Solutions to midterm 2
dy by implicitly di erentiating the following: x2 + y = 1:
1. (a) Find dx
4
9
x + 2y dy = 0. Solve
Solution: By implicitly di erentiating we get
2
9
dx
dy
9
x
this equation to get dx = , 4y .
2
(b) Find the equation of the tangent line to the ellipse x4 + y9 = 1: at the
p
3
point (,1; 2 3 ).
Solution: The slope at the point is obtained by substituting x =
p
,1; y = 3 2 3 in to your formula for dy=dx, hence slope m = 3=2.
Now use the point-slope formula to get
2
2
p
p
2.
T
T
T
F
F
p
p
p
p
y , 3 2 3 = 23 (x + 1) =) y = 23 x + ( 23 + 3 2 3 )
p
p
which can be simpli ed to y = ( 3=2) x + 2 3.
\A cubic function always has one point of in ection."
\The linear approximation to the function f (x) = x2 at x = 0 is the
line y = 0."
\A point (c; f (c)) on a graph y = f (x) can be both a maximum value
and a minimum value."
f (x) = 6x3 + 12 is never zero.
If f (x) is a quadratic function then f (x) ! 1 as x ! 1.
3. Consider the function y = f (x) = x2 x+ 1 on the interval ,2 x 2.
(a) Find f (x).
0
Solution:
By the quotient rule we have
2
1 , x (2x) = 1 , x2 :
f (x) = (x + (1)x2 +
1)2
(x2 + 1)2
0
(b) Find all of the critical points (we use only real numbers!).
Solution: Critical points are (i) endpoints, so x = 2 are endpoints; (ii)
stationary points (i.e., points where f (x) = 0) so x = 1 are stationary
points; and nally (iii) singular points (i.e., where f (x) is unde ned),
but the denomenator of f (x) is never zero, so there are no singular
points. Thus, the critical points occur at x = 1; 2.
(c) Compute the maximum and minimum values that f (x) attains on the
interval ,2 x 2.
Solution: Substitute all four critical points into f (x) are choose the
margest, namely f (1) = 1=2, and the smallest, namely f (,1) = ,1=2.
0
0
0
1
4. A light in a lighthouse 1 mile o shore from a straight shoreline is rotating
at 2 revolutions per minute, i.e., at 4 radians per minute! How fast is the
beam moving along the shoreline when it passes the point half a mile from
the point opposite the lighthouse?
Lighthouse
1
h
shore
Solution: We know d=dt = 4 , we want dh=dt, so we must relate and h.
The picture shows that tan = h. Implicitly di erentiate to give
dh = sec2 () d :
dt
dt
We want the value when h = 1=2, i.e., when tan = 1=2. There are several
ways to compute sec2 (). One is to compute cos() using Pythagorus, another
is to use the identity sec2 () = 1 + tan2 () = 1 + 1=4 = 5=4. Sub in to the
5
above to give dh
dt = 4 4 = 5 .
5. Let f be a continuous function and let its derivative f have the graph shown:
y
0
(0; 1) bc
b
b
(,2; 0)
(2; 0)
(0; ,1)
x
c
b
(a) Where is f increasing/decreasing ?
Solution: Increasing when f (x) > 0, i.e., when x < ,2, 0 < x < 2 and
x > 2. It's decreasing when f (x) < 0, i.e., when ,2 < x < 0.
(b) Where is f concave up/down ?
Solution: Concave up when f (x) is increasing, i.e., when x > 2. Concave down f (x) is decreasing, i.e., when x < ,2, ,2 < x < 0 and when
0 < x < 2.
(c) Assume that f (0) = 0. Draw the graph of y = f (x).
Solution: I'll do it in class. The point to watch out for is that the slope
of the graph moving to the right from the origin is +1, while the slope
of the graph moving to the left from the origin is ,1
0
0
0
0
2