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Math 105a Solutions Final Exam
- (a) If f (x) =
sin x x
then f ′(x) =
x cos x − sin x x^2
(b) If g(t) = ln(sin(t^2 )) then g′(t) =
cos(t^2 ) · 2 t sin(t^2 ) = 2t cot(t^2 )
(c)
t^3 + 7t t^4
dt =
t
dt = ln |t| −
2 t^2
+ C
(d)
∫ (^) π/ 4
0
cos x dx = sin x
π/ 4 0
= sin (π/4) − sin 0 =
(e) lim x→ 0
cos x − 1 ex^ − x − 1
L =′H
lim x→ 0
− sin x − 1 ex^ − 1
L =′H
lim x→ 0
− cos x ex^
- Use the limit definition of the derivative to find f ′(x) for f (x) =
2 + x
f ′(x) = lim h→ 0
f (x + h) − f (x) h
= lim h→ 0
1 2+x+h −^
1 2+x h
= lim h→ 0
(2 + x) − (2 + x + h) h(2 + x)(2 + x + h)
= lim h→ 0
−h h(2 + x)(2 + x + h) = lim h→ 0
(2 + x)(2 + x + h)
(2 + x)^2
- Consider f (x) =
− cos x, if x < 0 x^2 − a, if x ≥ 0
(a) lim x→ 0 −^
f (x) = lim x→ 0 −
(− cos x) = − 1
(b) lim x→ 0 +
f (x) = lim x→ 0 +
(x^2 − a) = −a
(c) In order for f (x) to be continuous at x = 0 we require lim x→ 0 −^
f (x) = lim x→ 0 +^
f (x) = f (0), note: f (0) = −a. lim x→ 0 −^
f (x) = lim x→ 0 +^
f (x) ⇐⇒ −1 = −a ⇐⇒ a = 1
(d) Let f be defined with a = 1. Use a graphical approach to zoom in on the graph of f near x = 0.
From (c) we know the function is continuous. Close inspection of the graph reveals no sharp corners or vertical tangents at x = 0. More specifically, we can see that as x approaches zero from both the left and right that the slope of the function is zero. Therefore, f ′(0) = 0. Note: this can be shown more formally using the limit definition of derivative.
- A 216 m^2 rectangular pea-patch is to be enclosed by a fence and divided into two equal parts by another fence parallel to one of the sides. What dimensions for the outer rectangle will require the smallest total length of fence? How much fencing material will be needed?
We want to minimize the total length of fence. Let x represent the width of the rectangle. Let y represent the length of the rectangle. We need to minimize P = 3x + 2y. The area of the rectangle given by A = xy is fixed at 216 m^2. Substitute y =
x
for y into the formula for perimeter
P = 3x + 2
x
Therefore, dP dx
x^2
3 x^2 − 432 x^2
= 0 when x = ±12.
Given the context of the problem, x = 12 is the only appropriate critical value.
At x = 12, d^2 P dx^2
(12)^3
0, therefore the total length of fence is minimized when x = 12 meters.
The dimensions of the outer rectangle are x = 12 meters and y = 18 meters. The amount of fence material is P = 3(12) + 2(18) = 72 meters.
- The tangent line to the curve x^2 /^3 +y^2 /^3 = 8 at the point (8, 8) and the coordinate axes form a triangle, as shown in the picture below. What is the area of the triangle?
Find the equation of the tangent line in order to locate the x and y-intercepts to help determine the base and height of the triangle. To find the slope of the tangent line, implicitly differentiate both sides with respect to x, solve for dy dx
, and then substitute in the x and y-values at (8, 8).
x−^1 /^3 +
y−^1 /^3 ·
dy dx
dy dx
x^1 /^3 y^1 /^3
x y
The slope of the tangent line at (8, 8) is given by
dy dx
3
The equation of the tangent line is y = −x + 16. Therefore, the x and y-intercepts are (16, 0) and (0, 16), respectively. The area of the triangle is A =
- Recall the Mean Value Theorem: if f is continuous on [a, b] and differentiable on (a, b), then there
exists a number c, with a < c < b, such that f ′(c) =
f (b) − f (a) b − a
The function f (x) = x^2 + 2x − 1 satisfies the conditions of the Mean Value Theorem on [0, 1] because f is continuous on [0, 1] and f ′(x) = 2x + 2 exists on (0, 1). Therefore, it follows that there exists a number c, with 0 < c < 1, such that f ′(c) =
f (1) − f (0) 1 − 0
f ′(c) = 2c + 2 = 3 =⇒ c =
(b) Inflection points occur when the function changes concavity. The graph of f ′(x) has sharp corners at x = 1, x = 3, and x = 4, so f ′′(x) does not exist at those values. Therefore, f has possible inflection points at x = 1, x = 3, and x = 4. We can use a sign chart to see that f ′′(x) < 0 for 1 < x < 3 since f ′(x) is decreasing on that interval. Similarly, f ′′(x) > 0 for 3 < x < 4 and 4 < x < 5 since f ′(x) is increasing on each interval. Therefore, the graph of f is concave down on (1, 3), and concave up on (3, 4) and (4, 5).
(c) The graph of f (x)
- is linear for 0 < x < 1 since the graph of f ′^ is constant on that interval
- has a local maximum at (2, 3 .5) since f ′^ changes from positive to negative at x = 2
- has a local minimum at (3. 5 , 2 .75) since f ′^ changes from negative to positive at x = 3. 5
- changes from concave down to concave up at x = 3 The graph is provided below.
- Suppose the rate of growth of a tumor is given by dr dt
= 0. 3 et^ mm/month.
(a) The integral
0
- 3 et^ dt represents the total amount, in mm, that the tumor has grown through- out the first three months.
(b) The integral
0
- 3 et^ dt represents the average rate of growth, in mm/month, of the tumor for the first three months.
(c) The general solution is r(t) = 0. 3 et^ + C.
(d) Use r(0) = 5 to find C: 5 = 0. 3 e^0 + C =⇒ C = 4. 7 The particular solution is r(t) = 0. 3 et^ + 4.7.
(e) Evaluate r(3) to find the size of the tumor after 3 months r(3) = 0. 3 e^3 + 4. 7 ≈ 10 .73 mm.
OR use part (a) along with the initial value in (c): the total size of the tumor is equal to the initial size at t = 0 plus the amount of growth in the first three months
0
- 3 et^ dt = 5 +
- 3 et
3 0
- 3 e^3 − 0. 3
≈ 10 .73 mm.
