At what points does the curve r(t) = ti + (5t − t2)k intersect the paraboloid z = x2 + y2? (if an answer does not exist, enter d

Question

bogdanovich [222]

3 years ago

11

At what points does the curve r(t) = ti + (5t − t2)k intersect the paraboloid z = x2 + y2? (if an answer does not exist, enter d

ne.)

Mathematics

2 answers:

Sergeu [11.5K] · Answer 1 · 2022-06-14T06:23:33+03:00

Answer : curve r(t) = t i + (2t - t 2)k intersect the paraboloid z = x 2 + y 2

By the equation of r(t), x = t, y = 0, and z = $5t - t^{2}$ .

Thus, z = $x^{2} + y^{2}$

on solving we get,

$5t - t^{2}$ = $t^{2} + 0^{2}$

Now, we can solve for t

$t^{2} + t^{2} -5t$ = 0

So, $2t^2 - 5t$ = 0

2t ( t - 5/2) = 0

t = 0 , t = 5

plugging these values in t = 0 into r(t)

r(0) = <0 , 0 , 0>

r(5) = < 5, 0 , 25 >

Reptile [31] · Answer 2 · 2022-06-14T08:08:38+03:00

The curve $r\left( t \right) = ti + \left( {5t - {t^2}} \right)k$ intersect the paraboloid $z = {x^2} + {y^2}$ at $\boxed{t = 0}$ and $\boxed{t = \frac{5}{2}}.$

Further explanation:

Given:

The equation of the curve $r\left( t \right) = ti + \left( {5t - {t^2}} \right)k.$

The equation of the paraboloid $z = {x^2} + {y^2}.$

Explanation:

From theequation of the curve $r\left( t \right) = ti + \left( {5t - {t^2}} \right)k.$

The value of $x$ is $t$ and the value of $y$ is $0$ and the equation of $z$ is $z = 5t - {t^2}.$

Substitute $t$ for $x$ , $0$ for $y$ and $5t - {t^2}$ in equation of the paraboloid

$\begin{aligned}5t - {t^2} &= {t^2} + 0\\0&= {t^2} + {t^2} - 5t\\0&= 2{t^2} - 5t\\0&= 2t\left( {t - 5} \right)\\t&= 0\;{\text{or}}\;t - 5 &= 0\\t &= 0\;{\text{or }}t&= 5\\\end{aligned}$

The value of $r\left( 0 \right)$ can be obtained as follows,

$\begin{aligned}r\left( 0 \right)&= 0 + \left( {0 - {0^2}} \right)k\\&= 0\\\end{aligned}$

The value of $y$ is $0$ and the value of z can be obtained as follows,

$\begin{aligned}z&= {5^2} + {0^2}\\&=25\\\end{aligned}$

The curve $r\left( t \right) = ti + \left( {5t - {t^2}} \right)k$ intersect the paraboloid $z = {x^2} + {y^2}$ at $\boxed{t = 0}$ and $\boxed{t = \frac{5}{2}}.$

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