gradient fields and exact differentials

Question

A. let f(x,y)= xe^xy, and C be the path y=1/x from (1,1) to (0,∞)

      ^{1. calculate F=∇f}

      ^{2. Calculate the integral ∫_cF*dr}

          ^{- directly}

         ^{- by using the fundamental theorem for line integrals}

 

B. Left f(x,y) = sinxcosx

      1. calculate F=∇f

 

 

Answer 1

A) The gradient  is defined as ∇f = <f_x, f_y>

 

1) Taking the partials we have F = <xye^xy + e^xy, x²e^xy>

 

2) The line integral is defined as ∫cF•dr = ∫F(r(t))•r`(t)dt

 

To find r(t) parameterize y = 1/x, from (1,1) to (0,0)

 

Let x = t then y = 1/t where 1≤t≤0 then r(t) = <t, 1/t> and r'(t) = <1, -1/t²>.

 

To find F(r(t)) substitute the x and y values of r into F we have F(r(t)) = <t(1/t)e(^t(1/t)) + e^(t(1/t)), t²e^(t(1/t))> = <2e, t²e>.

 

∫F(r(t))•r`(t)dt = ∫ from 1 to 0 of <2e, t²e>•<1, -1/t²>dt = ∫ from 1 to 0 of (2e - e)dt = e*t from 1 to 0 = -e

 

Using The Fundamental Theorem of Calculus for Line Integrals we have

 

f(rb)) - f(r(a)) = f(0,0) - f(1,1) = 0(0)e(0(0)) - 1(1)e(1(1)) = -e

 

You should be able to get part B now.  If you have any questions just ask.

 

Hi Sue, great that you got first part of B.  This is the first time I have come across a question like you asked.   I have looked into three different calculus books and cannot find any problems similar to the last two questions. However I will give it my best shot and you can decide if this make sense.

 

We know that F = ∇f = <cos²(x) - sin²(x), 0>.  Notice that our gradient does not depend on y since f_y = 0 so this problem reduces down to a single variable.  From the Fundamental Theorem for Line integrals we know that the value of the line integral can be evaluated knowing just the starting and endpoints of C so ∫cF•dr = f(x₂,y₂) - f(x₁, y₁) = sin(x₂)cos(x₂) - sin(x₁)cos(x₁), where (x1,y1) is the starting point and (x2,y2) is the ending point of C.  The value of this function maximum when sin(x2)cos(x2) is its maximum and sin(x1)cos(x1) is its minimum.  We can find the critical number by setting f_x = 0 and solving for x.

 

cos²(x) - sin²(X) = 0 → cos²(x) = sin²(x) → 1 = sin²(x)/cos²(x) → tan²(x) = 1 → tan(x) = ± 1 therefore

 

x = π/4 + πk or x = - π/4 + πk

 

This gives us x = {- π/4, π/4, 3π/4, 5π/4,....}

 

At x = - π/4 we have f(- π/4, y) = sin(- π/4)cos(- π/4) = -0.5, the minimum value, ending point

At x = π/4 we have f(π/4, y) = sin(π/4)cos(π/4) = 0.5, the maximum value, starting point

 

So the maximum value is 0.5 - (-.05) = 1

 

Now for the path it will be along the tangent curve y = tan(x) from (π/4, 1) to (- π/4, -1).  Since tangent is periodic in π we can write y = tan(x) from (π/4 + πk, 1) to (- π/4 + πk, -1) which gives all paths.