Due: 02/18/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #5 Solution
Name:
Question:
1
2
3
4
5
Total
Points:
20
15
16
16
8
75
Score:
Instructions: Please show all of your work as partial credit will be given where appropriate, and there may
be no credit given for problems where there is no work shown. All answers should be completely simplified,
unless otherwise stated. No calculators or electronics of any kind are allowed.
1. (Implicit differentation) Consider the ellipse x2 − xy + y 2 = 4 centered at the origin.
(a) (4 points) Find the equation of the tangent lines, where the ellipse intersects the x-axis.
Solution:
2x − y − x
dy
dy
+ 2y
=0
dx
dx
dy
2x − y
=−
dx
−x + 2y
The ellipse intersects the x-axis at (±2, 0). So the tangent lines are 2x ∓ 4.
(b) (4 points) Find the equation of the tangent lines, where the ellipse intersects the y-axis.
Solution: The ellipse intersects the y-axis at (0, ±2). So the tangent lines are 21 x ± 2.
(c) (4 points) Find all the points on the ellipse, where the tangent line is horizontal.
dy
2x−y
Solution: The tangent line is horizontal, if dx
= − −x+2y
= 0. So the points, where the
tangent line is horizontal, is the intersection of the line y = 2x with the ellipse. These are the
points (± √23 , ± √43 ).
(d) (4 points) Find all the points on the ellipse, where the tangent line is vertical.
dy
Solution: The tangent line is vertical, if the slope is infinity, i.e. the denominator of dx
is 0.
x
So the points, where the tangent line is vertical, is the intersection of the line y = 2 with the
ellipse. These are the points (± √43 , ± √23 ).
(e) (4 points) Draw the ellipse with all the tangent lines of the parts before. Mark the major and
minor axis.
1
MATH 1210, Spring 2016
Lab #5 Solution
2. (Deriving trigonometric identities with differentation)
(a) (5 points) Differentiate the identity cos 2x = cos2 x − sin2 x about x. Derive a formula for sin 2x.
Solution:
−2 sin 2x = −2 cos x sin x − 2 sin x cos x = −4 sin x cos x
So
sin 2x = 2 sin x cos x
(b) (5 points) Differentiate the identity sin 2x = 2 sin x cos x about x. Derive a formula for cos 2x.
Solution:
2 cos 2x = 2 cos2 x − 2 sin2 x
So
cos 2x = cos2 x − sin2 x
(c) (5 points) We know that sin 3x = 3 sin x − 4 sin3 x. Differentiate it about x, and apply the identity
cos2 x + sin2 x = 1, to get a formula for cos 3x.
Solution:
3 cos 3x = 3 cos x − 12 sin2 x cos x = −9 cos x + 12 cos3 x
Therefore
cos 3x = 4 cos3 x − 3 cos x
2
Due: 02/18/2016 in class
Instructor: Janina Letz
MATH 1210, Spring 2016
Lab #5 Solution
Name:
3. (All roads lead to Rome)
We’d like to differentiate y = f (x) =
1
(x2 +1)(x2 +2)
in four different ways,
(a) (4 points) (Composite function approach I) Let a(x) = x1 , b(x) = (x+1)(x+2) = x2 +3x+2, c(x) =
x2 , so you can see f (x) = a◦b◦c(x). (Chain rule for three functions:f 0 (x) = a0 (b(c(x)))·b0 (c(x))·c0 (x))
Solution:
f 0 (x) = a0 (b(c(x))) · b0 (c(x)) · c0 (x) =
−1
· (2x2 + 3) · (2x)
((x1 + 1)(x2 + 2))2
(b) (4 points) (Composite function approach II) Since (x2 + 1)(x2 + 2) = x4 + 3x2 + 2, We consider
A(x) = x1 , B(x) = x4 + 3x2 + 2, and f (x) = A ◦ B(x).
Solution:
f 0 (x) = A0 (B(x)) · B 0 (x) =
(c) (4 points) (Product rule) Let C(x) =
1
x2 +1 ,
−1
· (4x3 + 6x)
(x4 + 3x2 + 2)2
D(x) =
1
x2 +2
and f (x) = C(x) · D(x).
Solution:
f 0 (x) = C(x)D0 (x) + C 0 (x)D(x) =
1
−1
1
−1
·
· (2x) + 2
·
· (2x)
x2 + 1 (x2 + 2)2
x + 2 (x2 + 1)2
1
2
(d) (4 points) (Implicit differentiation) Set y = (x2 +1)(x
+ 1) · (x2 + 2) = 1. Now
2 +2) , so y · (x
0
differentiate with respect to x and solve for y .
Solution:
y 0 (x2 + 1)(x2 + 2) + y2x(x2 + 2) + y(x2 + 1)2x = 0
Solve for y 0 to get
y0 =
(x2
y 0 (x2 + 1)(x2 + 2) +
−2x
−2x
+ 2
2
2
+ 1) (x + 2) (x + 1)(x2 + 2)2
3
2x
2x
+
=0
x2 + 1 x2 + 2
MATH 1210, Spring 2016
Lab #5 Solution
4. (Derivatives) Find formulas for the following derivatives, where n, m a non-negative integer.
(a) (4 points)
dn n
dxn x
Solution:
dn−1
dn−1
dn n
x = n−1 nxn−1 = n n−1 xn−1 = · · · = n · (n − 1) · · · 2 · 1 = n!
n
dx
dx
dx
(b) (4 points)
dn 1
dxn x
Solution:
(c) (4 points)
dn 1
1
dn−2 1
1
dn−1
−
=
2 3 = · · · = (−1)n n! n+1
=
n
n−1
2
n−2
dx x
dx
x
dx
x
x
dn
m
dxn (am x
+ am−1 xm−1 + · · · + a1 x + a0 )
Solution:
=
(d) (4 points)
dn
dxn
(
0
am m! m−n
(m−n)! x
+
am−1 (m−1)! m−1−n
(m−1−n)! x
+ ··· +
an+1 (n+1)!
x
1!
+
an n!
0!
m<n
m≥n
sin(x) (Hint: Differ between even and odd n)
Solution:
dn
π
sin(x) = sin(x + n ) =
dxn
2
(
n
n even
(−1) 2 sin(x)
n−1
(−1) 2 cos(x) odd
5. (Coefficients of polynomials) Find the coefficients of the functions, such that the given properties are
statisfied.
(a) (4 points) f (x) = ax3 + bx2 + cx + d with f (0) = 4, f 0 (0) = 7, f 00 (0) = 24, f 000 (0) = −30
Solution:
f 0 (x) = 3ax2 + 2bx + c
f (0) = d = 4
f 00 (x) = 6ax + 2b
f 00 (0) = 2b = 24
f 0 (0) = c = 7
f 000 (x) = 6a
f 000 (0) = 6a = −30
So a = −5, b = 12, c = 7, d = 4.
(b) (4 points) g(x) = ax2 + bx + c with g(1) = 12, g 0 (2) = 63, g 00 (3) = 30
Solution:
g(1) = a + b + c = 12
g 0 (x) = 2ax + b
g 00 (x) = 2a
So a = 15, b = 3, c = −6.
4
g 0 (2) = 4a + b = 63
g 00 (3) = 2a = 30