Question

Given x=t³-3t and y=3t², find the value of dy/dx at t=√2​

Answer 1

The value of
`\frac{\mathrm{d} y}{\mathrm{d} x}` at t = √2 is
`(1)/(2 * √(2))` .

Explanation:

Given as :

x = t³ - 3 t

y = 3 t²

We have to calculate
`\frac{\mathrm{d} y}{\mathrm{d} x}` at t = √2

Now,

`\frac{\mathrm{d} x}{\mathrm{d} t}` =
`\frac{\mathrm{d} (t³ - 3 t)}{\mathrm{d} t}`

Or,
`\frac{\mathrm{d} x}{\mathrm{d} t}` =
`\frac{\mathrm{d}  t³}{\mathrm{d} t}` - 3
`\frac{\mathrm{d}  t}{\mathrm{d} t}`

Or,
`\frac{\mathrm{d} x}{\mathrm{d} t}` = 3 t² - 3 ......A

Again

`\frac{\mathrm{d} y}{\mathrm{d} t}` =
`\frac{\mathrm{d} (3 t²)}{\mathrm{d} t}`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} t}` = 3 × 2 t

Or,
`\frac{\mathrm{d} y}{\mathrm{d} t}` = 6 t ......B

So,

`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`((\partial y)/(\partial t))/((\partial x)/(\partial t))`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`(3 t^(2) - 3 )/(6 t)`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`(3(t^(2)-1))/(6 t)`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`((t^(2)-1))/(2 t)`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} x}` at t = √2

i.e
`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`((√(2)^(2)-1))/(2 * √(2))`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`((2-1))/(2 * √(2))`

Or,
`\frac{\mathrm{d} y}{\mathrm{d} x}` =
`(1)/(2 * √(2))`

Hence, The value of
`\frac{\mathrm{d} y}{\mathrm{d} x}` at t = √2 is
`(1)/(2 * √(2))` . Answer