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Fall 2004 Math 151
3 Derivatives
3.4 Limits and Derivatives of
Trigonometric Functions
c
Mon, 27/Sep
2004,
Art Belmonte
Solution
We have
cos θ − 1 cos θ + 1
=
lim
sin θ
cos θ + 1
θ →0
=
Summary
=
Four limits
lim sin θ = 0
sin θ
=1
θ →0 θ
lim cos θ = 1
cos θ − 1
=0
θ
θ →0
θ →0
θ →0
lim
lim
d
(cos x) = − sin x
dx
d
(tan x) = sec2 x
dx
d
(cot x) = − csc2 x
dx
d
(sec x) = sec x tan x
dx
d
(csc x) = − csc x cot x
dx
Find the limit lim
h→0
sin 5h
.
tan 3h
As h → 0, we have
5 sin 5h
sin 5h
5h
5
sin 5h
5h
3
5h
=
→ .
=
1
1
sin 3h
sin 3h
tan 3h
3
3h
3h
cos 3h
3h
cos 3h
185/22
Six antiderivatives
(Refer to Overview of Thu, 15/Jul.)
Z
Z
cos x d x = sin x + C
sin x d x = − cos x + C
Let y = x csc x. Find d y/d x.
Z
sec2 x d x = tan x + C
Solution
csc2 x = − cot x + C
Z
sec x tan x d x = sec x + C
csc x cot x d x = − csc x + C
dy
= (1) csc x + x (− csc x cot x) = csc x (1 − x cot x).
dx
We have
185/26
Hand Examples
Let y =
185/4
Find the limit lim x 2 sec x.
tan x − 1
. Find y 0 .
sec x
Solution
x→π
Rewrite y as y =
Solution
We have lim
x→π
!
Solution
d
(sin x) = cos x
dx
Z
− sin2 θ
lim
θ →0 sin θ (cos θ + 1)
− sin θ
= 0.
lim
θ →0 cos θ + 1
!
185/14
Six derivatives
Z
lim
θ →0
cos2 θ − 1
sin θ (cos θ + 1)
sin x
− 1 cos x = sin x − cos x. Then
cos x
π2
π2
x2
=
=
= −π 2 .
cos x
cos π
−1
y 0 = cos x + sin x.
186/44
185/10
cos θ − 1
.
sin θ
θ →0
1 − cos x
.
x→0
2x 2
Find the limit lim
Find lim
1
Solution
353/14
Find the most general antiderivative of f (θ ) = θ + sec θ tan θ .
We have
1 − cos x 1 + cos x
=
lim
1 + cos x
x→0
2x 2
x→0
=
lim
lim
1 − cos2 x
2x 2 (1 + cos x)
sin2 x
x→0 2x 2 (1 + cos x)
=
=
lim
x→0
sin x
x
2
1
2 (1 + cos x)
!
Solution
Z
Z
We have
f (θ ) dθ = θ + sec θ tan θ dθ = 12 θ 2 + sec θ + C.
MATLAB Examples
1
.
4
s185x32
Find an equation of the tangent line to y = 2 sin x at ( π6 , 1).
186/51
Find the limit lim
x→0
sin (sin x)
.
x
Solution
The slope of the tangent line is y 0 ( π6 ). Then use the point-slope
√ formula. The tangent line is y = 1 + 3 x − 16 π .
Solution
We have lim
x→0
sin (sin x) sin x
sin x
x
%-------------------------------------------------% Stewart 185/32
%
syms x
y = 2*sin(x);
yp = diff(y,x); pretty(yp)
= 1.
185/37
2 cos(x)
m = subs(yp, x, sym(pi/6)); pretty(m)
For what values of x does the graph of f (x) = x + 2 sin x have a
horizontal tangent line?
1/2
3
TL = 1 + m*(x - pi/6); pretty(TL)
Solution
1/2
1 + 3
(x - 1/6 pi)
%
h = 1;
x = linspace(pi/6 - h, pi/6 + h);
y = eval(y); TL = eval(TL);
plot(x,y, x,TL,’r--’); grid on; hold on
legend(’function’, ’tangent line’, ...
’Location’, ’NorthWest’)
plot(pi/6, 1, ’go’, ’MarkerFaceColor’, ’g’, ...
’MarkerSize’, 7)
xlabel(’x’)
title(’Stewart 185/32’)
set(gca, ’Ytick’, -1:3)
%
Solve f 0 (x) = 1 + 2 cos x = 0 for x. In other words, where is
cos x = − 12 ? There are infinitely many solutions.
n
o n
o
x ∈ 23 π + 2nπ : n ∈ Z ∪ 43 π + 2nπ : n ∈ Z
Here n is an element of Z, the set of all integers. (Also see
MATLAB Examples.)
353/12
echo off; diary off
√
Find the most general antiderivative of f (t) = sin t − 2 t.
Stewart 185/32
3
function
tangent line
Solution
2
(If needed, review the Summary and also the Overview of 15/Jul.)
Rewrite f as f (t) = sin t − 2t 1/2 . Then
Z
Z
f (t) dt =
sin t − 2t 1/2 dt
=
2 3/2
t
+C
− cos t −
3/2
=
− cos t − 43 t 3/2 + C.
1
0
−1
−0.5
0
0.5
1
x
2
1.5
2
s185x37 [185/37 revisited]
Stewart 186/57
s
For what values of x does the graph of f (x) = x + 2 sin x have a
horizontal tangent line?
d
Solution
r
θ
r
In the corresponding hand example, we obtained
n
o n
o
x ∈ 23 π + 2nπ ∪ 43 π + 2nπ .
Let’s illustrate this with a graph.
%-------------------------------------------------% Stewart 185/37
%
x = linspace(-10, 10);
f = x + 2*sin(x);
plot(x,f); grid on; hold on
plot([-10 10], [0 0], ’g’)
plot([0 0], [-15 15], ’g’)
xlabel(’x’); ylabel(’y’)
title(’Stewart 185/37’)
%
%-------------------------------------------------% Stewart 186/57
%
t1 = d2r(linspace(40, 140));
t2 = d2r(linspace(140, 400));
x1 = cos(t1); y1 = sin(t1);
x2 = cos(t2); y2 = sin(t2);
x3 = cos(t1(1)); y3 = sin(t1(1));
x4 = cos(t2(1)); y4 = sin(t2(1));
%
plot(x1,y1,’r--’)
grid on; hold on
plot(x2,y2,’k’)
plot([x3 x4], [y3 y4], ’b-.’)
plot([0 x3], [0 y3], ’k’)
plot([0 x4], [0 y4], ’k’)
axis equal
axis([-1.2 1.2 -1.2 1.2])
axis off
%
echo off; diary off
Stewart 185/37
15
10
echo off; diary off
y
5
0
−5
−10
−15
−10
−5
0
x
5
10
s186x57
The figure at top right shows a circular arc of length s and a chord
s
of length d, both subtended by a central angle θ . Find lim .
+
θ →0 d
Solution
Let r be the radius of the circle. Recall that arc length is s = r θ .
Drop a perpendicular from the center of the circle to the chord.
1
+
Then sin 12 θ = d/2
r , whence d = 2r sin 2 θ . Hence as θ → 0 ,
1
θ
θ
s
we have =
= 2 1 → 1, as one would expect
1
d
2 sin 2 θ
sin 2 θ
(since for small angles the arc is almost a straight line)!
3