**Computer Aided Geometric Design** Class Exercise #4 Differential Geometry of Curves 1

**Q.1** Let be an arc-length parameterized curve. Prove: is a straight line. is a plane curve. 2

**Solution - Part 1 (** Recall that is the magnitude of the curvature vector: Therefore, and by integration: for some real constants 3

**Solution - Part 1 (** This already means: The direction of the curve is in fact a constant direction. To verify that this is a line, we integrate again and: 4

**Solution - Part 1 (** and we obtained the parameterization of the straight line passing through with the direction (Conversely – the same with differentiation!) 5

**Solution - Part 1 (** Question: Where did we use the fact that the curve was arc-length parameterized? Is a general parameterization with not necessarily a straight line? 6

**Solution - Part 1 (** Answer: This should also be true for non arc-length parameterization… Recall that for non arc-length: 7

**Solution - Part 1 (** if then: is the zero vector. When does a cross product vanish? 8

**Solution - Part 1 (** when and are the same (or opposite) directions. We understand this as follows: The tangent never changes direction, only magnitude. 9

**Solution - Part 2 (** How can we show that a curve is planar? (without the knowledge of this being equivalent to which we are trying to prove….) 10

**Solution - Part 2 (** We must show that the curve is contained in some plane. A general (implicit) representation of a plane: where is a constant (non-zero) vector, and . 11

**Solution - Part 2 (** In our case: we show that: where is the bi-normal at some point. 12

**Solution - Part 2 (** let for some . since: we have: which means is a constant vector. 13

**Solution - Part 2 (** Now we look at: and therefore: and is planar. 14

**Solution - Part 2 (** conversely: If a curve is planar then it is in fact contained in the (constant) osculating plane. This means that the bi-normal is constant: 15

**Solution - Part 2 (** Points to think about: Is this the case for non arc-length parameterizations? What if ? Is the torsion well defined? 16

**Q.2** Let be an arc-length parameterized curve. Prove that if all normal vectors to the curve pass through a point, then is an arc of a circle. 17

**Solution** It suffices to show that: is planar. has constant curvature. 18

**Solution** The assumption on the normal vectors means there exists a scalar function such that: Where is a constant point in . 19

**Solution** Differentiating gives: Which means: 20

**Solution** or: 21

**Solution** We expressed the zero vector as a linear combination of at every . Therefore: 22

**Solution** In the case , we get , and the entire curve is a point. In the other case: , as required. 23

**Solution** Some points to think about: Any space curve contained in a sphere can have the radius from any point pass through the center of the sphere, and yet it is not necessarily a circle. So is our proof right or not?? 24

**Solution** Shouldn’t something bother us about writing: 25

**Q.3** Let be a parameterized quadratic curve. Prove that is planar. 26

**Solution** Again, it suffices to show that the bi-normal is constant. This shall imply that the osculating plane is constant, and the curve never leaves it. 27

**Solution** is quadratic. We may assume: 28

**Solution** Differentiating gives: 29

**Solution** We know that has the same direction as the unit tangent . Does has the same direction as the unit normal ? 30

**Solution** Not necessarily. It is true for arc-length parameterizations. (in fact for any constant speed…) But: is contained in the plane spanned by . 31

**Solution** This can be shown analytically. Geometrically, it is explained by the fact that changes the tangent in direction and in magnitude. Therefore: the direction is in the bi-normal direction! 32

**Solution** It suffices to check that is constant. For example, the first component of this cross product is: which does not depend on . Others are shown similarly. 33

**Solution** Thus, the bi-normal direction is constant – and the curve is planar. 34

**Q.4** Consider the curves: and: 35

**Q.4** Find rigid motions on such that it will be pieced together with and the resulting curve is: Continuous (). Geometrically . . 36

**Solution** The curves are the following: 37

**Solution** has a tangent: and at 38

**Solution** is the lower right quarter of the unit circle, inflated to a radius of length . For continuity we simply translate by: 39

**Solution** This gives: 40

**Solution** For Geometrically we need first to rotate such that the tangents have the same direction. Since: and: 41

**Solution** The angle between these tangents is given by: 42

**Solution** Which gives: or: It remains to rotate and translate: 43

**Solution** We have the rotation angle, but what is , the translation vector? After the rotation we have: 44

**Solution** The image of under the rotation is: Then: 45

**Solution** Finally: 46

**Solution** In order for the resulting curve to be , we need to change the magnitude of the tangent. Denote by the speed ratio: 47

**Solution** Re-parameterize as follows: Instead of: take: This results in the same trace, but the speed is factored by , as required. 48

**Solution** Re-parameterize as follows: Instead of: take: This results in the same trace, but the speed is factored by , as required. 49

**Solution** Some points to think about: How do we re-parameterize as a single curve? In practice – how do we rotate without explicitly invoking ? 50