INSTRUCTOR’S SOLUTIONS MANUAL
SECTION 2.3 (PAGE 115)
y 1 At x D 2 we have y D 3 and y 0 D 2. Thus, the equation yD x of the tangent line is y D 3 2.x 2/, or y D 2x C 7. The normal line is y D 3 C 12 .x 2/, or y D 12 x C 2. Full file at https://testbankuniv.eu/Calculus-Several-Variables-Canadian-9th-Edition-Adams-Solutions-Manual b
1 a
a;
1 1 0 , y D1 x x2 1 For horizontal tangent: 0 D y 0 D 1 so x 2 D 1 and x2 x D ˙1 The tangent is horizontal at .1; 2/ and at . 1; 2/
43. y D x C
44. If y D x 2 .4 0
y D 2x.4
x
x 2 /, then Fig. 2.3-47 2
2
3
x / C x . 2x/ D 8x
4x D 4x.2
2
x /:
48.
The slope of a horizontal line must be zero, so p 4x.2 x 2 / D 0, which impliesp that x D 0 or x D ˙ 2. At x D 0; y D 0 and at x D ˙ 2; y D 4. Hence, there are two horizontal lines that are tangent to the curve. Their equations are y D 0 and y D 4. 45. y D
1 , y0 D x2 C x C 1
Thus 2x C 1 D 0 and x D
1 2
The tangent is horizontal only at
46. If y D
ˇ dy ˇˇ ˇ dx ˇ
xD1
2x C 1 . .x 2 C x C 1/2
49.
1 4 ; . 2 3
˙ 12 ,
2
or .x C 2/ D
50.
1. Hence, the
3a2 C 4 D .a
2/2 .a C 1/:
The tangent to y D x 2 =.x mD
1 4:
xD1
1 : 2
D
The two tangent lines to y D x 3 passing through .2; 8/ correspond to a D 2 and a D 1, so their equations are y D 12x 16 and y D 3x C 2.
.x C 2/.1/ .x C 1/.1/ 1 D : .x C 2/2 .x C 2/2
1 D 4; .x C 2/2
D
ˇ ˇ ˇ ˇ
3=2 ˇ
The tangent to y D x 3 at .a; a3 / has equation y D a3 C 3a2 .x a/, or y D 3a2 x 2a3 . This line passes through .2; 8/ if 8 D 6a2 2a3 or, equivalently, if a3 3a2 C 4 D 0. Since .2; 8/ lies on y D x 3 , a D 2 must be a solution of this equation. In fact it must be a double root; .a 2/2 must be a factor of a3 3a2 C 4. Dividing by this factor, we find that the other factor is a C 1, that is, a3
In order to be parallel to y D 4x, the tangent line must have slope equal to 4, i.e.,
1 x 2
The product of the slopes is .2/ 12 D two curves intersect at right angles.
xC1 , then xC2 y0 D
.x
1/ at .a; a2 =.a ˇ 1/2x x 2 .1/ ˇˇ a2 D ˇ 2 ˇ .x 1/ .a xDa
1// has slope 2a : 1/2
The equation of the tangent is 3 2
5 . 2
Hence x C 2 D and x D or At x D y D 1, and at x D 52 , y D 3. Hence, the tangent is parallel to y D 4x at the points 3 ; 1 and 52 ; 3 . 2 47.
xD1
2x C 1 .x 2 C x C 1/2
For horizontal tangent we want 0 D y 0 D
1 Since p D y D x 2 ) x 5=2 D 1, therefore x D 1 at x theˇ intersection point. The slope of y D x 2 at x D 1 is ˇ 1 2x ˇˇ D 2. The slope of y D p at x D 1 is x
3 , 2
Let the point of tangency be .a; a1 /. The slope of the tanb a1 1 2 gent is D . Thus b a1 D a1 and a D . 2 a 0 a b b2 b2 Tangent has slope so has equation y D b x. 4 4
y
a2 a
1
D
a2 .a
2a .x 1/2
a/:
This line passes through .2; 0/ provided 0
a2 a
1
D
a2 .a
2a .2 1/2
a/;
or, upon simplification, 3a2 4a D 0. Thus we can have either a D 0 or a D 4=3. There are two tangents through .2; 0/. Their equations are y D 0 and y D 8x C 16.
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Full file at https://testbankuniv.eu/Calculus-Several-Variables-Canadian-9th-Edition-Adams-Solutions-Manual
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