Question

X=a[t+1/t], y=a[t-1/t] where "a" is constant than prove that dy/dx= x/y

Answer 1

[tex] \frac{dx}{dt} = a(1- \frac{1}{ t^{2} } ) \\ \frac{dy}{dt} = a(1+ \frac{1}{ t^{2} } ) \\ \frac{dy}{dx} = (\frac{t ^{2}+ 1}{ t^{2}-1} ) = \frac{x}{y} [/tex]

Answer 2

$\frac{dy}{dx}=\frac{x}{y}$ is proved for given values of x and y where "a" is a constant.

Step-by-step explanation:

• Given that $x=a[t+\frac{1}{t}]$ and $y=a[t-\frac{1}{t}]$

where "a" is constant.

• To find prove that $\frac{dy}{dx}=\frac{x}{y}$

• That is to prove LHS=RHS

• Let us take LHS $\frac{dy}{dx}$
• The above differentiation can be written as
• $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

• Now find $\frac{dy}{dt}$ from $y=a[t-\frac{1}{t}]$ we get
• $y=a[t-\frac{1}{t}]$
• Differentiating with respect to "t" we get
• $\frac{dy}{dt}=a[1-(-\frac{1}{t^2})]$
• $=a[1+\frac{1}{t^2}]$
• $=a[\frac{t^2+1}{t^2}]$

• Therefore $\frac{dy}{dt}=a[\frac{t^2+1}{t^2}]$
• Similarly for x $x=a[t+\frac{1}{t}]$
• Differentiating with respect to "t" we get
• $\frac{dx}{dt}=a[1+(-\frac{1}{t^2})]$
• $=a[1-\frac{1}{t^2}]$

• $=a[\frac{t^2-1}{t^2}]$
• Therefore $\frac{dx}{dt}=a[\frac{t^2-1}{t^2}]$
• Now $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$ becomes
• $\frac{dy}{dx}=a[\frac{t^2+1}{t^2}]\times \frac{1}{a[\frac{t^2-1}{t^2}]}$
• $=a[\frac{t^2+1}{t^2}]\times \frac{t^2}{a[t^2-1]}$
• $=\frac{t^2+1}{t^2-1}$

• Therefore  $\frac{dy}{dx}=\frac{t^2+1}{t^2-1}$=LHS
• Now taking RHS $\frac{x}{y}$
• $\frac{x}{y}=\frac{a[t+\frac{1}{t}]}{a[t-\frac{1}{t}]}$
• $=a[\frac{t^2+1}{t^2}]\times (\frac{t^2}{a(t^2-1)})$
• $=\frac{t^2+1}{t^2-1}$
• Therefore $\frac{x}{y}=\frac{t^2+1}{t^2-1}$=RHS

Therefore LHS=RHS

$\frac{dy}{dx}=\frac{x}{y}$

Hence proved