rectangular, flat- bottomed boat, 6 feet wide by 12 feet long. Suddenly, he hits a rock and gashes a hole in the boat. Water is now pouring into the boat at a rate of 1/2 cubic foot per minute. Huck is doing the best he can to bail out the boat, but he only has a tin cup to pitch the water, and can only remove 1/3 ft^3/min.
Try to picture this dire situation in your mind, and draw a diagram or two to help you understand what's going on here.
At what rate is the volume of water inside the boat increasing?
At what rate is the depth of the water in the boat in increasing?
If the boat is half a foot high, when will the boat completely fill with water?
3. Your esteemed professor has just finished teaching and needs to lift his spirits. He goes to the nearest bar and orders a gin martini, served up in a conical cocktail glass, and downs it. The bartender has filled it to the brim, though, so knowing his clumsiness he decides it's better to drink the first centimeter of it through a straw rather than move the glass and spill any of it. The straw is kind of narrow, so he can only drink at a rate of 10 ml/sec. The cocktail glass is a right angle cone, and it holds 133 ml.
Picture this situation, and then draw a diagram. A delicious, gin-and- vermouth-flavored diagram.
Given that its height and radius are the same, and its volume is 133 ml, what is its height in centimeters? (the volume of a cone is (1 /3)πr2h, where r is the radius and h is the height)
At what rate is the depth of the liquid in the glass decreasing, as a function of time?
How long does it take for the depth of liquid to drop 1 centimeter, so he can safely pick up his glass?
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Solution Summary
Related rates and the solutions are explained for all questions. All calculus and trigonometry terms are explained. Solutions are provided in attached files with illustrative diagrams for real life situations. FOR PROPER VIEW OF DIAGRAMS AND ENTIRE SOLUTION PLEASE SEE ATTACHED FILES.
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FOR PROPER VIEW OF DIAGRAMS AND ENTIRE SOLUTION PLEASE SEE ATTACHED FILES
Let us call person I as 'A' and person my friend as 'B'. Person A is moving in horizontal direction (towards west) in X-direction and Person B is heading towards y-axis (North direction). They takes of at time t=0 s. Velocity of A is 2.5 m/s and that of B is 3 m/s. After 12 s person A & B are at position Y0 and X0 and having speed of dx/dt and dy/dt respectively. 'h' is distance between them after 12 s.
This can be solved using Pythagorean Theorem. Here, x^2 +y^2 = h^2
Taking derivatives w. r. to time,
2x(dx/dt) + 2y(dy/dt) = 2h(dh/dt)
Therefore, dh/dt = [x(dx/dt) + y(dy/dt)]/h
Now at t = 12 s,
dh/dt = [x0 (dx/dt) + y0 (dy/dt)]/h,
{Where, x0 = speed * time, evaluate x0 & y0 by yourself, it's very easy! h = sqrt(x^2+y^2 )}
Put these values in above equation and find dh/dt, which is rate of increment in distance between A & B. Your answer should be 3.9 m/s.
Now, after 12.8 minutes 'A' hits the river and he has stopped running. 'B' continues with his constant speed of 3 m/s. Position of A would be at x1 and will be remain steady for after 12.8 minutes. So position of A after 13 minutes will be X1= 12.8*60*
2.5 m and that of B is 13*60*3 m. From figure you can observe that angle between them would always be θ1 and it increases
with time. Relation between θ1 & persons can be given by tan(θ1).
tan〖(θ1)〗= y1/x1 (here, X1 is constant, X1= 12.8 *60*2.5 ...
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