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:

hope it helps

Step-by-step explanation:

The equation of the curve is given as,

6y=x  

3

+2

Differentiating both sides w.r.t. t

6  

dt

dy

​  

=3x  

2

 

dt

dx

​  

+0

⇒2  

dt

dy

​  

=x  

2

 

dt

dx

​  

 

It is given that, (  

dt

dy

​  

=8  

dt

dx

​  

),

Thus we have,

2(8  

dt

dx

​  

)=x  

2

 

dt

dx

​  

 

⇒16  

dt

dx

​  

=x  

2

 

dt

dx

​  

 

⇒(x  

2

−16)  

dt

dx

​  

=0

⇒x  

2

−16=0⇒x=±4

When x=4,y=  

6

4  

3

+2

​  

=  

6

66

​  

=11

when  x=(−4),y=  

6

(−4)  

3

+2

​  

=−  

6

62

​ =−  

3

31

​Hence, the points required on the curve are (4,11) and (−4,  

3 −31  ).

Answer 2

Step-by-step explanation:

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