Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q11E

Expert-verified
Essential Calculus: Early Transcendentals
Found in: Page 780
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

(a) Find a function \(f\) such that \({\bf{F}} - \nabla f\) and (b) use part (a) to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\) along the given curve\(C\).

\({\bf{F}}(x,y) - x{y^2}{\bf{i}} + {x^2}y{\bf{j}}\).

(a)

The potential function \(f\) of vector field \({\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}\) is\(f(x,y) = \frac{1}{3}{x^3}{y^3}\)

(b)

The value of \(\int_C V f - dr\) along the curve \(C\) is \( - 9\).

See the step by step solution

Step by Step Solution

Step 1: Given information

The given equation is \({\bf{F}}(x,y) - \left( {y{e^x} + \sin y} \right){\bf{i}} + \left( {{e^x} + x\cos y} \right){\bf{j}}\)

Step 2: Definition of conservative vector field

Consider vector function \({\bf{r}}(t),a \le t \le b\) with a smooth curve\(C\). Consider \(f\) is a differentiable function two or three variables of gradient function \(\nabla f\) and is continuous on curve\(C\). Then,

\(\int_c \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))\)

Step 3: find the function f of conservative vector field

(a)

Vector field is \({\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}\) -

Consider\(\nabla f = {f_s}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}\).

Write the relation between the potential function \(f\) and vector field \({\bf{F}}\).

\(\nabla f = {\bf{F}}\)

Substitute \({f_x}(x,y){\rm{i}} + {f_y}(x,y){\bf{j}}\) for \(\nabla {f_{{\rm{, }}}}\),

\({\bf{F}} = {f_x}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}\)

Compare the equation \({\bf{F}} = {f_x}(x,y){\bf{i}} + {f_y}(x,y){\bf{j}}\) with \({\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}\).

\(\begin{aligned}{l}{f_x}(x,y) = {x^2}{y^3}(1)\\{f_y}(x,y)& = {x^3}{y^2}(2)\end{aligned}\)

Integrate equation (1) with respect to\(x\).

\(\begin{align} & f(x,y)=\int{\left( {{x}^{3}}{{y}^{3}} \right)}dx\\ & ={{y}^{3}}\int{{{x}^{2}}}dx\\ & ={{y}^{3}}\left( \frac{{{s}^{3}}}{3} \right)+g(y)\left\{ \because \int{{{t}^{2}}}dt=\frac{{{t}^{3}}}{3} \right\}\\ & f(x,y)=\frac{1}{3}{{x}^{3}}{{y}^{3}}+g(y)(3)\end{align}\)

Apply partial differentiation with respect to y on both sides of equation (3).

\(\begin{align} & {{f}_{y}}(x,y)=\frac{\partial }{\partial y}\left( \frac{1}{3}{{x}^{3}}{{y}^{3}}+g(y) \right) \\ & =\frac{\partial }{\partial y}\left( \frac{1}{3}{{x}^{3}}{{y}^{3}} \right)+\frac{\partial }{dy}(g(y)) \\ & =\frac{1}{3}{{x}^{3}}\frac{\partial }{\partial y}\left( {{y}^{3}} \right)+{{g}^{\prime }}(y) \\ & =\frac{1}{3}{{x}^{3}}\left( 3{{y}^{2}} \right)+{{g}^{\prime }}(y)\left\{ \because \frac{\partial }{id}{{t}^{n}}=n{{l}^{n-1}} \right\} \\ & {{f}_{y}}(x,y)={{x}^{3}}{{y}^{2}}+{{g}^{\prime }}(y)(4) \end{align}\)

Compare the equations (2) and (4).

\({g^\prime }(y) = 0\)

Apply integration on both sides of equation.

\(\begin{align} & \int{{{g}^{\prime }}}(y)dy=\int{0}dy \\ & g(y)=K\left\{ \because \int{0}dt=K \right\} \end{align}\)

Here,

\(K\) is constant.

Substitute \(K\) for \(g(y)\) in equation (3),

\(f(x,y) = \frac{1}{3}{x^3}{y^3} + K(5)\)

Consider the value of \(K\) as 0 .

Substitute 0 for \(K\) in equation (5),

\(\begin{array}{c}f(x,y) = \frac{1}{3}{x^3}{y^3} + 0\\ = \frac{1}{3}{x^3}{y^3}\end{array}\)

Thus, the potential function \(f\) of vector field \({\bf{F}}(x,y) = {x^2}{y^3}{\bf{i}} + {x^3}{y^2}{\bf{j}}\) is \(f(x,y) = \frac{1}{3}{x^3}{y^3}\).

Step 4: Evaluate the given function along curve c

Consider vector function \({\bf{r}}(t),a \le t \le b\) with a smooth curve\(C\). Consider \(f\) is a differentiable function two or three variables of gradient function \(\nabla f\) and is continuous on curve\(C\). Then,

\(\int_c \nabla f \cdot d{\bf{r}} = f({\bf{r}}(b)) - f({\bf{r}}(a))(8)\)

Write the expression of curve.

\({\bf{r}}(t) = \left\langle {{t^3} - 2t,{t^3} + 2t} \right\rangle (7)\)

Substitute 0 for \(t\) in equation\((7)\),

\(\begin{aligned}r(0) & = \left\langle {{0^3} - 2(0),{0^3} + 2(0)} \right\rangle \\ & = \langle 0,0\rangle \end{aligned}\)

Hence, the initial point of curve \((a)\) is \((0,0)\).

Substitute 1 for \(t\) in equation\((7)\),

\(\begin{aligned}{\bf{r}}(1) & = \left\langle {{1^3} - 2(1),{1^3} + 2(1)} \right\rangle \\ & = \langle 1 - 2,1 + 2\rangle \\ & = \langle - 1,3\rangle \end{aligned}\)

Hence, the terminal point of curve \((b)\) is\(( - 1,3)\).

Find the common value \(\int_C \nabla f \cdot dr\) by using equation\((6)\).

\(\begin{aligned} & \int_{C}{\nabla }f\cdot dr=f(-1,3)-f(0,0) \\ & =\left[ \frac{1}{3}{{(-1)}^{3}}{{(3)}^{3}} \right]-\left[ \frac{1}{3}{{(0)}^{3}}{{(0)}^{3}} \right]\ \ \ \left\{ \because f(x,y)=\frac{1}{3}{{x}^{3}}{{y}^{3}} \right\} \\ & =-9-0 \\ & =-9 \end{aligned}\)

Thus, the value of \(\int_C V f - dr\) along the curve \(C\) is \( - 9\).

Most popular questions for Math Textbooks

Icon

Want to see more solutions like these?

Sign up for free to discover our expert answers
Get Started - It’s free

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.