Derivatives - Examples on Parametric Differentiation

• Introduction to parametric differentiation
• Brief explanation of parametric equations
• Rules for finding derivatives of parametric equations
• Example 1: Finding the derivative of x = 2t^2 and y = 3t
  • Step 1: Find dx/dt and dy/dt
  • Step 2: Use the chain rule to find dy/dx
  • Step 3: Simplify the expression for dy/dx
• Example 2: Finding the derivative of x = sin(t) and y = cos(t)
  • Step 1: Find dx/dt and dy/dt using trigonometric identities
  • Step 2: Use the chain rule to find dy/dx
  • Step 3: Simplify the expression for dy/dx
• Example 3: Finding the derivative of x = ln(t) and y = e^t
  • Step 1: Find dx/dt and dy/dt using logarithmic and exponential rules
  • Step 2: Use the chain rule to find dy/dx
  • Step 3: Simplify the expression for dy/dx
• Example 4: Finding the derivative of x = 3t^2 + 2t and y = 4t^3 + 6t^2
  • Step 1: Find dx/dt and dy/dt
  • Step 2: Use the chain rule to find dy/dx
  • Step 3: Simplify the expression for dy/dx
• Example 5: Finding the derivative of x = e^t cos(t) and y = e^t sin(t)
  • Step 1: Find dx/dt and dy/dt using the product rule and chain rule
  • Step 2: Use the chain rule to find dy/dx
  • Step 3: Simplify the expression for dy/dx

• Summary of key points covered so far

Slide 11: Parametric Derivatives - Example 1

• Given parametric equations: x = 2t^2 and y = 3t
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = 4t
  • dy/dt = 3
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (3) / (4t)

Slide 12: Parametric Derivatives - Example 2

• Given parametric equations: x = sin(t) and y = cos(t)
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = cos(t)
  • dy/dt = -sin(t)
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (-sin(t)) / (cos(t))

Slide 13: Parametric Derivatives - Example 3

• Given parametric equations: x = ln(t) and y = e^t
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = 1/t
  • dy/dt = e^t
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (e^t) / (1/t)

Slide 14: Parametric Derivatives - Example 4

• Given parametric equations: x = 3t^2 + 2t and y = 4t^3 + 6t^2
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = 6t + 2
  • dy/dt = 12t^2 + 12t
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (12t^2 + 12t) / (6t + 2)

Slide 15: Parametric Derivatives - Example 5

• Given parametric equations: x = e^t cos(t) and y = e^t sin(t)
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = e^t cos(t) - e^t sin(t)
  • dy/dt = e^t sin(t) + e^t cos(t)
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (e^t sin(t) + e^t cos(t)) / (e^t cos(t) - e^t sin(t))

Slide 16: Summary

• Parametric differentiation involves finding the derivatives of parametric equations.
• To find the derivative of y with respect to x, we use the chain rule and find dy/dx.
• Examples covered:
  • Example 1: x = 2t^2, y = 3t
  • Example 2: x = sin(t), y = cos(t)
  • Example 3: x = ln(t), y = e^t
  • Example 4: x = 3t^2 + 2t, y = 4t^3 + 6t^2
  • Example 5: x = e^t cos(t), y = e^t sin(t)
• Simplify the expression of dy/dx for a clear understanding of the derivative.

Slide 21: Parametric Derivatives - Review

• Parametric differentiation involves finding the derivatives of parametric equations.
• The derivatives of the x and y components of the parametric equations are calculated separately.
• To find the derivative of y with respect to x, we use the chain rule and find dy/dx.
• The derivative dy/dx represents the rate of change of y with respect to x.
• The derivative dy/dx can be positive, negative, or zero, indicating the direction and steepness of the curve.

Slide 22: Parametric Derivatives - Generalized Formula

• Given parametric equations: x = f(t) and y = g(t).
• To find dy/dx, we can use the formula: dy/dx = (dy/dt) / (dx/dt).
• The numerator represents the rate of change of y with respect to t.
• The denominator represents the rate of change of x with respect to t.
• Simplify the expression of dy/dx by dividing the numerator and denominator by a common factor if possible.

Slide 23: Parametric Derivatives - Example 6

• Given parametric equations: x = t^3 - 3t and y = t^2 + 1.
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = 3t^2 - 3
  • dy/dt = 2t
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (2t) / (3t^2 - 3)

Slide 24: Parametric Derivatives - Example 7

• Given parametric equations: x = e^(-t) and y = t^2 + 2t + 3.
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = -e^(-t)
  • dy/dt = 2t + 2
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (2t + 2) / (-e^(-t))

Slide 25: Parametric Derivatives - Example 8

• Given parametric equations: x = cos(t) and y = sin(t).
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = -sin(t)
  • dy/dt = cos(t)
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (cos(t)) / (-sin(t))

Slide 26: Parametric Derivatives - Example 9

• Given parametric equations: x = 2t^3 + 3t^2 - t and y = t^2 + 2t + 1.
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = 6t^2 + 6t - 1
  • dy/dt = 2t + 2
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (2t + 2) / (6t^2 + 6t - 1)

Slide 27: Parametric Derivatives - Example 10

• Given parametric equations: x = t^3 - t^2 + t and y = t^2 - 4.
• To find the derivative of y with respect to x, we need to find dy/dx.
• Step 1: Find dx/dt and dy/dt
  • dx/dt = 3t^2 - 2t + 1
  • dy/dt = 2t
• Step 2: Use the chain rule to find dy/dx
  • dy/dx = (dy/dt) / (dx/dt)
• Step 3: Simplify the expression for dy/dx
  • dy/dx = (2t) / (3t^2 - 2t + 1)

Slide 28: Parametric Derivatives - Further Practice

• Practice more examples to strengthen your understanding of parametric differentiation.
• Try different types of parametric equations, such as combining trigonometric functions, logarithmic functions, or exponential functions.
• Use the generalized formula dy/dx = (dy/dt) / (dx/dt) to find the derivative of y with respect to x.
• Simplify the expression for dy/dx to make it more simplified and clear.

Slide 29: Parametric Derivatives - Applications

• Parametric differentiation has various real-life applications, such as physics, engineering, and economics.
• It helps in finding the rate of change of quantities in motion or dynamic systems.
• It can be applied to calculate velocity, acceleration, and other physical quantities in mechanics.
• It is useful in optimizing processes in engineering and determining optimal solutions in economics.

Slide 30: Parametric Derivatives - Summary

• Parametric differentiation involves finding the derivatives of parametric equations.
• Use the chain rule and the generalized formula dy/dx = (dy/dt) / (dx/dt).
• Simplify the expression for dy/dx to make it more simplified and clear.
• Practice more examples to strengthen your understanding.
• Parametric differentiation has various real-life applications in physics, engineering, and economics.