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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

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter6: Applications Of The Derivative
Section6.CR: Chapter 6 Review
Problem 20CR
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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
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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