Question

By setting one variable constant, find a plane that intersects the graph of z=(x^2+1)sin(y)+xy^2 in a:

(a) parabola
(b) straight line
(c) sine curve

Answer 1

@ikram002p

Answer 2

try setting x= a constant and see what you get, then see what you would have to set x to to get a sine curve.
then try setting y= a constant and see what you get

Answer 3

x=2

5sin(2)+2y^2
4.5+2y^2

Answer 4

idk what u mean by a sine curve

Answer 5

im sorry i typed that in wrong

Answer 6

x=2

5sin(y)+2y^2

Answer 7

if u put x=0 then
\[\large z=(x^2+1)\sin y+xy ^2 \]
becomes
\[\large z=(0+1)\sin y+0\cdot y^2=\sin y \]
which is a sine curve

Answer 8

see what u get with
\[\large y=\pi/2 \]
and
\[\large y=\pi \]

Answer 9

when y=pi/2

x^2+1+x*(pi^2/4)

Answer 10

when y=pi

xpi^2

Answer 11

thank you so much for your help!!

Answer 12

right. if \(y=\pi/2\) then
\[\large z=(x^2+1)\sin y+xy^2 \]
becomes
\[\large z=(x^2+1)\sin(\pi/2)+x(\pi/2)^2 \]
\[\large z=(x^2+1)\cdot1+x(\pi^2/4) \]
\[\large z=x^2+1+(\pi^2/4)x \]
\[\large z-1=x^2+(\pi^2/4)x+(\pi^2/8)^2 \]
\[\large z-1=(x+\pi^2/8)^2 \]
is the parabola on the xz-plane with vertix \((-\pi^2/8,0,1)\).

Answer 13

if \(y=\pi\) then
\[\large z=(x^2+1)\sin\pi+x\pi^2 \]
\[\large z=(x^2+1)\cdot0+x\pi^2 \]
\[\large z=\pi^2x \]
which is not a line, it is actually a plane (remember u r in R^3). [I made a mistake].

I don't think u can get a line from that equation (at least i don't see how)