Question Statement
Prove that:
ydxdyβ+x=0
if:
x=1+t21βt2β,y=1+t22tβ
Background and Explanation
To solve this, we need to:
- Differentiate x and y with respect to t to calculate dxdyβ using the chain rule.
- Substitute the values of x, y, and dxdyβ into the given equation to verify it.
The chain rule states that:
dxdyβ=dx/dtdy/dtβ
Understanding parametric equations like x(t) and y(t) is necessary, as well as simplifying rational expressions.
Solution
Step 1: Differentiate y with respect to t
Given:
y=1+t22tβ
Differentiate with respect to t:
dtdyβ=(1+t2)2(1+t2)(2)β2t(2t)β
Simplify the numerator:
dtdyβ=(1+t2)22+2t2β4t2β=(1+t2)22(1βt2)β
Step 2: Differentiate x with respect to t
Given:
x=1+t21βt2β
Differentiate with respect to t:
dtdxβ=(1+t2)2(1+t2)(β2t)β(1βt2)(2t)β
Simplify the numerator:
dtdxβ=(1+t2)2β2tβ2t3β2t+2t3β=(1+t2)2β4tβ
Step 3: Find dxdyβ using the chain rule
The chain rule gives:
dxdyβ=dtdxβdtdyββ
Substitute the values of dtdyβ and dtdxβ:
dxdyβ=(1+t2)2β4tβ(1+t2)22(1βt2)ββ
Simplify:
dxdyβ=β4t2(1βt2)β=β2t1βt2β
Step 4: Verify the given equation
We need to show:
ydxdyβ+x=0
Substitute y=1+t22tβ and dxdyβ=β2t1βt2β:
ydxdyβ=(1+t22tβ)(β2t1βt2β)
Simplify:
ydxdyβ=β(1+t2)(1βt2)β
Now, add x=1+t21βt2β:
ydxdyβ+x=β(1+t2)(1βt2)β+1+t21βt2β
Combine the terms:
ydxdyβ+x=1+t21βt2β(1βt2)β=1+t20β=0
Thus, the equation is verified.
- Chain rule:
dxdyβ=dtdxβdtdyββ
- Differentiation of rational functions.
- Simplifying expressions to verify the given equation.
Summary of Steps
- Differentiate y with respect to t to get dtdyβ.
- Differentiate x with respect to t to get dtdxβ.
- Use the chain rule to find dxdyβ.
- Substitute x, y, and dxdyβ into the equation ydxdyβ+x and simplify to verify it equals 0.