MT1 Tutorial Exercises, Sheet 3 : Solutions
Dr. Will Sutherland – Oct 2015
1) Find the solutions for all values of θ between 0 and 2π for the following:
√
(a) sin−1 ( 3/2) : θ = π/3, 2π/3
√
(b) cos−1 (1/ 2) : θ = π/4, 7π/4
√
(c) tan−1 ( 3) : θ = π/3, 4π/3
(Note: we can get the second solutions from the first ones by reflecting around the y−axis for sin,
reflecting around the x−axis for cos, and rotating by 180 deg or π radians for tan:
this is equivalent to the properties sin(π − θ) = sin θ; cos(2π − θ) = cos(−θ) = cos θ, and
tan(π + θ) = tan θ, which follow from the addition formulae.
2) Derive the expression for the derivative of cos−1 (x).
y
dy
− sin y
dx
dy
dx
dy
dx
= cos−1 x, so cos y = x
= 1
1
sin y
1
= −√
, using sin2 y + cos2 y = 1
2
1−x
= −
3) Differentiate the function y = x sin−1 (x) with respect to x.
Using the product rule,
y = x sin−1 (x)
x
dy
= sin−1 (x) + √
dx
1 − x2
4) Calculate the radius of curvature of the function y = x − x2 . Given this, evaluate the radius of
curvature at the point with x = 1.
h
R =
1 + ( dy
)2
dx
i3/2
d2 y
dx2
dy
2
= 1+ 2
dx
x
dy 2
4
4
( ) = 1+ 2 + 4
dx
x
x
4
d2 y
= − 3
x
dx2
h
R =
−x3 2 +
4
x2
+
4
x4
i3/2
4
1
√
At x = 1, R = −(2 + 4 + 4)3/2 /4 = − 1000/4 ' −7.91.
Note, negative R is allowed, and means that the centre of curvature is below the curve.
5) Evaluate the radius of curvature of the function y = 3x4 − x sin(x) at the point with x = 0.5.
h
)2
1 + ( dy
dx
R =
i3/2
d2 y
dx2
dy
= 12x3 − sin x − x cos x
dx
d2 y
= 36x2 − 2 cos x + x sin x
dx2
At x = 0.5 this evaluates to: y 0 ' 0.582 and y 00 ' 7.484, so R ' 0.207.
6) The function y = e−3x sin(x) describes a damped oscillator. Evaluate the radius of curvature of
this function at the point with x = 1.
h
R =
1 + ( dy
)2
dx
i3/2
d2 y
dx2
y 0 = e−3x cos x − 3e−3x sin x
y 00 = −3e−3x cos x − e−3x sin x − 3e−3x cos x + 9e−3x sin x
y 00 = e−3x (8 sin x − 6 cos x) (collecting terms)
At x = 1, y 0 = −0.0988 and y 00 = 0.1737, so R ' 5.840.
2
7) Find the positions of the turning point(s) of the function y = x2 + 2x + 1, and identify the
2
, and ddxy2 , noting the turning points.
nature of the turning points. Sketch the functions y, dy
dx
y 0 = 2x + 2
y 00 = 2
The turning points are at positions that satisfy y 0 = 0, so there is a single turning point at x = −1.
The second derivative is positive at this value of x, so this turning point is a minimum. See Figure 1
for graphs of y, y 0 and y 00 .
Figure 1: Graphs of y (solid),
dy
dx
(dashed), and
3
d2 y
dx2
(dotted) for Question 7.
8) Find the positions of the turning points of the function y = e−x cos(x) between x = 0 and
x = 2π.
2
, and ddxy2 noting the turning
Identify the nature of the turning points, and sketch the functions y, dy
dx
points.
y 0 = −e−x (sin(x) + cos(x))
y 00 = 2e−x sin(x)
The turning points are defined by y 0 = 0, which is satisfied for sin(x) = − cos(x), hence tan x = −1.
A calculator will give us arctan(−1) = −π/4, but we want a positive x so add π or 2π, giving 3π/4
rad (135 deg), and 7π/4 rad (315 deg) as the (two) solutions we want.
We now want the sign of y 00 at these; 2e−x is always positive, so the sign of y 00 is the same as the
sign of sin x; at 3π/4, sin x is positive, so this is a minimum; and at 7π/4, sin x is negative so a
maximum.
The points of inflection are defined by y 00 = 0, so when sin(x) = 0. These occur at 0 and π rad.
See Figure 2 for graphs of y, y 0 and y 00 .
Figure 2: Graphs of y (solid),
dy
dx
(dashed), and
4
d2 y
dx2
(dotted) for Question 8.