source : physicsforums.com

## Finding the point of intersection between two curves. (Vectors)

## Homework Statement

At what point do the curves r1(t) = (t, 4-t, 63+t^2) and r2(s)= (9-s, s-5, s^2) intersect?Answer in the form: (x,y,z) = ____

Find the angle of intersection theta to the nearest degree.

## Homework Equations

## The Attempt at a Solution

i: t=9-sj: 4-t=s-5k: 63+t^2=s^2

i/j: t-9sk: 63+(9-s)^2=s^2″Solving for “s””s=8t=1…I know not what to do from here. 🙁

[Solved] At what point do the curves… | Course Hero – (LR-2) Plot the points (x, y) to obtain a scatterplot. Use an appropriate scale on the horizontal and vertical axes and be sure to label carefully. V.4. (1 point) Find and sketch the projections of the curve on the three coordinate planes. r(t) = t, sin(t), 2 cos(t). The curves might intersect, but we need to know whether the objects are in the same position at at the same time. Suppose the trajectories of two particles are given by the vector functionsthe problem this onto what point? There was a curves are won t He won Manistee re past his squire on DH attitude us. It's the culture three months us as ministry as squared in this act wind in your angle of intersection corrupt to At what points does the curve $ r(t) = ti + (2t – t^2) k $ intersect the par…

PDF SOLUTIONS – Students have improved their scores by an avg. 106 points using this exhaustive course that comes with an integrated study plan, and more. One way to do this is to manipulate the original stem from r-s > 2t (given) into r-t > t-s (another way to see the stem info).Thus, the lines intersect at point(2, 3). Now we have to find the gradient of a line perpendicular to the line in Equation 3 Since the points of intersection are located where one curve equals the other, the following equations describing those curves must also equal each other, when [math]x[/math] equals…Solution: Using time in seconds and distances in feet, we have x = (115 cos 50o)t and y = 3 + (115 sin 50o)t − 16t2. The height at time tˆ is y(tˆ) ≈ 11.20 feet, so it will clear the fence. 4. Given the position vector r(t) = (3t − t3)i + 3t2j − 3k, nd the tangential and normal compo-nents of acceleration.

SOLVED:At what point do the curves r_1 (t) = \lan – wizard123 wizard123. Set components equal: From this we find: Substitute and solve system: The point where they intersect isHere are two paths r1(t) and r2(t) intersect if there is a point P lying on both curves. We say that r1(t) and r2(t) collide if r1(t0) = r2(t0) at some time t0. If u(t) = (sin t, cos t, t) and v(t) = (t, cos t, sin t), use Formula 4 of Theorem 3 to find View Answer. If a curve has the property that the position vector r(t) is…The curve C1 is through the point P. If r(t) = ⟨x(t) , y(t) , z(t)⟩ is parametrization for the curve C1 with r(t0) = P, then since the points of C1 are on the surface, we have This shows that the vector ∇F(x(t) , y(t) , z(t)) is perpendicular to the vector r′(t) = ⟨x′(t) , y′(t) , z′(t)⟩. Especially, at t = t0. we will have that…