Sketch the level curves for f(x,y) = x^2-y together with

garnentas3m

garnentas3m

Answered question

2022-01-04

Draw the gradient f and level curves for f(x,y)=x2y at a few common locations. For the point where the level curve f(x,y)=1 meets the tangent line, write an equation (2,1).

Answer & Explanation

Shannon Hodgkinson

Shannon Hodgkinson

Beginner2022-01-05Added 34 answers

Step 1
If y=f(x) , then the independent variable x is called the input of the function and the dependent variable y is called the output of the function. The derivative of the function is the measure of how fast the function is changing with respect to x. A multi variable function f(x,y,z) contains more than one inputs.
Step 2
The gradient of a multivariable function f(x,y) is defined as ffxi+fyj. To find gradient of f(x,y)=x2y, substitute it in ffxi+fyj. According to standard results in partial differentiation xnx=nxn1 and ynx=nyn1, where n is an integer.
f=fxi+fyj
=ix(x2y)+jy(x2y)
=2ξj
To find the gradient at a particular point (x0,y0), substitute it in f=2ξj and find the corresponding vectors. Thus the value of gradient corresponding to the points that lies on the y axis can be calculated by substituting y=0 in f=2ξj
f=2ξj
=0ij
=j

temnimam2

temnimam2

Beginner2022-01-06Added 36 answers

Second part:
The equation of the tangent line to the function y=f(x) at x=x0 is given by y=f(x0)+f(x0)(xx0). In this problem the f(x,y)=1, since f(x,y)=x2y the equation of curve in the x-y plane becomes x2y=1 or y=x21.
To find the equation of the tangent line to the curve y=x21 at (2,1), find the value of f(x)=x21 and its derivative at x=2 and then substitute obtained values in y=f(x0)+f(x0)(xx0)
f(x)=x21
f(2)=(2)21
=1
f(x)=2x
f(2)=22
y=f(2+f(2)(x2)
=1+22(x2)
=22x222+1
=22x3

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