2006 Test 2 solution

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10 Apr 2012
School
Department
Course
Professor
MAT 137Y, 2006–2007 Winter Session, Solutions to Term Test 2
1.
(9%) (i) Find the equation of the tangent line to the graph of y=secx2cosxat the point (π
3,1).
y0=secxtanx+2sinx, so the slope of the tangent line is
y0(π
3) = sec π
3tan π
3+2sin π
3=23+2·3
2=33.
Thus the equation of the tangent line is y1=33(xπ
3)or y=33xπ3+1.
(9%) (ii) For the equation x2+4xy +y3+5=0, find d2y
dx2at the point (2,1).
Differentiating implicitly, we have 2x+4xdy
dx +4y+3y2dy
dx =0.Plugging in x=2,y=1 gives
us
4+4dy
dx 4+3dy
dx =0=dy
dx (2,1)
=0.
Differentiating implicitly again, we have
2+4xd2y
dx2+4dy
dx +4dy
dx +6ydy
dx 2
+3y2d2y
dx2=0.
Plugging in x=2,y=1,dy
dx =0, we have
2+8d2y
dx2+3d2y
dx2=0=d2y
dx2=2
11.
2.
(10%) (i) A balloon is rising at a constant speed of 5
3meters per second. A boy is cycling along a straight
road at a speed of 5 meters per second. When he passes under the balloon, it is 15 meters above
him. How fast is the distance between the boy and the balloon changing three seconds later?
Let ybe the height of the balloon and xbe the horizontal distance between the balloon and the
bicycle. If zis the distance between the boy and balloon, then
x2+y2=z2=2xdx
dt +2ydy
dt =2zdz
dt =xdx
dt +ydy
dt =zdz
dt .
Three seconds after the bicycle passes the balloon, we have x=15 and y=20, so by the
Pythagorean Theorem z=25. Since dx
dt =5 and dy
dt =5
3, we get
15(5) + 20(5
3) = 25dz
dt =dz
dt =13
3,
so the distance between the cyclist and the balloon is increasing at 13
3meters per second.
(10%) (ii) Consider the function f(x) = (x2sin 1
x,x6=0,
0,x=0.
Prove that fis differentiable at 0, and find f0(0).
1
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