University of Newcastle Calculus Assignment Solution

This assignment solution addresses problems related to calculus, specifically focusing on parametric equations, tangent vectors, and plane equations. The solution starts by finding the tangent vector of a given parametric curve. The solution then proceeds to find the equation of the tangent line to the curve at a specific point. The assignment also includes solving for the parametric equation of the line through two points, finding the equation of a plane, and determining the intersection of the line and the plane. The solution demonstrates the application of calculus concepts to solve these problems, providing detailed steps and explanations. The solution is a valuable resource for students studying calculus, especially those enrolled in courses like MATH2310 at the University of Newcastle, providing a clear understanding of the concepts and problem-solving techniques.

3a.
Find the tangent vector of the parametric curve r (s)=( x , y , z)
Solution
r ( s)=(2 cos s , 3 sin s , 2 π −s )
This can also be represented as the unit vector as below
r͢ ͢ (s)=¿
Finding the derivative of r
r ' (s)= δ x
δ s ¿
¿ (−2 sin s ) i+¿
Now putting S=o and substituting in the derivative obtained
¿ (−2 sin( 0) ) i +¿
=¿ { ( 0 ) i , ( 3 ) j ,−k }
Therefore, the tangent vector is ¿ 0,3 ,−1> ¿
3b.
Find the equation of the tangent line to the curve at point (
0 ,−3 , π /2 ¿
Solution

Given that x=2 cos s , y=3 sin s∧¿ z=2 π−s on coordinates 0 ,−3 , π (¿ 2)
Finding the parametric values of of s that corresponds to the
coordinates (0 ,−3 , π /2 ¿
0=2 cos s , −3=3 sin s and z=2 π−s
Now, we find the parametric value for s that satisfy all the three
equations.
We find that S=270, all the three equations are satisfied.
r ( s )=¿
Now finding the derivative of r
r ' ( s )=¿)
Substituting the value of s i.e. s=270 or s= 1.5 π
r ' ( s )=¿)
r ' ( s )=¿)
Applying the point formula
¿ ¿)*(x-0, y+3, z- π /2 ¿
¿ ¿)
¿ 2 x+ 0−z + π /2
Therefore, the equation of the tangent is

F(x,y,z)¿ 2 x−z+ π /2