Download Solutions for the Practice for Exam #2

Practice Problems for Exam #2
Solutions
1. a) Evaluate lim
x→0
tan x
4x
Use your limit laws:
tan x
1
tan x
= lim
x→0
4x
4
x
sin x
1
= lim cos x
4 x→0 x
1 limx→0 sinx x
=
4 limx→0 cos x
1
11
= .
=
41
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Here we also use the fact that the cosine function is continuous and cos 0 = 1, so limx→0 cos x =
lim
x→0
cos 0 = 1.
b) Find dy/dx and d2 y/dx2 if y = x tan x − sin2 x.
Nothing tricky here; just use our derivative rules and formulas. On the exam BE SURE TO
SHOW ALL OF THE STEPS!!
dy
dx tan x d sin2 x
=
−
dx
dx
dx
dx
d tan x du2 d sin x
= tan x
+x
−
dx
dx
du
dx
(u = sin x)
= tan x + x sec2 x − 2u cos x
= tan x + x sec2 x − 2 sin x cos x.
d2 y
d tan x dx sec2 x d2 sin x cos x
=
+
−
dx2
dx
dx
dx
2
dx
d sec x
d sin x cos x
= sec2 x + sec2 x
+x
−2
dx
dx
dx
2
d sin x
d cos x
du d sec x
2
2
= sec x + sec x + x
− 2 cos x
+ sin x
du dx
dx
dx
(u = sec x)
= 2 sec2 x + x(2u) sec x tan x − 2(cos2 x − sin2 x)
= 2 sec2 x + x2 sec2 x tan x − 2 cos2 x + 2 sin2 x.
2. A 13 foot ladder leaning up against a wall is sliding down the wall at 1 foot per second.
The ladder makes an angle θ with the ground. How is this angle θ changing when the top of the
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ladder is five feet from the ground?
Draw an appropriate picture, which in this case will be a right triangle. The base x is the
distance from the bottom of the ladder to the wall, the opposite side y is the wall (just from the
ground up to the top of the ladder), and the hypotenuse 13 is the ladder. Then sin θ = y/13.
Differentiating with respect to time t gives
dy/13
d sin θ
=
dt
dt
d sin θ dθ
1 dy
=
dθ dt
13 dt
dθ
−1
cos θ
=
,
dt
13
since we’re told that dy/dt is −1 foot per second (negative since y is getting smaller with time).
Now when y = 5, then by the Pythagorean Theorem x = 12 (using x2 + y 2 = 132 ). Since
cos θ = x/13, we get cos θ = 12/13 when y = 5. Plugging that in above, we see that dθ/dt = −1/12
when y = 5. This is in radians per second, and is negative since this angle is getting smaller with
time.
3. Find an equation for the tangent line to the curve y 2 = 2x2 − xy at the point (1, 1)
As always, the derivative gives the slope of the tangent line. To find dy/dx we use implicit
differentiation:
dy 2
d2x2 − xy
=
dx
dx
2
dy dy
d2x2
dxy
=
−
dy dx
dx
dx
2
dx
dx
dy
0
2yy = 2
− y
+x
dx
dx
dx
2yy 0 = 4x − (y + xy 0 ).
At the point x = 1, y = 1 we have 2y 0 = 4 − (1 + y 0 ), so y 0 = 1. That’s the slope of the tangent
line, so in point-slope form an equation for the tangent line is
(y − 1) = 1(x − 1).
4. Use the linearization to approximate sin(π/2 − .01).
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Here we are approximating the graph of y = sin x when x is near π/2. So first we find the
linearization of sin x at x = π/2. The derivative is d sin x/dx = cos x, which is cos(π/2) = 0 at
x = π/2. The function value is sin(π/2) = 1 at x = π/2. Thus, an equation for the tangent line at
x = π/2 is
(y − 1) = 0(x − π/2),
so the linearization (just solve for y) is
L(x) = 1.
This means that sin(π/2 − .01) ≈ L(π/2 − .01) = 1.
5. Locate the absolute extrema for f (x) = x3 + 3x2 − 24x + 4 on [−5, 5].
We first find any critical points within the interval. For this function, f 0 (x) = 3x2 + 6x − 24 =
3(x2 + 2x − 8) = 3(x − 2)(x + 4). There are two critical points; one at x = 2 and the other at
x = −4. The absolute extrema thus occur at x = ±5, 2 or −4. Plugging these values into our
function, we have
f (−5) = −125 + 75 + 120 + 4 = 74,
f (2) = 8 + 12 − 48 + 4 = −24,
f (5) = 125 + 75 − 120 + 4 = 84,
f (−4) = −64 + 48 + 96 + 4 = 84.
The absolute minimum is −24 and occurs at x = 2. The absolute maximum is 84 and occurs at
x = −4 and at x = 5.
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