Solutions For Calculus Quiz #2
1. (a) f0(x) = (x3e3)0+ (3e)0 = [(x3)0e3+ x3(e3)0] + 0 = [3x2e3+ 0] + 0 = 3x2e3 (b) f (x)0 = (√
3x3+ 3x)0cos√
3x3+ 3x = 9x2+ 3 2√
3x3+ 3xcos√
3x3+ 3x
2. Let y = f (x) = x2. Then f0(x) = 2x. Note that the curve f (x) = x2 does not go through the point (0,−a2). Suppose that the tangent line and the curve intersect at the point (t, t2).
Then the slope of the tangent line at (t, t2) is f0(t) = 2t. Furthermore, (t, t2) and (0,−a2) are two points on the tangent line, then the slope of the line is −a2− t2
0− t = a2+ t2 t . Solve 2t = a2+ t2
t , i.e. t2 = a2,
we get t =±a , (t, t2) = (±a, a2) and f0(±a) = ±2a.
Therefore, the equation of the tangent lines is y− a2 =±2a(x − ±a).
3. We differentiate both sides of the equation x2+ y2 = 1 with respect to x.
We need to remember that y is a function of x.
d
dx(x2+ y2) = d dx(1) d
dx(x2) + d
dx(y2) = d
dx(1) (the derivative of a sum is the sum of the derivatives) 2x + 2ydy
dx = 0 (the power rule and the chain rule ) We find that dy
dx =−2x
2y =−x y
Differentiateing both sides of this equation with respect to x, we have d
dx[dy
dx] = d dx[−x
y] d2y
dx2 = −y + xdydx y2 Substituting−xy for dxdy, we obtain
d2y
dx2 = −y + x(−xy)
y2 = −y − xy2
y2 = −y2− x2
y3 = −(y2 + x2)
y3 =− 1 y3 since x2+ y2 = 1
4. (a) Let y = f (x).
Solve √
xf (x) = x2+ 1 for f (x).
xf (x) = (x2 + 1)2 xf (x) = x4+ 2x2+ 1
f (x) = x4+ 2x2+ 1 x Then
dy
dx = f0(x) = x(4x3+ 4x)− (x4+ 2x2+ 1)· 1
x2 = 3x4+ 2x2− 1 x2 1
(b)
√xy = x2+ 1 (xy)0
2√
xy = 2x 1
2√
xy(y + xdy
dx) = 2x y + xdy
dx = 4x√ xy xdy
dx = 4x√ xy− y dy
dx = 4√
xy− y x
5. (a) Suppose that f is continuous on the closed interval [a, b]. If L is any real number with f (a) < L < f (b) or f (b) < L < f (a), then there exists at least one number c on the open interval (a, b) such that f (c) = L.
(b) A solution to e−x = x is a value of x for which the function f (x) = e−x− x = 0. Now, f (1) = e−1 − 1 is negative, since e−1 = 1e is less than 1, while f (0) = e0 − 0 = 1 is positive. Since the function is continuous as the difference of two continuos functions and f (0) > 0 > f (1), we can apply the Intermediate Value Theorem to learn that there is a value c in (0, 1) such that f (c) = 0, that is e−c = c.
(c) We define the function f (x) = e−x− x, which is continous.
We are looking for a zero of this funciton. We have f (0) > 0 andf (1) < 0 by (b).
Therefore there must be a zero in (0, 1) by the intermediate value theorem.
Using the bisection method,
x 0 1 0.5 0.75 0.625
f (x) 1 -0.632 0.1065 -0.278 -0.0897
sign + - + - -
narrow the interval to (0,1) (0,0.5) (0.5,0.75) (0.5,0.625)
x 0.5625 0.59375 0.578125
f (x) 0.00728 -0.0414 -0.01717
sign + - -
narrow the interval to (0.5625,0.625) (0.5625,0.59375) (0.5625,0.578125)
x 0.5703125 0.56640625
f (x) -0.004964 0.001155
sign - +
narrow the interval to (0.5625,0.5703125) (0.56640625,0.5703125)
Any number within (0.56640625,0.5703125) can be written as 0.57 to two decimal places. Therefore our solution is 0.57.
6. (a) We will need to multiply and divide an inequality by x and sin x. Since multiplying or dividing an inequality by a negative number reverses the inequality sign, we will split the proof into two cases, one in which 0 < x < π/2, the other in which −π/2 < x < 0.
In the former case, both x and sin x are positive. In the latter, both x and sin x are negative. (Since we are interested in the limit as x → 0, we can restrict the values of x to values close to 0.) We start with the case 0 < x < π/2. In the figure, we draw the unit circle together with the triangles OAD and OBC. The angle x is measured in
2
O 1
x A
C
D
B 1
radians. Since OB = OD = 1, we find
arc length of BD = x OA = cos x AD = sin x BC = tan x Furthermore,
area of4OAD ≤ area of sector OBD ≤ area of4OBC
The area of a sector of central angle x (measured in radians) and radius r is 12r2x.
Therefore,
1
2OA· AD ≤ 1
2OB2· x ≤ OB · BC or
1
2cos x sin x≤ 1
2· 12· x ≤ 1
2· 1 · tan x
Dividing this by 12sin x (and noting that 12sin x > 0 for 0 < x < π/2) yields cos x≤ x
sin x ≤ 1 cos x On the right part, we used the fact that tan x = sin xcos x. Taking reciprocals and reversing the inequality signs yields
1
cos x ≥ sin x
x ≥ cos x We can now take the limit as x→ 0+.
(Remember, we assumed that 0 < x < π/2, so we can only approach 0 from the right.) Note that
lim
x→0+cos x = 1 and
lim
x→0+
1
cos x = 1
limx→0+cos x = 1 3
We now apply the sandwich theorem and find lim
x→0+
sin x x = 1
We only showed that limx→0+ sin xx = 1, but a similar argument can be carried out when
−π/2 < x < 0. In this case, limx→0− sin x
x = 1 Then left-hand and the right-hand limits are the same and we conclude that
xlim→0
sin x x = 1 (b) Note that −1 ≤ sin x ≤ 1 for any x.
Multiplicaton of these inequalities by x1 gives −1
x ≤ sin x x ≤ 1
x for x > 0 Since lim
x→∞−1
x = 0 and lim
x→∞
1 x = 0, we obtain lim
x→∞
sin x
x = 0, thanks to the sandwich theorem.
7. The volume V of a sphere of radius r is given by V = 4 3πr3
Note that V is a function of r. Since r is increasing at a certain rate, we think of r as a function of t, that is, r = r(t).
Since the volume V depends on r, it changes with time t as well.
We therefore consider V also as a function of time t.
Differentiating both sides of V = 4
3πr3 with respect to t, we find dV
dt = 4 3πr3dr
dt = 4πr2dr dt When r = 8 cm and dr/dt = 3 cm/s, then
dV
dt = 4π82(cm2)3(cm/s) = 768π(cm3/s)
4
