WW2.1
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School
Thornlea Secondary School**We aren't endorsed by this school
Course
MATHEMATICS 244
Subject
Mathematics
Date
Jan 10, 2025
Pages
1
Uploaded by GrandStorkPerson790
Problem 1. (1 point) If 7= (x+y)e and x = 3t and y = 1 —#2, find the following de- rivative using the chain rule. Enter your answer as a function of ¢. s _ dt Correct Answers: o (3-21)e!™" — <3t—|— 1 —t2) el = 0t Problem 2. (1 point) A bison is charging across the plain one morning. His path takes him to location (x,y) at time ¢, where x and y are functions of ¢, and north is in the direction of increasing y. The temperature is always colder farther north. As time passes, the sun rises in the sky, sending out more heat, and a cold front blows in from the east. At time ¢ the air temperature H near the bison is given by H = f(x,y,t). The chain rule expresses the derivative dH /dt as a sum of three terms: dH 3f ox dH _dfox dfdy of dt Ox ot dyor ot Identify the term that gives the contribution to the change in tem- perature experienced by the bison that is due to: 1. The bison’s change in latitude . The rising sun 3. The coming cold front of ot of ox ox ot of dy Jy ot Correct Answers: o C o A e B A. B. C. Problem 4. (1 point) Find the directional derivative of f(x,y) = x%y> 4+ 2x*y at the point (3, -1) in the direction 6 = 27 /3. The gradient of f is: Vf={ , ) Vi(3,—1)=( ) The directional derivative is: Correct Answers: o 2xy’ +8x%y o 3xxyy+2x* o —222 e 189 o 274.678801315259 Problem 6. (1 point) The temperature at a point (x,y,z) is given by T(x,y,z) = 200e_"2 —*/4=2/ 9, where T is measured in degrees Celsius and Xy, and z in meters. There are lots of places to make silly errors in this problem; just try to keep track of what needs to be a unit vector. Find the rate of change of the temperature at the point (-1, -1, 2) in the direction toward the point (-1, 2, -3). In which direction (unit vector) does the temperature increase the fastest at (-1, -1, 2)? ( , ) What is the maximum rate of increase of T at (-1, -1, 2)? Correct Answers: e 23.4533542088982 0.948354065592898 0.237088516398225 —0.210745347909533 77.4821403113886
