Review for Midterm 2
1. Find y 0 for
• y = cos(x)x
• y = 10tan(πθ)
y 0 = (ln(cos(x)) − x tan(x)) cos(x)x
y 0 = (ln(10))π sec2 (πθ)10tan(πθ)
2. A waterskier skis over a ramp with length 15 ft and height 4 ft at a speed of
30 ft/s. How fast is she rising as she leaves the ramp?
120
√
ft/sec ≈ 7.7ft/sec
241
3. Find the linearization of f (x) = (1+3x)1/3 at a = 0 and use it to approximate
(1.03)1/3 . Do the same with differentials.
L(x) = 1 + x
L(0.01) = 1.01
4. Find the extreme values of the function f (x) = 6x2 + x − 4 on [−1, 3].
Abs Max: 53
Abs Min: −4.04
5. State the Mean Value Theorem (what are the hypotheses?). For f (x) = 1/x,
find all c in [1, 3] which satisfy the conclusion of the Mean Value Theorem.
The MVT says that if a function is continuous and differentiable on an
f (b) − f (a)
.
interval (a, b) then there is a point c ∈ (a, b) such that f 0 (c) =
b−a
√
c= 3
1
, find the domain, intercepts, asymptotes, intervals of
x(x − 3)2
increase and decrease, local maximums and minimums, intervals of concavity, and inflection points. Sketch f (x).
6. For f (x) =
1
Review for Midterm 2
7. Evaluate
tan(4x)
4
=
x→0 x + sin(2x)
3
• lim
• lim+ sin(x) ln(x) = 0
x→0
8t − 5t
= ln 8 − ln 5
t→0
t
• lim
8. How many critical points can the function g(x) = ax3 − 21 bx2 + cx − d have
if b2 = 12ac?
1
9. State Newton’s Method. (For more Newton’s Method practice see problem
15.)
xn+1 = xn −
f (xn )
f 0 (xn )
10. Find the dimensions of the largest isosceles triangle that can be inscribed in
a circle of radius r.
First we draw the following picture:
2
Review for Midterm 2
We have the equation x2 + y 2 = r2 (from the little triangle). And the area
of the larger triangle is
1
A = (2x)(r + y)
2
p
Now solving the first equation for x gives x = r2 − y 2 . Then the area of
the larger triangle becomes
1 p
A = (2 r2 − y 2 )(r + y)
2
p
= r2 − y 2 (r + y)
So
p
−2y
r2 − y 2 + (r + y) p
2 r2 − y 2
r2 − y 2
−ry − y 2
=p
+ 2
r − y2
r2 − y 2
A0 =
r2 − ry − y 2
= p
r2 − y 2
Now find the critical points. Obviously the points r and −r are not going to
be the points we are looking for so we must use the quadratic formula on to
solve r2 − ry − 2y 2 = 0. This yields the solutions
√
r ± 9r2
y=
−4
r
= −r or
2
So now we have found the dimension for y. Looking back at our picture this
3r
tells us the the height of the triangle is y + r = . Plugging the solution for
√2
3
2
2
2
y into the equation x + y = r tells us x =
r so the base of the triangle
2
√
is 3r.
√
3r
So our final answer is that the triangle has height
and base 3r.
2
11. Find f for
• f 0 (x) = 2ex + sec(x) tan(x) f (x) = 2ex + sec(x) + C
1
• f 0 (x) = cos(x) − √
f (x) = sin(x) − arctan(x) + C
1 − x2
3
Review for Midterm 2
√
x2 + x
• f (x) =
and f (0) = 2
x
0
√
f (x) = 12 x2 + 2 x + 2
12. A paper cup has the shape of a cone with height 10 cm and radius 3 cm at
the top. If water is poured into the cup at a rate of 2 cm3 /s, how fast is the
water level rising when the water is 5 cm deep?
8
dh
=
dt
9π
13. Calculate y 0 for
• sin(xy) = x2 − y
2x − y cos(xy)
dy
=
dx
x cos(xy) + 1
•
√
x + y = 1 + x2 y 2
√
dy
2xy 2 − x + y
=√
dx
x + y − 2x2 y
14. A Norman window has the shape of a rectangle surmounted by a semicircle.
(Thus the diameter of the semicircle is equal to the width of the rectangle).
If the perimeter of the window is 30 ft, find the dimensions of the window
with the largest area.
60
Width =
4+π
30
15π
Height = 15 −
+
4π 4 + π
15. Find an equation of the tangent to y = 4 sin2 (x) at the point (π/6, 1).
√ π
y−1=2 3 x−
6
16. Find the point on the curve y =
√
x that is closest to the point (3, 0).
r !
5
5
,
2
2
17. Use Newton’s Method with x1 = −1 to find x4 , the fourth approximation to
the root, for x7 + 4 = 0.
x4 ≈ −1.23
4