Find the unit tangent, normal and binormal vectors T N B , and the curvature of the curve?
Find the unit tangent, normal and binormal vectors T, N, B , and the curvature of the curve x=2t y=t^2 z=4t^3 at t=1. T=( )i +( )j+ ( )k N=( )i +( )j+ ( )k B=( )i +( )j +( )k k=_____________ K so it's second time i ask this question...the last answer by KB unfortunately is wrong...i know it's wrong because the first three values of the problem i.e the T(t) vector are 38/sqrt38; 38/sqrt38 and 12/3sqrt38.. KB factored incorrectly the 4 from the sqrt on the denominator and it made the whole answer wrong....there's a link to the question from yest....please someone help me solve that problem i have been nesting on it for several hours from diff angles and i cant get the right answers...thank you http://answers.yahoo.com/question/index;_ylt=Ajwo5U4V8Rd9H2StUwgTSxEazKIX;_ylv=3?qid=20121023182944AANm7DL
Tangent, Normal, Binormal Vectors. Will give 50 points for help!?
First of all, a straight line has zero curvature or torsion. The tangent is the line itself. The normal vector is the function's derivative. The binormal is the cross-product of the unit tangent and unit normal vectors (divide by their lengths). Your vector function is simply (1, 1, 1) + a(1, 2, 3). Can you find the derivative from there?
Help finding unit tanget vector, unit normal vector, binormal vector, vector equation?
r(t) = <3sin(t), 3cos(t), 4t> r'(t) = <3cos(t), -3sin(t), 4> |r'(t)| = √13 T(t) = r'(t) / |r'(t)| T(t) = <3cos(t), -3sin(t), 4> / 5 T'(t) = <-3sin(t), -3cos(t), 0> / 5 |T'(t)| = 3/5 N(t) = T'(t) / |T'(t)| N(t) = <-3sin(t), -3cos(t), 0> / 3 N(t) = <-sin(t), -cos(t), 0> B = T × N B = 3/5
What are applications of the unit tangent, unit normal, and binormal vectors?
Consider how you would define directions in an arbitrary place out in space. Almost anywhere there is going to be a dominant gravitational field that defines the up-down axis, and the matter swirling around it will have a dominant direction of motion which defines east-west, and then north-south is perpendicular to both (with a handedness rule defining which is north). Following one particular bit of matter, the unit tangent is the "east" (direction it is presently moving toward), the unit normal is "down" (component of the second derivative that is perpendicular to the first derivative) and the binormal is "north". These provide a shifting framework of 3D axes with standard meaning.
Principal Normal vector and Binormal vector problem?
Suppose that the position vector of a moving particle is given by r(t) = e^(6t) cos(4t) i + e^(6t) sin(4t) j + (sqrt(4/3)) e*(6t) k. (a) Find the principal normal vector N (b) Find the binormal vector B Please Help!!
How do you find the equation of a tangent/normal line of a curve at a given point?
To find the equations of the tangent and the normal (perpendicular) lines at a given point on a curve, you need to know two things: 1.) the slope of the curve at the given point and 2.) a point (x, y) that each line passes through. Once we know this information, we can use it to determine the desired equation for each line.We can use the above slope and point information to find the desired equations by plugging this information into the “point-slope” form of the equation of a straight line, i.e., y ‒ y1 = m(x – x1), where m is the slope of the line and (x1, y1) is a specific point that the line passes through. We know that both the tangent and normal lines pass through and intersect at a given point (x1, y1) (a point of tangency) on the given curve. To find the slope of a curve and of the tangent line to the curve at a given point (x1, y1) on the curve, we need to know the equation, y = f(x), which algebraically represents the given curve and which defines a function f. Once we know y = f(x), we can then find the derivative f ′(x) of the function f. Graphically, the derivative f ′(x) is the slope of a curve at any point (x, y) on the curve, as well as the slope of the tangent line to the curve at the same point (x, y), provided f ′(x) exists; therefore, the slope of a given curve and of the tangent line to the curve at a given point (x1, y1) on the curve is given by m = f ′(x) = f '(x1), provided the derivative f ′(x) at x = x1 exists. Therefore, the equation of a tangent line to a curve at a given point (x1, y1) is found as follows:y ‒ y1 = m(x ‒x1)y ‒ y1 = [f '(x1)](x ‒ x1) y ‒ y1 + y1 = [f '(x1)](x ‒ x1) + y1 y + 0 = [f '(x1)](x ‒ x1) + y1 y = [f '(x1)](x ‒ x1) + y1 We know that the slopes of two perpendicular lines are negative reciprocals of one another, i.e., if m is the slope of one, then the slope of the other is equal to ‒1/m, i.e., their product, m(‒1/m), equals ‒1.Since the line normal to a curve at a given point (x1, y1) is also the same line that is perpendicular to the tangent line to the curve at the same point (x1, y1), then the slope of the normal line = ‒1/m = ‒1/f ′(x).Therefore, the equation of a normal (perpendicular) line to a curve at a given point (x1, y1) is found as follows: y ‒ y1 = m(x ‒x1)y ‒ y1 = [‒1/ f '(x1)](x ‒ x1)y ‒ y1 + y1 = [‒1/ f '(x1)](x ‒ x1) + y1y + 0 = [‒1/ f '(x1)](x ‒ x1) + y1y = [‒1/ f '(x1)](x ‒ x1) + y1
Needs Calculus 3 Help With Normal and Binormal Vectors?
First of all, r simplifies to r(t) = (cos t, -sin t, -2t). Taking derivatives, r'(t) = (-sin t, -cos t, -2). Since ||r'(t)|| = √5, T(t) = r'(t) / ||r'(t)|| = (1/√5) * (-sin t, -cos t, -2). At t = π/6, T = (1/√5) * (-1/2, -√3/2, -2). Next, T'(t) = (1/√5) * (-cos t, sin t, 0). Since ||T'(t)|| = 1/√5, N(t) = T'(t)/||T'(t)|| = (-cos t, sin t, 0). At t = π/6, N = (1/√5) * (-√3/2, 1/2, 0). Therefore, B(π/6) = T(π/6) x N(π/6) = (1/√5) * (1, √3, -1) κ = ||T'(t)|| / ||r'(t)|| = (1/√5) / (√5) = 1/5 (this actually holds for all t, including π/6) I hope this helps!
Binormal vector help.?
I've started working on a problem to determine the binormal vector at 2 different points (t = 0 and t = 1) but it seems to have gotten incredibly complicated and I'm not sure if I messed up or if there is a shortcut I'm missing. I have r(t) = (t, 2t, t^3) I found r'(t) = (1, 2, 3t^2) Using T(t) = r'(t)/||r'(t)|| I got that T(t) = (1/sqrt(5+9t^4), 2/sqrt(5+9t^4), 3t^2/sqrt(5+9t^4)) As you can see finding T'(t), finding N(t), and eventually finding the cross product of the two is just ridiculous so I feel like I'm missing something here. So If I did something silly early on or if there is a much more simplified method to finding binormals please let me know.