Answer to Question #246033 in Calculus for Njabulo

Question

1. L1 and L2 are lines in R
3
. A parametric equation for L1 is
(x, y, z) = (1, 0, 2) + t(2, 4, −1); t ∈ R.
L2 is parallel to L1 and contains the point (3, 4, 3). V is the plane that contains both the lines L1 and L2. Find
an equation for V . [7]
2. Prove from first principles that
lim
(x,y)→(2,2)
(x − y) = 0.
[8]
3. Prove that lim
(x,y)→(0,0)
x
2 + y
x − y
does not exist. [8]
4. Consider the R
2 − R function f defined by
f (x, y) = x
2 − 2xy + y
2
.
Let V be the plane that is tangent to the graph of f at the point (x, y, z) = (1, −1, 4).
(a) Find a vector that is perpendicular to the plane V . (4)
(b) Find an equation for the plane V .

Answer 1

1,

$\overline{a}=(2,4,-1)||V$

$A(1,0,2)\isin L_1$

$B(3,4,3)\isin L_2$

$\overline{AB}=(2,4,1)||V$

$\overline{n}=[\overline{a},\overline{AB}]=\begin{vmatrix} i & j&k \\ 2 & 4&-1\\ 2&4&1 \end{vmatrix}=8i-4j=(8,-4,0)$

equation for V:

$8(x-1)-4y=0$

$2x-y-2=0$

2.

 Let $\varepsilon$ > 0 be given.

triangle inequality:

$|x-y|=|(x-2)-(y-2)|\le|x-2|-|y-2|$

Also:

$|x-2|\le\sqrt{|x-2|^2}\le|(x-2),(y-2)|$

and

$|y-2|\le|(x-2),(y-2)|$

 Therefore:

$|x-y|\le2|(x-2),(y-2)|$

 Set $\delta=\varepsilon/2$ , then if

$0<2|(x-2),(y-2)|<\delta$

we have

$|y-2|\le|(x-2),(y-2)|<2\delta=\varepsilon$

3.

 On the curve C₁: x=0

$\displaystyle{\lim_{(x,y)\to(0,0)}}\frac{x^2+y}{x-y}=\displaystyle{\lim_{(x,y)\to(0,0)}}\frac{y}{-y}=-1$

On the curve C₂: y=0

$\displaystyle{\lim_{(x,y)\to(0,0)}}\frac{x^2+y}{x-y}=\displaystyle{\lim_{(x,y)\to(0,0)}}\frac{x^2}{x}=0$

 As the limits are not equal, $\frac{x^2+y}{x-y}$  cannot have a limit at (0,0).

4.

a)

Let

$g(x,y,z)=x^2-xy-y^2-z$

$gradg$ is normal to the graph of f.

$gradg(x,y,z)=(2x-2y,2y-2x,-1)$

Therefore $gradg(1, −1, 4) = (4, −4, 1)$ will be normal to the tangent plane V .

b)

$(x,y,z)(4,-4,-1)=(1,-1,4)(4,-4,-1)$

$4x-4y-z=4$