Continuity and Differentiability Exercise 06
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=acosθ,y=bcosθ
Answer:

Differentiate both sides of the equation with respect to x: dx/dx = d(acosθ)/dx and dy/dx = d(bcosθ)/dx

Use the Chain Rule to differentiate both sides of the equation: dx/dx = a(sinθ)dθ/dx and dy/dx = b(sinθ)dθ/dx

Solve for dy/dx: dy/dx = b(sinθ)dθ/dx / a(sinθ)dθ/dx dy/dx = b/a
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=a(cost+logtan t/2),y=asint
Answer:
 Differentiate both sides of the equation with respect to x:
dy/dx = (d/dx)(asint)
 Use the chain rule to expand the derivative on the right side:
dy/dx = a(cos t)(d/dx)(sint)
 Use the chain rule again to expand the derivative on the right side:
dy/dx = a(cos t)(sin t)(d/dx)(t)
 Differentiate the variable t with respect to x:
dy/dx = a(cos t)(sin t)(1/2)(1/cos^2 t)
 Simplify the expression:
dy/dx = a(sin t)/2cos^2 t
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=sin^3t/√cos2t ,y=cos^3t/√cos2t
Answer:

Take the derivative of x with respect to t: dx/dt = 3sin²tcos t/√cos2t  (2sin^3tcos2t)/(2cos³t√cos2t)

Take the derivative of y with respect to t: dy/dt = 3cos²t/√cos2t + (2cos^3tcos2t)/(2sin³t√cos2t)

Divide the two derivatives: dy/dx = (3sin²tcos t/√cos2t  (2sin^3tcos2t)/(2cos³t√cos2t))/(3cos²t/√cos2t + (2cos^3tcos2t)/(2sin³t√cos2t))

Simplify the equation: dy/dx = (sin²tcos t√cos2t + 2sin^3tcos2t)/(cos²t√cos2t  2cos^3tcos2t)
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=a(cosθ+θsinθ),y=a(sinθ−θcosθ)
Answer:
Step 1: Differentiate x and y with respect to θ
dx/dθ = a(−sinθ + θcosθ)
dy/dθ = a(cosθ − θsinθ)
Step 2: Substitute the values of dx/dθ and dy/dθ in the equation
dy/dx = (dy/dθ)/(dx/dθ)
Step 3: Simplify the equation
dy/dx = (a(cosθ − θsinθ))/(a(−sinθ + θcosθ))
Step 4: Answer
dy/dx = −tanθ
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=sint,y=cos2t
Answer:

Differentiate both sides of the equation with respect to t: dx/dt = cos t and dy/dt = 2sin2t

Calculate dy/dx using the quotient rule: dy/dx = (dx/dt) / (dy/dt) = (cos t) / (2sin2t)

Simplify the expression: dy/dx = (1/2)cot2t
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=a(θ−sinθ),y=a(1+cosθ)
Answer:
 Differentiate both sides with respect to x:
d/dx [x] = d/dx [a(θ−sinθ)]
 Use the chain rule to differentiate:
d/dx [x] = d/dθ [a(θ−sinθ)] * d/dx [θ]
 Differentiate both sides with respect to y:
d/dy [y] = d/dy [a(1+cosθ)]
 Use the chain rule to differentiate:
d/dy [y] = d/dθ [a(1+cosθ)] * d/dy [θ]
 Set the derivatives equal to each other:
d/dx [x] = d/dy [y]
 Substitute the derivatives from steps 2 and 4:
d/dθ [a(θ−sinθ)] * d/dx [θ] = d/dθ [a(1+cosθ)] * d/dy [θ]
 Simplify:
d/dx [θ] = d/dy [θ]
 Solve for dy/dx:
dy/dx = 1
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=4t,y=4/t
Answer:
Answer: Step 1: Differentiate both sides of the equation with respect to x.
dx/dt = 4
Step 2: Substitute dx/dt in the equation dy/dx = dy/dt * dt/dx
dy/dx = dy/dt * (dt/dx)
Step 3: Substitute the given equations in the equation dy/dx
dy/dx = (dy/dt) * (dt/dx)
= (4/t) * (4)
Step 4: Simplify the equation
dy/dx = 16/t
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx. x=asecθ,y=btanθ
Answer:
Step 1: Differentiate both sides of the equation with respect to x
dx/dt = asecθ * dθ/dt
Step 2: Substitute the values of x and y in the equation
dx/dt = asecθ * dθ/dt
Step 3: Differentiate both sides of the equation with respect to θ
d²x/dθ² = asecθ * d²θ/dt² + (asecθ)² * (dθ/dt)²
Step 4: Substitute the values of x and y in the equation
d²x/dθ² = asecθ * d²θ/dt² + (asecθ)² * (btanθ)²
Step 5: Differentiate both sides of the equation with respect to t
d²x/dt² = asecθ * d³θ/dt³ + (asecθ)² * (btanθ)³ * dθ/dt
Step 6: Substitute the values of x and y in the equation
d²x/dt² = asecθ * d³θ/dt³ + (asecθ)² * (btanθ)³ * dθ/dt
Step 7: Solve for dy/dx
dy/dx = (asecθ * d³θ/dt³ + (asecθ)² * (btanθ)³ * dθ/dt) / (asecθ * dθ/dt)
dy/dx = d³θ/dt³ + (asecθ * btanθ)³ * dθ/dt
Question:
If x and y are connected parametrically by the given equation, then without eliminating the parameter, find dy/dx x=2at^2,y=at^4
Answer:
Step 1: Differentiate the given equation with respect to ’t'
x = 2at^2
⇒ dx/dt = 4at
y = at^4
⇒ dy/dt = 4at^3
Step 2: Substitute the value of ‘x’ in the equation of dx/dt
dx/dt = 4at
⇒ dx/dt = 4a(2at^2)
⇒ dx/dt = 8a^2t^2
Step 3: Substitute the value of ‘y’ in the equation of dy/dt
dy/dt = 4at^3
⇒ dy/dt = 4a(at^4)
⇒ dy/dt = 4a^2t^4
Step 4: Divide the equations of dx/dt and dy/dt
⇒ dy/dx = (4a^2t^4)/(8a^2t^2)
⇒ dy/dx = (1/2)t^2
Step 5: Substitute the value of ‘x’ in the equation of dy/dx
dy/dx = (1/2)t^2
⇒ dy/dx = (1/2)(2at^2)^2
⇒ dy/dx = 2a^2t^4
Hence, the required derivative is dy/dx = 2a^2t^4.