MAT137Y, 2010-2011 Winter Session, Solutions to Term Test 2

1.Evaluate the following expressions.

(a) (7%) d

dx p1 + sec(x)

1 + cos2(x) + sin4(x)!

d

dx p1 + sec(x)

1 + cos2(x) + sin4(x)!

=1

(1 + cos2(x) + sin4(x))2×1

2(1 + sec(x))−1/2sec(x) tan(x)(1 + cos2(x) + sin4(x))

−p1 + sec(x)−2 cos(x) sin(x) + 4 sin3(x) cos(x)

(b) (9%) Find dy

dt when t= 1 if y=1

1 + x2, x =1−u

1 + u, u =t−3

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From the Chain Rule we have

dy

dt =dy

dx

dx

du

du

dt ,

and

dy

dx =−2x

(1 + x2)2,

dx

du =(−1)(1 + u)−(1 −u)(1)

(1 + u)2=−2

(1 + u)2,

du

dt =−3

2t−5/2.

Also, when t= 1 we have u= 1 and x= 0. Therefore

dy

dxx=0 = 0,dx

duu=1 =−1

2,and du

dt t=1 =−3

2.

It follows that

dy

dt t=1 = (0) −1

2−3

2= 0.

(c) (7%) lim

x→0

2−x2−2 cos(x)

x4

1

lim

x→0

2−x2−2 cos(x)

x4= lim

x→0−2x+ 2 sin(x)

4x3(by L’Hopital’s Rule)

= lim

x→0−2 + 2 cos(x)

12x2(by L’Hopital’s Rule)

= lim

x→0−2 sin(x)

24x(by L’Hopital’s Rule)

= lim

x→0−2 cos(x)

24 (by L’Hopital’s Rule)

=−1

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2. (12%) Consider the function

f(x) = x+ 2x2sin(1/x),if x6= 0

0,if x= 0.

Find f0(0) if it exists, or explain why it does not exist.

Recall that f0(0), if it exists, is given by

f0(0) = lim

h→0

f(0 + h)−f(0)

h

= lim

h→0

h+ 2h2sin 1

h

h

= lim

h→01+2hsin 1

h.

Now notice that lim

h→02hsin 1

h= 0. To see this, we recall that

−1≤sin 1

h≤1 =⇒ −2h≤2hsin 1

h≤2h.

Since limh→0−2h= 0 = limh→02h, the Squeeze Theorem implies that

lim

h→02hsin 1

h= 0.

Therefore,

f0(0) = lim

h→01+2hsin 1

h= 1.

3. (12%) Water is pouring into a leaky tank at a rate of 10 m3/h. The tank is a right

circular cone with the vertex pointing down, 9 m in depth and 6 m in diameter at the top.

The water level in the tank is rising at a rate of 1

5m/h when the depth of the water is 6 m.

How fast is the water leaking out at that time?

Note: Volume of cone = 1

3πr2h, where h= height of the cone, and r= radius of the cone.

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