Homework Statement

grad f = (3x2 + y2)i + (2xy - 3)j

a) find f

b) evalue the integral of grad f dot dr where c is any path from 1,2 to 2,1

c)evaluate the integral of gradf dot dr where c is the curve given by 4x2 + 9y2 = 36

The Attempt at a Solution

a) f = x3 + xy2 - 3j

b) integral grad f dot dr = f(Q) - f(P) = (1 + 4 - 12) - (8 + 2 - 3) = -14

c) integral grad f dot dr = integral F(r(t)) dot r'(t) is this the correct method

r(t) = 4cost i + 9sint j
r'(t) = -4sint i + 9cost j

F(r(t)) = (3(4cost)2 + (9sint)2)i + (2(4cost)(9sint) - 3)j

F(r(t)) dot r'(t) = (3(-4sint)(4cost)2 + (-4sint)(9sint)2) + (2(9cost)(4cost)(9sint) - 3(9cost)

= (3(-4sint)(4cost)2 + (-4sint)(9sint)2) + (2(9cost)(4cost)(9sint) - 3(9cost)

=$$\int$$ -192(cost)2sint + 324(sint)3 + 648(cost)2(sint)2 - 27cost dt from 0 to pi

is this correct thus far

gabbagabbahey
Homework Helper
Gold Member
a) f = x3 + xy2 - 3j

Be more mindful of typos; I'm sure you meant to write " f = x3 + xy2 - 3y", right?

b) integral grad f dot dr = f(Q) - f(P) = (1 + 4 - 12) - (8 + 2 - 3) = -14

I see two problems here:

(1) $3*2\neq 12$

(2) You seem to be going from P=(2,1) to Q=(1,2), but the problem statement you posted suggests that the curve goes from P=(1,2) to Q=(2,1)

c) integral grad f dot dr = integral F(r(t)) dot r'(t) is this the correct method

Careful, $\vec{r}'(t)=\frac{d\vec{r}}{dt}\neq d\vec{r}$. Intstead you should have written $d\vec{r}=\vec{r}'(t)dt$

r(t) = 4cost i + 9sint j

This parameterization does not describe the correct curve. Using your parameterization, x(t)=4cos(t) and y(t)=9sin(t)....That would mean that $4x^2+9y^2=64\cos^2 t+ 729\sin^2 t=64+665 sin^2 t\neq 36$

b) f(Q) - f(P) = (8+2-3) - (1+4-6) = 5

c) dont i have to keep them as vectors

r(t) = 64cos2t i + 729sin2t j

r'(t) = -64sin2t i - 729sin2t j

gabbagabbahey
Homework Helper
Gold Member
b) f(Q) - f(P) = (8+2-3) - (1+4-6) = 5

$(8+2-3)-(1+4-6)=(7)-(-1)=8\neq 5$

c) dont i have to keep them as vectors

r(t) = 64cos2t i + 729sin2t j

r'(t) = -64sin2t i - 729sin2t j

I'm not exactly sure what you are trying to ask here, but you do understand that any 2D curve can be parameterized as $\vec{r}(t)=x(t)\hat{i}+y(t)\hat{j}$ right?

So when you write $\vec{r}(t)=4\cos(t)\hat{i}+9\sin(t)\hat{j}$, you are essentially saying that at every point along your curve, $x(t)=4\cos(t)$ and $y(t)=9\sin(t)$. This isn't true though, because you are told that $4x^2+9y^2=36$ at every point on your curve; and $x(t)=4\cos(t)$ and $y(t)=9\sin(t)$ don't satisfy that equation.

You need to choose an $x(t)$ and $y(t)$ that do satisfy the equation $4x(t)^2+9y(t)^2=36$ for all values of t.

r(t) = 3cost i + 2sint j
r'(t) = -3sint i + 2cost j

f = x3 + xy2 - 3y

f(r(t)) = (3cost)3 + (3cost)(2sint)2 - 3(2sint)

I then have to dot it with r'(t), but my f(r(t)) is not in vector form, there are no i,j,z compnents, how do i proceed

gabbagabbahey
Homework Helper
Gold Member
r(t) = 3cost i + 2sint j
r'(t) = -3sint i + 2cost j

f = x3 + xy2 - 3y

f(r(t)) = (3cost)3 + (3cost)(2sint)2 - 3(2sint)

I then have to dot it with r'(t), but my f(r(t)) is not in vector form, there are no i,j,z compnents, how do i proceed

I thought the question wanted you to compute Grad(f) dot dr....

you right

$$\int$$ (27cos2 + 4sin2)i + (12sincos - 3)j dot (-3sin i + 2cos j) dt =

$$\int$$ -81cos2sin - 12sin3 + 24sincos2 - 6cos dt

how do i integrate that

gabbagabbahey
Homework Helper
Gold Member
you right

$$\int$$ (27cos2 + 4sin2)i + (12sincos - 3)j dot (-3sin i + 2cos j) dt =

$$\int$$ -81cos2sin - 12sin3 + 24sincos2 - 6cos dt

how do i integrate that

The easiest way is to write $\sin^3(t)=\sin(t)(1-\cos^2(t))=\sin(t)-\sin(t)\cos^2(t)$, then separate everything into three integrals of the form

$$c_1\int \sin(t)\cos^2(t)dt+c_2\int \sin(t)dt+c_3\int\cos(t)dt$$

Which I'm sure you know how to integrate.

By the way, are you supposed to calculate this over the entire ellipse, or just some section of it?

evaluate grad f dot dr, where c is the curve given by 4x^2 + 9y^2 = 36