**Q. (14.4 #30) I’m not sure if 4N is a parameter in physics that has its own meaning or it’s the normal vector or it’s just simply a letter.**

Thanks for asking, as others might be wondering about this too. The letter stands for “Newton”, which is a unit of force. So this is wind blowing with 4 units of force. A Newton is the same as one (a kilogram metre per second squared). Remember that force is mass times acceleration and this makes some sense: a kilogram is mass, and metres per second squared is a unit of acceleration. Since the gravitational acceleration of the earth, is measured in , and the projectile’s weight is given in , all the units in this problem work out without needing any conversions.

**Q. (14.3 #60) I’m not sure how to use the knowledge we learned to get such a polynomial. I can guess by using graphing program but I can not associate it with the things when learned in class.**

This question asks you to find a polynomial of the form

so that the piecewise function in the problem is a) continuous; b) has continuous slope; and c) has continuous curvature. The relation to what we’ve done in class is mostly in part (c). We learned how to find the curvature of a function like (see Equation 11 and Example 5 in Section 14.3).

So my suggestion on how to start: leave as unknowns in the definition of and , find the slope and curvature of , and then check whether a) is continuous, b) its slope is continuous, and c) its curvature are continuous. (By check whether they are, I mean figure out what needs to be true about in order that they are.) There are two important places ( and ) to do the check, so that’s 6 conditions. From these conditions you should be able to find that will make all the necessary things continuous.

**Q. I was wondering if we could use simple T, N, and B vectors for simplicity when finding equations for normal and osculating planes. I did this method for 14.3 #46. I just found those vectors without making them as unit vectors (because unit vectors get complicated). I think this is still correct since the ratios between x,y,z are still the same without being unit vectors. Am I correct? And if so, can we just simplify thru this method for quizzes and midterm?**

Yes, you are correct. Suppose we have two vectors in the same direction,

and

If we form the equation of the plane with the normal vector and passing through , we get

.

But if we use , we get

.

The second equation can be written

.

And multiplying the entire equation by does not change the solutions, or the plane it represents. So these two planes are exactly the same! So yes, you can change the normal vector by any scalar multiple in order to make it more convenient to write the plane. In particular, you can just find instead of , because these two vectors are in the same direction.

**Q. I was curious about #46 of 14.3. In the textbook they calculate the normal plane using the vector T(t). Is it equivalent to use the vector N(t) to calculate the normal plane? The equations for the planes in each case are different but are they inherently the same?**

No, they are different. When you calculate a plane from a vector , you are finding the plane made up of all the vectors perpendicular to . Since and point in different directions (they are always perpendicular to each other!), the planes you make from them will be different.

Part of the confusion here might be the names. The word “normal” in calculus usually means “perpendicular.” But of course, this begs the question, “perpendicular to what?” The “normal vector” is perpendicular to . The “normal vector” of a plane (the vector you make the plane from) is perpendicular to the plane. So these are two different sorts of “normal vectors” and they have nothing to do with each other! Very frustrating, but we’re stuck with it for now.