# Calculating Tangent Vector to Curve: $\varphi$

• skrat
In summary, the tangent vector to a curve is given by the equation $$t=\frac{\partial }{\partial \varphi }s(\varphi) \vec e_r+s\frac{\partial }{\partial \varphi }\vec e_r. skrat ## Homework Statement Calculate the tangent vector to a curve$$r=R(1-\varepsilon \sin ^2 \varphi)$$as a function of ##\varphi## ## Homework Equations ## The Attempt at a Solution Ok, I tried like this: I defined a vector$$(f(\varphi),\varphi)=(R(1-\varepsilon \sin ^2 \varphi),\varphi)$$and than I calculated the derivative$$\frac{\partial }{\partial \varphi}$$therefore my result is$$(-\varepsilon R \sin(2\varphi),1).$$But I have a friend claiming that the result is$$(-\varepsilon R \sin(2\varphi),R(1-\varepsilon \sin ^2 \varphi)).$$So, my question to you is... which is right? :/ Looks like you've computed the normal. You need to find a vector such that your vector dotted with it give zero. I think both of you are wrong. First of all: What basis are you using to write down your vector? Unless you specify a basis, your two numbers mean nothing. What basis is your friend using? Edit: Second, how do you define the tangent vector of a curve? Orodruin said: First of all: What basis are you using to write down your vector? Unless you specify a basis, your two numbers mean nothing. What basis is your friend using? It's polar basis ##r## being the distance from the origin and ##\varphi ## the polar angle. The function$$R(1-\varepsilon \sin ^2\varphi)$$turns out to be an ellipse. ##R## and ##\varepsilon ## are two constants, where ##\varepsilon \ll 1##. skrat said: It's polar basis ##r## being the distance from the origin and ##\varphi ## the polar angle. And what is the definition of the tangent vector of a curve? Yeah that could be the reason why my result is wrong. If I am not mistaken, the vector is ##(f(\varphi),\varphi)## and normalized tangent vector should be$$t=\frac{(\frac{\partial f(\varphi)}{\partial \varphi},\frac{\partial }{\partial \varphi}\varphi)}{|(f(\varphi),\varphi)|}$$skrat said: Yeah that could be the reason why my result is wrong. If I am not mistaken, the vector is ##(f(\varphi),\varphi)## and normalized tangent vector should be$$t=\frac{(\frac{\partial f(\varphi)}{\partial \varphi},\frac{\partial }{\partial \varphi}\varphi)}{|(f(\varphi),\varphi)|}$$Your big problem is that you are trying to only deal with the components of a vector, and even the components you quote are wrong. For any curve ##\vec r(t)##, the tangent vector is given by ##\vec t = d\vec r/dt##. In polar coordinates, the position vector ##\vec r## is given by ##\vec r = r \vec e_r##, where ##\vec e_r## is the radial unit vector. There is no component in the angular direction. However, you do need to note that ##\vec e_r## is not a constant vector and in order to differentiate this with respect to time, you will also need to consider the time derivative of ##\vec e_r##. This is a big disadvantage of trying to write everything as a list of numbers only. The first step is to stop referring to vectors as a list of numbers without any reference as to what basis is being used. Ok, let's write$$\vec r=R(1-\varepsilon \sin ^2 \varphi)\vec e_r=s(\varphi)\vec e_r$$than tangent vector ##\vec t## is$$\vec t=\frac{\partial }{\partial \varphi }s(\varphi) \vec e_r+s\frac{\partial }{\partial \varphi }\vec e_r.$$The second term is$$\frac{\partial }{\partial \varphi }e_r=\frac{\partial }{\partial \varphi }(\cos\varphi, \sin \varphi)=(-\sin\varphi,\cos\varphi) $$but IF (and only if) I am not mistaken, than by definition$$\vec e_\varphi =s(-\sin\varphi,\cos\varphi).$$Meaning the tangent vector is$$t=\frac{\partial s(\varphi)}{\partial \varphi }\vec e_r + \vec e_\varphi .$$skrat said:$$\vec e_\varphi =s(-\sin\varphi,\cos\varphi).$$There is an ##s## too much in front here. Remember that ##\vec e_\varphi## is a unit vector. Meaning the tangent vector is$$t=\frac{\partial s(\varphi)}{\partial \varphi }\vec e_r + \vec e_\varphi .

And one too few in front of the ##\vec e_\varphi## here. Otherwise you are doing well now.

skrat

## 1. What is a tangent vector?

A tangent vector is a vector that is tangent to a curve at a specific point. It is a mathematical concept used to describe the direction and magnitude of change of a curve at a particular point.

## 2. How is a tangent vector calculated?

The tangent vector to a curve is calculated by taking the derivative of the curve's equation at a specific point. This derivative represents the slope of the curve at that point, which is the direction of the tangent vector. The magnitude of the tangent vector can be calculated using the Pythagorean theorem.

## 3. What is the purpose of calculating a tangent vector?

Calculating a tangent vector allows us to determine the direction and rate of change of a curve at a specific point. This information is useful in many scientific and mathematical fields, such as physics, engineering, and geometry.

## 4. How is a tangent vector used in real-world applications?

Tangent vectors have many practical applications, such as in physics to calculate the acceleration of an object in motion, in engineering to determine the direction of force on a structure, and in computer graphics to create smooth animations of moving objects.

## 5. Are there different types of tangent vectors?

Yes, there are two types of tangent vectors: the unit tangent vector and the normal tangent vector. The unit tangent vector has a magnitude of 1 and is used to describe the direction of the curve, while the normal tangent vector is perpendicular to the unit tangent vector and is used to describe the curvature of the curve.

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