What shape are possible cross section cylinder?
1 answer:
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This question is incomplete, the complete question is;
Find the gradient of the function g(x,y)= xy² at the point (5, - 1). Then sketch the gradient together with the level curve that passes through the point.
First find the gradient vector at (5, - 1).
Vg(5,- 1) = [ ]i - [ ]j (Simplify your answers.)
Choose the graph that shows the level curve and the gradient vector at (5, - 1).
options of the sketched graphs are uploaded along this answers
Answer:
Vg(5,- 1) = [ 1 ]i - [ 10 ]j
Option A is the correct graph for this Level
Step-by-step explanation:
Given that;
g(x,y) = xy²
we have to find gradient of this function at ( 5, -1 )
so
Δg(x,y) = dg/dx + dg/dy
d(g)/dx = y² , dg/dy = 2gx
therefore
Δg(x,y) = [y²]i + [2yx]j
Δ( 5, -1) = [-1²]i + [2×-1 ×5]j
=Δ(5, -1) = 1i - 10j
[ 1 ]i - [ 10 ]j
Therefore Option A is the correct graph for this Level
