2004 Test 2 solution

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10 Apr 2012
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MAT 137Y, 2004-2005, Solutions to Term Test 2
1. For the following, simplify your answers unless otherwise instructed.
(8%) (i) Evaluate lim
x0
cos5xcos3x
x2.
Let Lbe the limit above. Then by L’Hˆ
opital’s Rule,
LH
=lim
x05sin5x+3sin3x
2xH
=lim
x025cos5x+9cos3x
2=8.
(8%) (ii) Find the derivative of f(x) = x, where x<0, using the definition of derivative.
f0(x) = lim
h0
f(x+h)f(x)
h=lim
h0p(x+h)x
h·p(x+h) + x
p(x+h) + x
=lim
h0(x+h)(x)
h(p(x+h) + x)=lim
h0h
h(p(x+h) + x)=lim
h01
p(x+h) + x
=1
2x.
(8%) (iii) Find the equation of the tangent line to the curve 2(x2+y2)2=25(x2y2)at the point (3,1).
Implicitly differentiating gives us
4(x2+y2)·(2x+2yy0) = 50x50yy0.
Sticking in x=3, y=1 gives 40(6+2y0) = 15050y0or 130y0=90, so y0=9
13. Hence the
equation of the tangent line is y1=9
13(x3).
(8%) (iv) Use Newton’s Method with x1=2 to find the next approximation x2to the root of x420 =0.
Applying Newton’s Method to f(x) = x420, we have
x2=x1f(x1)
f0(x1)=22420
4·23=2+4
32 =17
8.
(12%) 2. A car leaves an intersection at noon and travels due south at a speed of 80 km/h. Another car has been
heading due east at 60 km/h and reaches the same intersection at 1:00 p.m. At what time were the cars
closest together? Verify that your answer is correct by showing that at that given time the distance is
minimized.
Let tbe the number of hours past noon. Let the intersection be the origin on the coordinate axes. Then
at any given time t, the car travelling south is at the point (0,80t), and the car travelling east is at the
point (60+60t,0). The distance Dbetween the two cars at any given time tis given by the equation
D2(t)=(60+60t)2+ (80t)2=10000t27200t+3600,t[0,1].
To minimize D(t), it is sufficient to minimize D2(t) = f(t) = 10000t27200t+3600. f0(t) =
20000t7200 =0 when t=9
25. Since f00(t) = 20000 >0, then t=9
25 is a local minimum, and
hence an absolute minimum (since it is the only critical point). Therefore the cars are closest together
9
25 hours past noon, or at 12:21:36 p.m.
1
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