2. The graph shown below is that of the bifolium: (x² + y²)² = 16 x²y (2) Equivalently, this bifolium is describable as the locus of all points: (x,y) = (rcos 0,r sin 0) where r = 16 cos² sin for 0°≤ ≤ 180°. in eq (2). (a) Apply the method of Implicit Differentiation to find (b) Ray OP makes an angle of 0 = 45° with the positive x-axis and |OP| = r = 4√2. So it follows that point P has the coordinates (x,y) where x = r cos 0 = 4 and y = rsin 0 = 4. Use these coordinates for P and your result from part (a) to show that the tangent line to the graph at P has zero slope. (c) Ray OQ makes an angle of 0 = 30° with the positive x-axis and |OQ| = r = 6. Thus, point has coordinates (x,y) where x = r cos 0 = 3√3 and y = r sin 0 = 3. Use these coordinates for Q and your result from part (a) to show that the tangent line to the graph at Q is vertical. 2√3 Ө 300 30° 45° (d) Ray OR makes an angle of 8 = 60° with the positive x-axis and |OR| = r = 2√3. Point R has coordinates (x, y) where x = r cos 0 = √√√3 and y = rsin 0 = 3. Find the equation of the tangent line to the bifolium at the point R as well as the angle of inclination, a, of this tangent line. (Recall that tan a = m.) 60° cos sin EENIN √3 2 √2 hyp=r 2 1 2 adj-x-r cose 2 √2 2 √3 2 (x,y) opp-y-r sin 0
2. The graph shown below is that of the bifolium: (x² + y²)² = 16 x²y (2) Equivalently, this bifolium is describable as the locus of all points: (x,y) = (rcos 0,r sin 0) where r = 16 cos² sin for 0°≤ ≤ 180°. in eq (2). (a) Apply the method of Implicit Differentiation to find (b) Ray OP makes an angle of 0 = 45° with the positive x-axis and |OP| = r = 4√2. So it follows that point P has the coordinates (x,y) where x = r cos 0 = 4 and y = rsin 0 = 4. Use these coordinates for P and your result from part (a) to show that the tangent line to the graph at P has zero slope. (c) Ray OQ makes an angle of 0 = 30° with the positive x-axis and |OQ| = r = 6. Thus, point has coordinates (x,y) where x = r cos 0 = 3√3 and y = r sin 0 = 3. Use these coordinates for Q and your result from part (a) to show that the tangent line to the graph at Q is vertical. 2√3 Ө 300 30° 45° (d) Ray OR makes an angle of 8 = 60° with the positive x-axis and |OR| = r = 2√3. Point R has coordinates (x, y) where x = r cos 0 = √√√3 and y = rsin 0 = 3. Find the equation of the tangent line to the bifolium at the point R as well as the angle of inclination, a, of this tangent line. (Recall that tan a = m.) 60° cos sin EENIN √3 2 √2 hyp=r 2 1 2 adj-x-r cose 2 √2 2 √3 2 (x,y) opp-y-r sin 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:2. The graph shown below is that of the bifolium:
(x² + y²)² = 16 x²y
(2)
Equivalently, this bifolium is describable as the locus of all points:
where r = 16 cos²0 sin for 0°≤ 0 ≤ 180°.
(x,y) = (rcos 0,r sin 0)
(a) Apply the method of Implicit Differentiation to find
in eq (2).
(b) Ray OP makes an angle of 0 = 45° with the positive x-axis and |OP| = r = 4√2.
So it follows that point P has the coordinates (x,y) where x = r cos 0 = 4
and y = r sin0 = 4. Use these coordinates for P and your result from
part (a) to show that the tangent line to the graph at P has zero slope.
(c) Ray OQ makes an angle of 0 = 30° with the positive x-axis and |OQ| =r= 6.
Thus, point Q has coordinates (x,y) where x = r cos 0 = 3√3 and
y = r sin 0 = 3. Use these coordinates for Q and your result from part (a)
to show that the tangent line to the graph at Qis vertical.
√₂-
0
600
(d) Ray OR makes an angle of 0 = 60° with the positive x-axis and |OR| = r = 2√3.
Point R has coordinates (x, y) where x = r cos 0 = √√3 and y = rsin0 = 3.
Find the equation of the tangent line to the bifolium at the point R
of inclination, a, of this tangent line. (Recall that tan a = m.)
as well as the angle
2√3
5.50
4√2
0
0
30°
45°
60°
cos sin
√3
NINANE
NONN
√2
hyp = r
adj-x-r cos 0
(x,y)
opp-y-r sin 0
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