STEM-Master
Calculus IIChapter 6: 3-Dimensional SpaceSection 6-6

Vector Functions — Question 2

PDF ↗

Question 2

For r→(t)=⟨3t,t2,2−t⟩\vec r(t)=\left\langle 3t,t^2,2-t\right\rangle, eliminate the parameter to describe the curve as the intersection of two surfaces. State its orientation as tt increases and locate the point corresponding to t=−1t=-1.

Original worksheet page 1: question and worked solution for 6-6-002
Show solutionHide solution

Question 2 – Solution

Strategy Solve the simplest component for tt and substitute into the others.

See the diagram in the original worksheet below.

Elimination Since x=3tx=3t, t=x/3t=x/3. Therefore y=x2/9,z=2−x/3.\boxed{y=x^2/9,\qquad z=2-x/3}. The curve is the intersection of a parabolic cylinder and a plane.

Orientation and point As tt increases, xx increases and zz decreases. At t=−1t=-1, the point is (−3,1,3)\boxed{(-3,1,3)}.

Verification Substituting these coordinates into both surface equations confirms membership.

Original worksheet page 2: question and worked solution for 6-6-002

Original worksheet layout. Use Enlarge or open the PDF for a closer view.

← Section questionsChapter contents