278
CHAPTER 34
34.27
Find an equation of the normal line to the curve x = a cos
4 0, y- a sin
4 0 at 0 = ir/4.
dx/dO = 4acos
3 0(-sin0), dy/dO =4asin
3 0(cos 0). At 0 = ir/4, dxldt=-a, dy/dO = a, giving as
tangent vector (-a, a) = —a(l,-1). So the normal line has equation x — y + c = 0. To find c, substitute
the values of x and y corresponding to 0 = Tr/4: - — ^ + c = 0 or c = 0.
34.28
Find the slope of the curve x = 3t-l, y = 9t
2 -3t when / = !.
dx/dt = 3, dy/dt = 18t-3. Hence, dy/dt = 6t-l = 5 when / = !.
34.29
For the curve of Problem 34.28. determine where it is concave upward.
A curve is concave upward where d
2 y/dx
2 > 0. In this case,
Hence, the
curve is concave upward everywhere.
34.30
Where is the curve x = In t, y = e' concave upward?
dxldt = \lt, dy/dt=e'. So, dy/dx = te',
= e't(t + I). Thus, d
2 y/dx
2 >Q if and only
if f(t + l)>0. Since t>0 (m order for x-\nt to be defined), the curve is concave upward everywhere.
34.31
Where does the curve x = 2t
2 — 5, y = t
3 + t have a tangent line that is perpendicular to the line x + y +
3 = 0?
dx/dt = 4t, dy/dt = 3t
2 + l. So, dy/dx = (3t
2 + l)/4f. The slope of the line *+y+3=0 is-1, and,
therefore, the slope of a line perpendicular to it is 1. Thus, we must have dy/dx = 1, (3t
2 + l)/4t = 1,
3t
2 + l = 4t, 3f
2 -4r+l=0, (3t- l)(f- 1) = 0, t=\, or t = l. Hence, the required points are (-f, $)
and (-3,2).
34.32
Find the arc length of the circle x = acosO, y = asin0, 0<0^2ir.
Recall that the arc length
dx/d0 = -asinO,
dy/d0 = acos0,
and
the standard formula for the circumference of a circle of radius a.
34.33 Find the arc length of the curve x = e' cos t, y = e' sin <, from t = 0 to t= IT.
dxldt = e'(~sin t) + e'(cos t) = e'(cos t - sin t),
(dx/dt)
2
= e'(cos
2 t - 2 sin t cos t + sin
2 t) = e
2 '(l -
2 sin t cos t). dy/dt = e' cos t + e' sin t = e'(cos t + sin (), (dy/dt)
2 = e
z
'(cos
2 t + 2 sin t cos t + sin
2 t) = e
2
'(l +
2 sin tcost). So, s = /; \/V(l - 2 sin t cos 0 + e
2 '(l + sin t cos 0 r df = V2e' ]„ = V2(e" - 1).
34.34
Find the arc length of the curve x = | In (1 + t2), y = tan 11, from t = 0 to t = l.
dx/dt =t/(l + t
2 ), (dxldt)
2 = t
2 l(\ + t
2 )
2 . dy/dt = \l(\ + t
2 ), (dyldt)
2 = l/(\ + t
2 )
2 . Hence,
The substitution r = tan0, dt = sec
2 0d0 yields Jo"'
4 sec 0 rf0 = In |sec 0 + tan 0\ Jo'
4 = In |V2+ 1| -In 1 =
ln(V2+l).
34.35
Find the arc length of x = 2 cos 0 + cos 20 + 1, y = 2 sin 0 + sin 20, for 0 < 0 < 27r.
djc/dfl = -2 sin 0 - 2 sin 20,
(dxlde)
2 = 4(sin
2 0 + 2 sin 0 sin 20 + sin
2 20).
dy/d0 = 2 cos 0 + 2 cos 20,
(rfy/d0)
2 = 4(cos
2 0 + 2cos0cos20 + cos
2 20). So,
[Note that sin 0 sin 20 + cos 0 cos 20 = cos (20 -0) = cos0.] Since 1 + cos 0 = 2cos
2 (0/2), VI + cos 0 =
V2 |cos (0/2)|.
Thus, we have:
2V2[J 0 " V2 cos (0/2) d0 + J
2lr - V2cos (0/2) d0] = 4{2sin (0/2) ]J -
2 sin (0/2) }
2 J} = 8[(1 - 0) - (0 - 1)] = 16.
du, where u is the parameter. In this case,