ENGR 233 Study Guide - Final Guide: Level Set, Unit Vector, Dot Product
31 May 2018
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Solutions to Midterm Exam Emat 233
February, 2006
Problem 1. Consider the surface given by the formula
1 = cos(zx)ez+y
(a) Find the equation of the tangent plane at the point (π, 0,0)
Solution: Let f(x, y, z) = cos(zx)ez+y. Then the surface given by 1 = cos(zx)ez+yis the level surface {(x, y, z) :
f(x, y, z)=1}. To find the tangent plane at a point we first need to find a normal vector to the surface at that point,
which is given by the gradient
~
∇f=∂f
∂x i+∂f
∂y j+∂f
∂z k.
Computing the partial derivatives (using the chain rule, the product rule and the fact that ez+y=ezey), we get:
∂f
∂x =−zsin(zx)ez+y
∂f
∂y = cos(zx)ez+y
∂f
∂z =−xsin(zx)ez+y+ cos(zx)ez+y.
Plugging in x=π, y = 0, z = 0 gives:
~
∇f(π, 0,0) = −0 sin(0)e0i+ cos(0)e0j+ (−πsin(0)e0+ cos(0)e0)k= 0i+ 1j+ (−0 + 1)k=j+k.
This is a normal vector to the plane. Therefore any vector in the plane, of the form (x−π, y −0, z −0) must be
perpendicular to this vector, i.e.
0=(j+k)·((x−π)i+yj+zk) = 0(x−π) + y+z.
This gives the equation of the plane:
y+z= 0.
(b) Find normal line in symmetric or in parametric form passing through the point (π, 0,0).
Solution: Since we already know a normal vector at that point, namely ~
∇f(π, 0,0) = j+k= (0,1,1),all we need to
do is give the equation of the line passing through (π, 0,0) in the direction of this vector. The parametric equation is:
~r(t) = (π, 0,0) + t(0,1,1),
or in terms of coordinates: x=π, y =t, z =t.
To get the symmetric equations we eliminate the tso that we have x=π, y =z.
Problem 2 Let r(t) = 3 cos(t)i+ 2 sin(t)j+tkbe the position vector of a moving particle.
(a) Find the velocity vector v(t)and the acceleration vector a(t)at any t.
Solution: The velocity is the derivative of the position vector:
v(t) = r0(t) = −3 sin(t)i+ 2 cos(t)j+ 1k,
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Document Summary
1 = cos(zx)ez+y (a) find the equation of the tangent plane at the point ( , 0, 0) Solution: let f (x, y, z) = cos(zx)ez+y. Then the surface given by 1 = cos(zx)ez+y is the level surface {(x, y, z) : f (x, y, z) = 1}. To nd the tangent plane at a point we rst need to nd a normal vector to the surface at that point, which is given by the gradient. Computing the partial derivatives (using the chain rule, the product rule and the fact that ez+y = ezey), we get: Plugging in x = , y = 0, z = 0 gives: ~ f ( , 0, 0) = 0 sin(0)e0i + cos(0)e0j + ( sin(0)e0 + cos(0)e0)k = 0i + 1j + ( 0 + 1)k = j + k. This is a normal vector to the plane.
