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Example 4

Find the angle between the two curves 2y2 = x3 and y2 = 32x at their point of intersection in the Ist quadrant

Solution : 2y2 = x3 ® (1) and

y2 = 32x ® (2)

Solving (1) and (2), we get

2 (32x) = x3

\ x3 - 64x = 0

\ x (x2 - 64) = 0

\ x = 0 and x = ± 8

As the point of intersection is in the Ist quadrant it must be +ve and ¹ 0

\ x = 8

Putting in y2 = 32x

we get y2 = 32(8)

\ y = 256

\ y = ±8

Accepting y = 16, we have the point of intersection is P = (8,16) which lies in the Ist quadrant


Now Differentiating (1) and (2) w. r. to x, we get

Let q be the acute angle between curves (1) and (2) at P.

 

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Index

5.1 Tangent And Normal Lines
5.2 Angle Between Two Curves
5.3 Interpretation Of The Sign Of The Derivative
5.4 Locality Increasing Or Decreasing Functions 5.5 Critical Points
5.6 Turning Points
5.7 Extreme Value Theorem
5.8 The Mean-value Theorem
5.9 First Derivative Test For Local Extrema
5.10 Second Derivative Test For Local Extrema
5.11 Stationary Points
5.12 Concavity And Points Of Inflection
5.13 Rate Measure (distance, Velocity And Acceleration)
5.14 Related Rates
5.15 Differentials : Errors And Approximation

Chapter 6





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