280
Fig. 34-13
34.46
Show that, if R(t) traces out a curve and the parameter trepresents time, then R(f) is the velocity vector, that is, its
direction is the direction of motion and its length is the speed.
By Problem 34.45, we already know that R'(') has the direction of the tangent vector along the curve. By
Problem 34.43, R'(t) = (dx/dt, dy/dt), since R(t) = (x(t), y(t)). Hence, |R'(Ol = ^J(dx/dt)2 + (dy/dt)2 =
dsldt, where s is the arc length along the curve (measured from some fixed point on the curve). But dsldt is the
speed. [The "speed" is how fast the end of the position vector R(f) is moving, which is the rate of change of its
position s along the curve.]
34.47
If R(s) is a vector function tracing out a curve and the parameter s is the arc length, show that the tangent vector
R'(s) is a unit vector, that is, it has constant length 1.
34.48
For the curve R(t) = (t, t
2 ), find the tangent vector v = R'(0» the speed, the unit tangent vector T, and the
acceleration vector a = R"(034.49 Find the unit tangent vector for the circle R = (a cos 6, a sin 6).
34.51
Find the tangent vector v and acceleration vector a for the ellipse R(f) = (a cos t, b sin t), and show that a is
opposite in direction to R(t) and of the same length.
v = R'(t) = (-a sin t, b cos t), a = R"(0 = (~a cos t, -b sin t) = -(a cos t, b sin () = -R(034.52
Find the magnitude and direction of the velocity vector for R(t) = (e
1 , e
2
' — 4e' + 3) at t = 0.
R'(t) = (e',2e
2 '-4e') = e'(l,2e' + 4). When f = 0, R'(0) = (l,-2), |R'(0)| = V5, and the vector
R'(0) is in the fourth quadrant, with an angle e = tan'
1 (-2) = -63°26'.
34.53 Find the velocity and acceleration vectors for R(t) = (2 - t, 2? - t) at f=l.
R'(0 = (-l,6f
2 ), R"(/) = (0,120- Hence, R'(l) = (-l,6) and R"(l) = (0,12).
34.54
If R(M)=/(M)F(M), show that R'(u) =f(u)F'(u) +f'(u)V(u), analogous to the product formula for ordinary
derivatives.
CHAPTER 34
v = (l,20 and a = (0,2). The speed is
and the unit tangent vector T =
R'(0) = (-a sin 8, a cos 6). \R'(0)\ =
= a. Hence, T = R'(0)/|R'(0)| = (-sin 6»,cos 0).
34.50
Find the unit tangent vector for the curve R(0) = (e", e ").
Hence, the unit tangent vector
Fig. 34-13
34.46
Show that, if R(t) traces out a curve and the parameter trepresents time, then R(f) is the velocity vector, that is, its
direction is the direction of motion and its length is the speed.
By Problem 34.45, we already know that R'(') has the direction of the tangent vector along the curve. By
Problem 34.43, R'(t) = (dx/dt, dy/dt), since R(t) = (x(t), y(t)). Hence, |R'(Ol = ^J(dx/dt)2 + (dy/dt)2 =
dsldt, where s is the arc length along the curve (measured from some fixed point on the curve). But dsldt is the
speed. [The "speed" is how fast the end of the position vector R(f) is moving, which is the rate of change of its
position s along the curve.]
34.47
If R(s) is a vector function tracing out a curve and the parameter s is the arc length, show that the tangent vector
R'(s) is a unit vector, that is, it has constant length 1.
34.48
For the curve R(t) = (t, t
2 ), find the tangent vector v = R'(0» the speed, the unit tangent vector T, and the
acceleration vector a = R"(034.49 Find the unit tangent vector for the circle R = (a cos 6, a sin 6).
34.51
Find the tangent vector v and acceleration vector a for the ellipse R(f) = (a cos t, b sin t), and show that a is
opposite in direction to R(t) and of the same length.
v = R'(t) = (-a sin t, b cos t), a = R"(0 = (~a cos t, -b sin t) = -(a cos t, b sin () = -R(034.52
Find the magnitude and direction of the velocity vector for R(t) = (e
1 , e
2
' — 4e' + 3) at t = 0.
R'(t) = (e',2e
2 '-4e') = e'(l,2e' + 4). When f = 0, R'(0) = (l,-2), |R'(0)| = V5, and the vector
R'(0) is in the fourth quadrant, with an angle e = tan'
1 (-2) = -63°26'.
34.53 Find the velocity and acceleration vectors for R(t) = (2 - t, 2? - t) at f=l.
R'(0 = (-l,6f
2 ), R"(/) = (0,120- Hence, R'(l) = (-l,6) and R"(l) = (0,12).
34.54
If R(M)=/(M)F(M), show that R'(u) =f(u)F'(u) +f'(u)V(u), analogous to the product formula for ordinary
derivatives.
CHAPTER 34
v = (l,20 and a = (0,2). The speed is
and the unit tangent vector T =
R'(0) = (-a sin 8, a cos 6). \R'(0)\ =
= a. Hence, T = R'(0)/|R'(0)| = (-sin 6»,cos 0).
34.50
Find the unit tangent vector for the curve R(0) = (e", e ").
Hence, the unit tangent vector