194 STEP 4. Review the Knowledge You Need to Score High
Step 3: Sincer=〈
s·√
2
2
t,3+s·√
2
2
t− 16 t^2〉
, the derivative isr′=〈
s·√
2
2,3+
s·√
2
2
− 32 t〉
, and the ball will attain its maximum height when the ver-tical component 3 +
s·√
2
2
− 32 t is equal to zero. Since s ≈ 105 .556,3 +105. 556 ·
√
2
2
− 32 t =0 producest ≈ 2 .462 seconds. For that valueof t, r =〈
105. 556√
2
2(2.462), 3+
105. 556
√
2
2(2.462)−16(2.462)^2
〉
≈
〈 183 .762, 89. 779 〉. The ball will reach a maximum height of 89.779 feet, when
it is 183.762 feet from home plate.
Example 3
Find the velocity, acceleration, tangent, and normal vectors for an object on a path defined
by the vector-valued functionr(t)=〈
etcost,etsint〉
whent=
π
2.
Step 1: v(t) =r′(t)=〈
et(cost−sint),et(sint+cost)〉. When evaluated at t =
π
2
,
v(
π
2)
=〈
−eπ/^2 ,eπ/^2〉
≈〈
− 4 .810, 4. 810〉. The velocity vector is
〈− 4 .810, 4. 810 〉.
Step 2: a(t)=
〈
− 2 etsint,2etcost〉. Evaluated att=
π
2
, this isa
(
π
2)
=〈
− 2 eπ/^2 ,0〉≈〈
− 9 .621, 0〉
.
Step 3: The tangent vector is given by T(t) =
r′(t)
‖r′(t)‖. Since r′(t) =
〈
−etsint+etcost,etsint+etcost
〉
, the tangent vector becomes T(t) =
〈
−etsint+etcost,etsint+etcost〉
√
(−etsint+etcost)^2 +(etsint+etcost)^2, which simplifies to T(t) =
〈
cost−sint
√
2,
sint+cost
√
2〉. When t =
π
2
, the tangent vector is
〈
− 1
√
2,
1
√
2〉
=〈
−√
2
2,
√
2
2〉
.Step 4: The normal vector N(t) =
T′(t)
‖T′(t)‖=
〈
cost−sint
√
2,
sint+cost
√
2〉′∥∥
∥∥
∥〈
cost−sint
√
2,
sint+cost
√
2〉∥∥
∥∥
∥=
〈
−cost−sint
√
2,
cost−sint
√
2〉
.Att =
π
2
,N(t) =〈
− 1
√
2,
1
√
2〉
=
〈
−√
2
2,
√
2
2〉. CheckT
(
π
2)
·N(
π
2)
=−
√
2
2·
−
√
2
2+
√
2
2·
−
√
2
2=
1
2
−
1
2
= 0
to be certain the tangent and normal vectors are orthogonal.