~Question:  How curved is a curve?

~Answer: It varies, depends on where we are on the curve.

~Question: Is there a number that measures the degree of "curvedness"?

~Answer: Yes, it's called curvature.

~Question: Is it a positive or negative number?

~Answer:  Could be positive or negative, depends on the shape of the curve at that point.   

~Question:  What do you mean by shape?

~Answer: If the curve is concave down at that point, curvature will be negative. If the curve is concave up there, the curvature will be positive.                    

~Question:  Why?

~Answer: The definition of curvature is given by the derivative (instantaneous rate of change) of the angle the tangent line makes with the positive directed horizontal axis (inclination angle) with respect to arc length along the curve (s), so, this will be negative when drawn to a point where the curve is concave down (inclination angle decreases) and positive where the curve is concave up (inclination angle increases).

~Note: There is an equivalent way of defining curvature using vectors. In calculus II and III, parametric representations of curves are mainly used to analyze the motion of a point on plane and space curves.

~Question: How does one calculate the curvature at a point?

~Answer: By use of a formula which I will derive.

~Let's formalize the definition in mathematical terminology.
 Curvature = k = dq/ds  (the rate of change of the inclination angle (q) wrt arc length)

~The curve can be in parametric form (x and y considered functions of time t) or by a direct relationship (i.e., y=f(x) or x=f(y)).

~In calculus II, you will derive the integral that gives arc length for all 3 cases.
In parametric form s = ∫ [(dx/dt)² + (dy/dt)²]^(1/2) dt.

~Then ds/dt = [(dx/dt)² + (dy/dt)²]^1/2  by the fundamental theorem.

~Since  dy/dx = tan(q) by definition,  q = tan^-1(dy/dx)

~But in parametric form dy/dx = [(dy/dt) / (dx/dt)] by the chain rule.

~So, q = tan^-1[(dy/dt) / (dx/dt)].  hang in there.

~So,  dq/dt =  {[(dx/dt)(d²y/dt² - (dy/dt)(d²x/dt²)]/(dx/dt)²]} divided by
{1+[(dy/dt)/(dx/dt)]²}....ugh!

~Finally, k = dq/ds = (dq/dt) / (ds/dt) = (the above mess) divided by
[(dx/dt)² + (dy/dt)²]^(1/2)....even worse!

~Let's make this easier to read. Let dx/dt = x*,  dy/dt = y*, d²x/dt² = x**, d²y/dt² = y**.

~So, our formula becomes:  k = (x*y**-y*x**) / (x*² + y*²)^(3/2)...a little better, but still quite involved. (Dots are usually placed over the x's and y's in conventional courses)

~So, there we have it. All we have to do is apply it & get our curvature number.  Remember, we need parametric form for the curve to use this formula.  There are other formulas for curvature that do not need parametric form (coming up).

~Let's do some examples:

~Example: Find the curvature of a straight line.  Guess first.

~Let x=t,  y=mt+b represent the line in parametric form. By the way,
there are many parametric forms of a curve that can be formulated,
this is just one.

~x*=1, y*=m, x**=0, y**=0,  so,  k = [(1)(0)-(m)(0)] / [(1²)+m²]^(3/2)=0
Did you guess right?  makes sense, yes?

~Question: Is there a formula for curvature if we have a direct relationship, y = f(x) or x = f(y)?

~Answer:  yes.  Let's derive it.

~We use arc length for these types of functions. For y=f(x) , ds/dx = [1+(dy/dx)²]^(1/2) and for x=f(y), ds/dy = [1+ (dx/dy)²]^(1/2).

~So, since k = dq/ds and q = tan^-1(dy/dx) or cot^-1(dx/dy), we get , k=(dq/dx)/(ds/dx) or (dq/dy)/(ds/dy) giving us the following;

k= d²y/dx² / [1+(dy/dx)²]^(3/2) or d²x/dy² / [1+(dx/dy)²]^(3/2)

~So, applying this to y=mx+b, using the first, we get, k = 0/ (1+m²)^(3/2) = 0, much easier.

~Example: Find the point on y=lnx where there is maximum curvature.

~Use the formula for y=f(x).  Since y' = 1/x, y''= -1/x².
So, k= (-1/x²)/[1+(1/x)²]^(3/2).
Note that for all x in the domain of this function, the curvature is negative. Makes sense since this curve is always concave down.

~We need to maximize this expression. So, find where dk/dx = 0. Use the quotient rule to find dk/dx and simplify the expression. Then find where the numerator of dk/dx = 0 and you will get x=1/√2. Then pick up the y value of (-1/2)ln2.  (I'll leave this as an exercise for you)...nice of me.

~Using a parametric representation for a circle of radius a, one can show that the curvature will be constant and equal to 1/a.  makes sense.

~If we draw a circle with "closest contact" to the curve at the point in question (on the concave side), it will be called the "circle of curvature" for that point. The radius of this circle is called the "radius of curvature" for that point & the center of this circle is called the "center of curvature for that point". Note that there are many circles that can be drawn tangent to the curve at that point, but only one has the "closest contact". This means the first & second derivatives of the curve agree with the first & second derivatives of the circle at that point.  (the more derivatives having the same value for two curves, the closer the curves approximate each other)

~So, a sharp turn on a curve, has a small circle of curvature, thus a small radius of curvature. On the other hand, a straighter curve has a very large circle of curvature & radius of curvature. The circle of curvature & radius of curvature become undefined for a straight line.

~Much of this is all tied together in calculus II & III when analyzing the motion of a particle on a plane or space curve. Tangent & normal Components of the vectors involved at a point on the curve include the velocity, speed, acceleration, and the curvature of the path. By analyzing these forces, one can see what it takes to keep a point on a given path (important when driving your car along a curved road).