Question

please help! hour and a half till the question is due. not sure what i'm doing wrong here. thanks

(1 pt) Suppose F⃗ (x,y)=−xyi⃗ +y^2j⃗ .

(a) Find a vector parametric equation for the parabola y=x^2 from the origin to the point (4,16) using t as a parameter.
r⃗ (t)=

(b) Find the line integral of F⃗ along the parabola y=x^2 from the origin to (4,16).

Answer 1

the first part i got as x=t and y=t^2. or <t,t^2>

Answer 2

tried \[\int\limits_{0}^{4}(-t)(t^2)+(t^2)^2 = 704/5\] for part b but it's not right.

Answer 3

thers a small mistake that is u forget to multiply r(t)=<1,2t)\[F=-xy \hat{i}+y^2 \hat{j}\\x=t,y=t^2\\F_n=(-t^3 \hat{i}+t^4 \hat{j})\\r=<t,t^2>\\dr=<1,2t>=\hat{i}+2t \hat{j}\]

Answer 4

so \[line~Integral=\int\limits_{C}^{}F_n.dr=\int\limits_{}^{}(-t^3 \hat{i}+t^4 \hat{j})*(\hat{i}+2t \hat{j})\]

Answer 5

you missed the second part

Answer 6

are the bounds still right? 0 to 4?

Answer 7

cause i'm still getting the wrong answer. 16448/15

Answer 8

whats the answer? is it 1301

Answer 9

ok, 1301 is right. i ws getting 16448/15 for the integral \[\int\limits_{0}^{4}(-t^3+t^4)(1+2t)\]

Answer 10

f(t)= (2t+1)⋅(t^4−t^3)
∫f(t)dt= 20t^6−12t^5−15t^4/60
∫04f(t)dt= 16448/15
Approximation 1096.533333333333

Answer 11

weird.

Answer 12

no
it will be
\[\int\limits_{}^{}(-t^3 \hat {i}+t^4 \hat{j})*(i+2t \hat{j})\\now ~remember ~that ~ \\i*i=1\\i*j=j*i=0\\j*j=1\\so ~then~\it~will~be \\ \int\limits_{0}^{4}(-t^3+2t^5)\]

Answer 13

oh, ok. i didn't know about that relationship. i'll make a note of that. thanks a bunch!

Answer 14

yw!!