Find equations fot the osculation, normal, and rectifying planes of the curve r(t)=ti+t3k at the point (1, 1, 1).

Question

Find equations fot the osculation, normal, and rectifying planes of the curve r(t)=ti+t3k at the point (1, 1, 1).

Oung1985 · Accepted Answer

Step1 Given: The curve r(t)=ti+t2j+t3k. The point (1, 1, 1). Step 2 To find: Osculating, normal, and rectifying planes of the curve r(t)=ti+t2j+t3k. If r(t)=(x(t),y(t),z(t)) be vector-valued function and p0(x0,y0,z0) be a point on a curve c generated by vector-valued function r(t0)→. Then The normal plane of P is perpendicular to r3(t0).→. The rectifying plane of P is perpendicular to [r3(t0)→×r(t0)→}×r3(t0)→. The osculating plane P is perpendicular to r3(t0)×r(t0). Now, if ∇f(x0,y0,z0) is perpendicular to the tangent vector at point P ( corresponding to t0 and passes though the point P(x0,y0,z0) then the equation of tangent plane at point P is, ∇f(x0,y0,z0). (x−x0,y−y0,z−z0)=0. Step3 To find a normal plane corresponding to the point (1,1,1) that is at t=1 first find the vector perpendicular to these planes Find r3(t)→ at t=1. Here r(t)→=ti+t2j+t3k ⇒r3(t)→=i+2tj+3t2k ⇒r3(t)→=(1,2t,3t2) ⇒r3(t)→=(1,2,3) Therefore the equation of the tangent plane is. 1(x−1)+2(y−1)+3(z−1)=0 x−1+2y−2+3z−3−0 x+2y+3z−6=0 Step 4 To find the osculating plane corresponding to the point (1,1,1) tht is at t=1 need to find a vector NS perpendicular to these planes. To find an osculating curve need to find

Find equations fot the osculation, normal, and rectifying planes of

Answered question

Answer & Explanation

New Questions in Calculus 2