Worksheet 14
April 18, 2016
1.
Find dy/dx and d2 y/dx2 of the curve given by x = et , y = te−t . For which values of t is
the curve concave upward?
Solution:
We are asked to nd the total derivatives dy/dx and d2 y/dx2 however we are given x
and y as parametric functions of t hence we must express the total derivative in terms
of the parametric derivatives. This can be done as
dy
dy/dt
=
dx
dx/dt
and
d dy
d dy
dy 0 (x)/dt
dy 0
d2 y
dt dx
=
=
=
=
dx2
dx dx
dx
dx/dt
dx/dt
With these equations relating the total derivatives to the parametric derivatives, we nd
that
−t
dy
(1 − t)e
=
dx
et
= (1 − t)e−2t
and
d2 y
=
dx2
d
dt
((1 − t)e−2t )
−2(1 − t)e−2t − e−2t
(2t − 3)e−2t
=
=
= (2t − 3)e−3t
dx/dt
et
et
Moreover, a curve is concave up when
d2 y
dx2
> 0 hence,
(2t − 3)e−3t > 0 ⇒ 2t − 3 > 0 ⇒ t >
1
3
2
2.
At what points on the curve x = 2t3 , y = 1 + 4t − t2 does the tangent line have slope 1?
Solution:
We are asked to nd the points where
= 1 thus our rst task is to nd
dy
dx
dy
dy/dt
4 − 2t
=
=
dx
dx/dt
6t3
which is 1 when
1=
4 − 2t
2
3
3
, −1
⇒
6t
=
4
−
2t
⇒
6t
+
2t
−
4
=
0
⇒
t
=
6t3
3
These values of t correspond to the x − y coordinates
t=
2
16 29
: (x, y) = ( , )
3
27 9
and
t = −1 : (x, y) = (−2, −4)
3.
Find the arc length of the curve given by x = t sin t, y = t cos t, 0 ≤ t ≤ 1.
Solution:
Arc length is given by
ˆ
a
and
b
s
ds =
L=
where,
ˆ
b
a
dx
dt
2
+
dx
= t cos t + sin t
dt
dy
= −t sin t + cos t,
dt
2
dy
dt
2
dt
dy
dx
:
So
ˆ
b
s
L =
a
ˆ
dx
dt
2
+
dy
dt
2
dt
1
p
(t cos t + sin t)2 + (−t sin t + cos t)2 dt
=
ˆ0 1 q
t2 cos2 (t) + 2t sin t cos t + sin2 t + t2 sin2 t − 2t sin t cos t + cos2 tdt
=
ˆ0 1 √
=
t2 + 1dt
√0
√
2 1
+ ln(1 + 2)
=
2
2
where the last integral is derived using trig. sub (t = tan θ).
4.
Find the area enclosed by the astroid x = a cos3 θ, y = a sin3 θ, 0 ≤ θ < 2π .
Solution:
Without a graph, it is dicult to deduce properties of the function to aid our integration.
However, observe that both x(θ) and y(θ) have the same period, namely P = 2π . From
here we can deduce that the function is symmetric about the 4 quadrants hence we can
solve the area of one quadrant and multiply it by 4:
ˆ
ˆ
a
0
(a sin3 θ)
ydx = 4
A = 4
π/2
0
ˆ
d
(a cos2 θ)dθ
dθ
0
(a sin3 θ)(−3a cos2 θ) sin θdθ
= 4
π/2
ˆ
ˆ
π/2
π/2
1
sin2 θ( sin 2θ)2 dθ
2
0
#
"ˆ0 π
ˆ π/2
ˆ π/2
/2
3 2
3 2
2
2
2
cos 2θ sin 2θdθ
=
(1 − cos 2θ) sin 2θdθ = a
sin 2θdθ −
a
2
2
0
0
0
"ˆ π
#
ˆ π/2
/2
3 2
1
=
a
(1 − cos 4θ)dθ −
cos 2θ sin2 2θdθ
2
2
0
0
π/2
3 2 θ 1
1
3
=
a
− sin 4θ − sin3 2θ
= πa2
4
2 4
6
8
0
= 12a
2
2
2
sin θ(sin θ cos θ) dθ = 12a
3
2