Math Problem Statement

solve

Solution

The image shows an exercise from a textbook related to using the chain rule in partial differentiation to find the total derivative dzdt\frac{dz}{dt} at a given point.

Problem Breakdown:

  1. Part (a): z=xcos(y),wherex=sin(t),y=cos(t)z = x \cos(y), \quad \text{where} \quad x = \sin(t), \quad y = \cos(t) Find dzdt\frac{dz}{dt}.

  2. Part (b): z=x2+2xy+y2,wherex=etcos(t),y=etsin(t)z = x^2 + 2xy + y^2, \quad \text{where} \quad x = e^t \cos(t), \quad y = e^t \sin(t) Find dzdt\frac{dz}{dt}.

Solutions:

Part (a)

  1. Expression for zz: z=xcos(y)z = x \cos(y)

  2. Given functions: x=sin(t),y=cos(t)x = \sin(t), \quad y = \cos(t)

    The total derivative dzdt\frac{dz}{dt} using the chain rule is given by: dzdt=zxdxdt+zydydt\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}

  3. Partial derivatives: zx=cos(y)\frac{\partial z}{\partial x} = \cos(y) zy=xsin(y)\frac{\partial z}{\partial y} = -x \sin(y)

  4. Derivatives of xx and yy with respect to tt: dxdt=cos(t)\frac{dx}{dt} = \cos(t) dydt=sin(t)\frac{dy}{dt} = -\sin(t)

  5. Substitute and simplify: dzdt=cos(y)cos(t)+(xsin(y))(sin(t))\frac{dz}{dt} = \cos(y) \cdot \cos(t) + \left(-x \sin(y)\right) \cdot (-\sin(t)) dzdt=cos(cos(t))cos(t)+sin(t)sin(cos(t))sin(t)\frac{dz}{dt} = \cos(\cos(t)) \cdot \cos(t) + \sin(t) \cdot \sin(\cos(t)) \cdot \sin(t) dzdt=cos(cos(t))cos(t)+sin2(t)sin(cos(t))\frac{dz}{dt} = \cos(\cos(t)) \cdot \cos(t) + \sin^2(t) \cdot \sin(\cos(t))

Part (b)

  1. Expression for zz: z=x2+2xy+y2z = x^2 + 2xy + y^2

  2. Given functions: x=etcos(t),y=etsin(t)x = e^t \cos(t), \quad y = e^t \sin(t)

    Again, use the chain rule: dzdt=zxdxdt+zydydt\frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt}

  3. Partial derivatives: zx=2x+2y\frac{\partial z}{\partial x} = 2x + 2y zy=2x+2y\frac{\partial z}{\partial y} = 2x + 2y

  4. Derivatives of xx and yy with respect to tt: dxdt=etcos(t)etsin(t)\frac{dx}{dt} = e^t \cos(t) - e^t \sin(t) dydt=etsin(t)+etcos(t)\frac{dy}{dt} = e^t \sin(t) + e^t \cos(t)

  5. Substitute and simplify: dzdt=(2x+2y)dxdt+(2x+2y)dydt\frac{dz}{dt} = (2x + 2y) \cdot \frac{dx}{dt} + (2x + 2y) \cdot \frac{dy}{dt} Simplify further by substituting the values of xx and yy and their derivatives.

Would you like further steps for the simplification, or do you have any specific questions?


Follow-up Questions:

  1. How does the chain rule apply when dealing with multiple variables?
  2. Can you derive the partial derivatives of a more complex function?
  3. What are the key differences between total and partial derivatives?
  4. How would you handle a similar problem with more variables?
  5. What happens if one of the functions is implicitly defined?

Tip: Always keep track of the variables being differentiated to avoid mistakes, especially when multiple variables depend on the same parameter.

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Math Problem Analysis

Mathematical Concepts

Calculus
Chain Rule
Partial Differentiation

Formulas

Chain rule for partial derivatives

Theorems

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Suitable Grade Level

Advanced College Level