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:

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

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

3. 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:

1. Differentiate both sides of the equation with respect to x:

dy/dx = (d/dx)(asint)

2. Use the chain rule to expand the derivative on the right side:

dy/dx = a(cos t)(d/dx)(sint)

3. Use the chain rule again to expand the derivative on the right side:

dy/dx = a(cos t)(-sin t)(d/dx)(t)

4. Differentiate the variable t with respect to x:

dy/dx = a(cos t)(-sin t)(1/2)(1/cos^2 t)

5. 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:

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

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

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

4. 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:

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

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

3. 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:

1. Differentiate both sides with respect to x:

d/dx [x] = d/dx [a(θ−sinθ)]

2. Use the chain rule to differentiate:

d/dx [x] = d/dθ [a(θ−sinθ)] * d/dx [θ]

3. Differentiate both sides with respect to y:

d/dy [y] = d/dy [a(1+cosθ)]

4. Use the chain rule to differentiate:

d/dy [y] = d/dθ [a(1+cosθ)] * d/dy [θ]

5. Set the derivatives equal to each other:

d/dx [x] = d/dy [y]

6. Substitute the derivatives from steps 2 and 4:

d/dθ [a(θ−sinθ)] * d/dx [θ] = d/dθ [a(1+cosθ)] * d/dy [θ]

7. Simplify:

d/dx [θ] = d/dy [θ]

8. 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.