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Math 105a Solutions Exam 2
- (a) dy dx
q 1 (x^2 )^2
2 x = 2 x p 1 x^4
(b) To nd dy dx given^ e
y (^) + ln y (^2) = xy + x, it is necessary to implicitly di erentiate both sides with respect to x.
ey^ dy dx +
y^2 ^2 y
dy dx =^ y^ +^ x
dy dx + 1 dy dx
ey^ +
y ^ x
= y + 1
dy dx =^
y + 1 ey^ + (^) y^2 x or simplify to get dy dx =^
y^2 + y yey^ + 2 xy
- (a) Apply l'H^opital's rule since the limit leads to the indeterminate form (I.F.) 0=0.
lim! 0
tan()
L =^0 H (^) lim ! 0
1 cos^2 () ^ 1 = lim ! 0
cos^2 ()
cos^2 (0)
(b) Apply l'H^opital's rule (twice) since the limit leads to the I.F. 1 = 1.
xlim!
ex^ + x x^2
L =^0 H (^) lim x!
ex^ + 1 2 x
L =^0 H (^) lim x!
ex 2
- Is the following argument valid?
Consider lim x! 0 +
x
sin x
. Because lim x! 0 +
x = 1 and lim x! 0 +
sin x
it must surely be true that the limit in question has the value 1 1 = 0.
The argument is not valid. Recall from our study of limits (Theorem 2.1) that
xlim!a (f^ (x) +^ g(x)) = lim x!a f^ (x) + lim x!a g(x) only if both limits on the right hand side of the equation exist. For our example, the individual limits lim x! 0 +
x and lim x! 0 +
sin x are in nite, i.e., do not exist. Therefore, we cannot break up the (original) limit of a sum into the sum of individual limits. To nd a legitimate method to evaluate the limit start by rewriting
x
sin x using algebra, and then reevaluate the limit by applying l'H^opital's rule.
lim x! 0 +
x
sin x
= lim x! 0 +
sin x x x sin x
L =^0 H
lim x! 0 +
cos x 1 sin x + x cos x
L =^0 H
lim x! 0 +
sin x 2 cos x sin x
- (a) The local linearization of f (x) = arctan(x) near x = 1 is given by f (x) f (1) + f 0 (1)(x 1).
Note: f (1) = arctan 1 =
4 since tan(=4) = 1;^ and^ f^
(^0) (x) = 1 1 + x^2 so^ f^
0 (1) =^1
Therefore, f (x)
(x 1).
(b) To approximate arctan(1:01), let x = 1: 01
then f (1:01) = arctan(1:01)
- A kite 100 feet above the ground moves horizontally at a speed of 8 ft/sec. At what rate is the angle between the string and the horizontal changing when 200 feet of string have been let out. Be sure to express your answer using the appropriate units.
We are given dx dt = 8 ft/sec, but need to^ nd^
d dt. Using trigonometry we can establish a relationship between the base x of the right triangle and the angle . tan =^100 x Implicitly di erentiating both sides with respect to t gives: 1 cos^2
d dt
x^2
dx dt We can use the Pythagorean theorem to nd x and trigonometry to nd cos when z = 12. In partic- ular, if z = 200 ft then x =
p 2002 1002 = 100
p 3 and cos =
p 3 2
Therefore, d dt = 100 cos
x^2 dx dt
(^) p 3 2
p 3
2 =^ ^
radians/sec.
In other words, the angle between the string and the horizontal is changing at a rate of
50 radians/sec.
- Suppose the function f is continuous at the point P when x = c. The graph of f has
a vertical tangent at P if lim x!c ^ f 0 (x) and lim x!c+^ f 0 (x) are either both + 1 or both 1.
a cusp at P if lim x!c ^ f 0 (x) and lim x!c+^ f 0 (x) are both in nite with opposite signs (one + 1 and the other 1).
If f (x) = 3 x^2 =^3 , then f 0 (x) =
p (^3) x. To determine if the graph of f has a vertical tangent or cusp at x = 0 we need to look at the behavior of f 0 (x) as x! 0 ^ and x! 0 +. In particular,
- Consider the function f with f 0 (x) = 4(x + 1) 3 3
p x^2
and f 00 (x) = 4(x 2) 9 3
p x^5 (a) f 0 (x) is unde ned when 3 3
p x^2 = 0 =) x = 0 f 0 (x) = 0 when 4(x + 1) = 0 =) x = 1 Use a sign chart to see that f 0 (x) > 0 for 1 < x < 0 and x > 0, and f 0 (x) < 0 for x < 1. There- fore, f is increasing on ( 1 ; 0) and (0; 1 ). The graph of f is decreasing on ( 1; 1). In other words, the graph of f has a local minimum at x = 1. However, there is no local extreme at x = 0.
(b) f 00 (x) is unde ned when 9 3
p x^5 = 0 =) x = 0 f 00 (x) = 0 when 4(x 2) = 0 =) x = 2 Use a sign chart to see that f 00 (x) > 0 for x < 0 and x > 2, and f 00 (x) < 0 for 0 < x < 2. Therefore, f is concave up on ( 1; 0) and (2; 1 ). The graph of f is concave down on (0; 2). In other words, the graph of f has in ection points at x = 0 and x = 2.
(c) The graph of f passing through ( 1 ; 3), (0; 0), (1; 5), and (2; 6 3
p
- is shown below.
- Consider f (x) = x^2 =^3 on [ 1 ; 1].
(a) The graph of f is provided below.
(b) Notice that f (1) f ( 1) 1 ( 1)
= 0, but there is no c in the interval ( 1 ; 1) such that f 0 (c) = 0. Does this contradict the Mean Value Theorem?
No, this does not contradict the Mean Value Theorem. The function f (x) = x^2 =^3 is not di eren- tiable on ( 1 :1). In particular, f 0 (x) =
3 x^1 =^3 is unde ned at x = 0. We can see this graphically by noticing that the graph of f has a cusp at x = 0. In other words, the function f does not satisfy the conditions of the Mean Value Theorem since the derivative does not exist at x = 0. Therefore, the conclusion of the Mean Value Theorem does not apply to f (x) = x^2 =^3 on [ 1 ; 1].
