Question

a particle moves along the curve 6y=x^3+2 . find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate?​

Answer 1

Answer:

ForSphere

\rm Radius \: (r) = 4.2 \: cmRadius(r)=4.2cm

\therefore \sf Volume \: of \: sphere = \frac{4}{3}\pi r {}^{2} ∴Volumeofsphere=

3

4

πr

2

\: \: \: \: \: \: \sf = \frac{4}{3}\pi (4.2) {}^{3} \: cm {}^{3} =

3

4

π(4.2)

3

cm

3

\bf {For \: Cylinder}ForCylinder

\rm Radius(R) = 6 \: cmRadius(R)=6cm

\sf{Let \: the \: height \: of \: the \: cylinder \: be \: H \: cm.}LettheheightofthecylinderbeHcm.

\rm Then, \: Then,

\sf{Volume \: of \: Cylinder = \pi r {}^{2} H = \pi(6) {}^{2} H \: cm {}^{3} }VolumeofCylinder=πr

2

H=π(6)

2

Hcm

3

\rm According \: to \: the \: Question,AccordingtotheQuestion,

\sf{Volume \: of \: the \: metallic \: sphere \: must \: be}Volumeofthemetallicspheremustbe

\rm = Volume \: of \: the \: cylinder=Volumeofthecylinder

\implies \: \: \: \: \sf\frac{4}{3} \: \pi(4.2) {}^{3} = \pi(6) {}^{2} h⟹

3

4

π(4.2)

3

=π(6)

2

h

\rm Dividing \: both \: sides \: by \: \pi \: and \: cross \: multiplying,Dividingbothsidesbyπandcrossmultiplying,

\: \: \: \: \: \: \: \: \sf{3(6) {}^{2} H = 4(4.2) {}^{3} }3(6)

2

H=4(4.2)

3

\begin{gathered} \implies \: \: \: \: \: \: \sf{ H = \frac{4(4.2) {}^{3}} {3(6) {}^{2}}} \\ \end{gathered}

⟹H=

3(6)

2

4(4.2)

3

\begin{gathered} \implies \: \: \rm H = \frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \\ \end{gathered}

⟹H=

3×6×6

4×4.2×4.2×4.2

\: \: \: \: \: \: \: \sf = 4 \times 1.4 \times 0.7 \times 0.7=4×1.4×0.7×0.7

\begin{gathered} \: \: \: \: \: \: \: \: \: \rm = 4 \times \frac{14}{10} \times \frac{7}{10} \times \frac{7}{10} \\ \end{gathered}

=4×

10

14

×

10

7

×

10

7

\begin{gathered} \: \: \: \: \: \: \sf = \frac{56 \times 49}{1000} = \frac{2744}{1000} \\ \end{gathered}

=

1000

56×49

=

1000

2744

\implies \rm \: \: \: \: \: \: \: \: H = 2.744⟹H=2.744

\begin{gathered} \sf Hence, \: the \: height \: of \: the \: cylinder \: is \: 2.744 \\ \: \sf \:cm.\end{gathered}

Hence,theheightofthecylinderis2.744

cm.

Answer 2

Given that,

A particle moving along the curve

6y = x³ + 2 ...(1)

[we need to find points on the curve at which y coordinate is changing 8 times as fast as the x coordinate]

i.e.

we need to find (x,y) for which

$\sf \frac{dy}{dt} = 8\frac{dx}{dt}$

From (1)

Diff. both sides w.r.t (t)

$\implies \sf\frac{d(6y)}{dt} = \frac{d( {x}^{3} + 2)}{dt}$

$\sf \implies6\frac{dy}{dt} = \frac{d( {x})^{3} }{dt} + \frac{d(2)}{dt}$

$\sf\implies6 \frac{dy}{dt} = \frac{d {(x)}^{3} }{dt} \times \frac{dx}{dx} + 0$

$\sf\implies6 \frac{dy}{dt} = \frac{d {(x)}^{3} }{dx} \times \frac{dx}{dt}$

$\sf\implies6 \frac{dy}{dt} = 3 {x}^{2} . \frac{dx}{dt}$

$\sf \implies\frac{dy}{dt} = \frac{3 {x}^{2} }{6} . \frac{dx}{dt}$

$\sf\implies\frac{dy}{dt} = \frac{ {x}^{2} }{2} . \frac{dx}{dt} \: \: \: \: \: ...(2)$

We need to find point for which

$\sf \frac{dy}{dt} = 8 \frac{dx}{dt}$

Putting $\sf\frac{dy}{dt} = \frac{ {x}^{2} }{2} . \frac{dx}{dt}$ from (2)

$\sf\implies\frac{ {x}^{2} }{2} . \frac{dx}{dt} = 8 \frac{dx}{dt}$

$\sf\implies \frac{ {x}^{2} }{2} = 8$

$\sf\implies \: {x}^{2} = 16$

$\sf\implies \: x = ± \sqrt{16}$

$\sf\implies \: \boxed{ x = ±4}$

$x = 4 , - 4$

Putting values of x in equation (1)

When, x= 4

we get, y = 11

point is (4,11)

When,x = -4

we get, y = -31/3

so point is $\sf( - 4, \frac{ - 31}{3} )$

$\bold{ \bigstar \: answer \bigstar}$

Hence, required points on the curve are $\sf(4,11)$ and $\sf( - 4, \frac{ - 31}{3} )$