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=secx−2cosxat 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=2√3+2·√3

2=3√3.

Thus the equation of the tangent line is y−1=3√3(x−π

3)or y=3√3x−π√3+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).

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