A particle following the curve traced out by the path $c(t) = left(sinleft[e^{t}right], t, 4-t^2right)$ flies...
$begingroup$
A particle following the curve traced out by the path $c(t) = left(sinleft[e^{t}right], t, 4-t^2right)$ flies off on a tangent at time $t = 1$. Find the position of particle at time $t = 2$.
Solution:
$$C(t) = left(sinleft[e^tright], t, 4-t^2right),qquad C'(t) = left(e^tcosleft[e^tright], 1, -2tright)$$
$$C(1) = big(sin[e], 1, 3big),qquad C'(t) = big(ecos[e], 1, -2big)$$
At $t = 2$ the parametric is
$$[x, y, z] = big(sin[e], 1, 3big) + big(ecos[e], 1, -2big)(2-1)$$
so
$$(x, y, z) = big(sin[e] + ecos[e], 2, 1big)$$
Is it correct?
multivariable-calculus
edited Jan 21 at 0:01
El borito
666216
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
$endgroup$
add a comment |
$begingroup$
A particle following the curve traced out by the path $c(t) = left(sinleft[e^{t}right], t, 4-t^2right)$ flies off on a tangent at time $t = 1$. Find the position of particle at time $t = 2$.
Solution:
$$C(t) = left(sinleft[e^tright], t, 4-t^2right),qquad C'(t) = left(e^tcosleft[e^tright], 1, -2tright)$$
$$C(1) = big(sin[e], 1, 3big),qquad C'(t) = big(ecos[e], 1, -2big)$$
At $t = 2$ the parametric is
$$[x, y, z] = big(sin[e], 1, 3big) + big(ecos[e], 1, -2big)(2-1)$$
so
$$(x, y, z) = big(sin[e] + ecos[e], 2, 1big)$$
Is it correct?
multivariable-calculus
edited Jan 21 at 0:01
El borito
666216
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
$endgroup$
3
$begingroup$
Yes, that's correct
$endgroup$
– caverac
Jan 20 at 22:42
add a comment |
$begingroup$
A particle following the curve traced out by the path $c(t) = left(sinleft[e^{t}right], t, 4-t^2right)$ flies off on a tangent at time $t = 1$. Find the position of particle at time $t = 2$.
Solution:
$$C(t) = left(sinleft[e^tright], t, 4-t^2right),qquad C'(t) = left(e^tcosleft[e^tright], 1, -2tright)$$
$$C(1) = big(sin[e], 1, 3big),qquad C'(t) = big(ecos[e], 1, -2big)$$
At $t = 2$ the parametric is
$$[x, y, z] = big(sin[e], 1, 3big) + big(ecos[e], 1, -2big)(2-1)$$
so
$$(x, y, z) = big(sin[e] + ecos[e], 2, 1big)$$
Is it correct?
multivariable-calculus
edited Jan 21 at 0:01
El borito
666216
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
$endgroup$
A particle following the curve traced out by the path $c(t) = left(sinleft[e^{t}right], t, 4-t^2right)$ flies off on a tangent at time $t = 1$. Find the position of particle at time $t = 2$.
Solution:
$$C(t) = left(sinleft[e^tright], t, 4-t^2right),qquad C'(t) = left(e^tcosleft[e^tright], 1, -2tright)$$
$$C(1) = big(sin[e], 1, 3big),qquad C'(t) = big(ecos[e], 1, -2big)$$
At $t = 2$ the parametric is
$$[x, y, z] = big(sin[e], 1, 3big) + big(ecos[e], 1, -2big)(2-1)$$
so
$$(x, y, z) = big(sin[e] + ecos[e], 2, 1big)$$
Is it correct?
multivariable-calculus
multivariable-calculus
edited Jan 21 at 0:01
El borito
666216
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
edited Jan 21 at 0:01
El borito
666216
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
edited Jan 21 at 0:01
El borito
666216
edited Jan 21 at 0:01
El borito
666216
edited Jan 21 at 0:01
El borito
666216
666216
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
asked Jan 20 at 22:38
first n lastnfirst n lastn
61
61
3
$begingroup$
Yes, that's correct
$endgroup$
– caverac
Jan 20 at 22:42
add a comment |
3
$begingroup$
Yes, that's correct
$endgroup$
– caverac
Jan 20 at 22:42
3
3
$begingroup$
Yes, that's correct
$endgroup$
– caverac
Jan 20 at 22:42
$begingroup$
Yes, that's correct
$endgroup$
– caverac
Jan 20 at 22:42
add a comment |
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Yes, that's correct
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– caverac
Jan 20 at 22:42