More Applications of Derivatives 199
8 7 6 5 4 3 2 1
0123456789
t
v
v(t)
(feet/sec)
(seconds)
Figure 9.7-1
- Find the Cartesian equation for the curve
defined byr=4 cosθ. - The motion of an object is modeled by
x=5 sint, y= 1 −cost. Find the
y-coordinate of the object at the moment
when itsx-coordinate is 5. - Calculate 4u− 3 vifu=
〈
6,− 1
〉
and
v=
〈
−4, 3
〉
.
- Determine the symmetry, if any, of the
graph ofr=2 sin(4θ). - Find the magnitude of the vector 3i+ 4 j.
9.8 Solutions to Practice Problems
Part A The use of a calculator is not
allowed.
- Equation of tangent line:
y=f(a)+f′(a)(x−a)
f′(x)=
1
4
( 1 +x)−^3 /^4 (1)=
1
4
( 1 +x)−^3 /^4
f′(0)=
1
4
and f(0)=1;
thus,y= 1 +
1
4
(x− 0 )= 1 +
1
4
x.
f(0.1)= 1 +
1
4
(0.1)= 1. 025
- f(a+Δx)≈ f(a)+ f′(a)Δx
Letf(x)=^3
√
xand f( 28 )=
f( 27 + 1 ).
Thenf′(x)=
1
3
(x)−^2 /^3 ,
f′( 27 )=
1
27
, and f( 27 )=3.
f( 27 + 1 )≈ f( 27 )+f′( 27 )( 1 )≈
3 +
(
1
27
)
( 1 )≈ 3. 037
- f(a+Δx)≈ f(a)+f′(a)Δx
Convert to radians:
46
180
=
a
π
⇒a=
23 π
90
and 1◦=
π
180
;
45 ◦=
π
4
.
Let f(x)=cosxandf(45◦)=
f
(
π
4
)
=cos
(
π
4
)
=
√
2
2
.
Thenf′(x)=−sinxand
f′( 45 ◦)=f′
(
π
4
)
=−
√
2
2
f( 46 ◦)= f
(
23 π
90
)
= f
(
π
4
+
π
180
)
f
(
π
4
+
π
180
)
≈f
(
π
4
)
+
f′
(
π
4
)(
π
180
)
≈
√
2
2
−
(√
2
2
)(
π
180
)
≈
√
2
2
−
π
√
2
360