Implicit Differentiation
Implicit differentiation is a powerful technique used to find the derivative of functions that are not explicitly defined in terms of one variable, given that a variable can be written as a function of the other. This method is particularly useful when dealing with equations where it's difficult or impossible to isolate one variable.
The Process of Implicit Differentiation
Assume $y$ is a function of $x$. Sometimes the explicit relationship $y=f(x)$ isn't given, yet we can follow the steps below to differentiate implicitly.
Assume you are interested in finding the differentiation in terms of $x$, hence $\frac{dy}{dx}$
- Differentiate both sides of the equation with respect to $x$.
- Since $y$ is a function of $x$, you have to apply the chain rule on $y$ such that you differentiate in terms of $y$ then multiply in terms of $\frac{dy}{dx}$.
- Similarly, you do the same for $x$, so you multiply with $\frac{dx}{dx}$ but this is just equal to 1 so this term can be omitted.
- Group terms containing $\frac{dy}{dx}$ on one side of the equation.
- Solve for $\frac{dy}{dx}$.
Let's consider the equation of a circle: $x^2 + y^2 = r^2$
To find $\frac{dy}{dx}$:
- Differentiate both sides with respect to x: $\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(r^2)$
- Apply the chain rule: $2x + 2y\frac{dy}{dx} = 0$
- Solve for $\frac{dy}{dx}$: $2y\frac{dy}{dx} = -2x$ $\frac{dy}{dx} = -\frac{x}{y}$
The term $\frac{dx}{dx}$ disappears above since it is just equal to 1.
- It is important to note that in implicit differentiation, you choose the variable to derive in terms
- Hence you could derive in terms of $y$ in which when you apply the chain rule $\frac{dy}{dy}$ would just be one hence disappear and you would be deriving $x$ in terms of $y$ too such that you end up with $\frac{dx}{dy}$
Hence make sure to derive $x$ and $y$ normally and apply the appropriate chain rule to each depending on what you are differentiating with respect to.
- Implicit differentiation can also be used when $x$ is a function of $y$
It can also be used when they aren't functions of each other,
- But this means that $\frac{dy}{dx} = \frac{dx}{dy} = 0$ hence you derive assuming the other is just a constant
Related Rates of Change
Related rates problems involve finding the rate of change of one quantity with respect to time when it's related to another quantity whose rate of change is known. These problems often require the use of implicit differentiation and the chain rule.
Steps to Solve Related Rates Problems
- Identify the known and unknown rates of change.
- Write an equation relating the variables involved.
- Differentiate both sides of the equation with respect to time.
- Substitute known values and solve for the unknown rate.
A conical water tank is being filled at a rate of 10 cubic meters per minute. The tank has a height of 12 meters and a base radius of 6 meters. At what rate is the water level rising when the water is 4 meters deep?
- Let r be the radius of the water surface and h be the height of the water. We know $\frac{dV}{dt} = 10$ and need to find $\frac{dh}{dt}$.
- The volume of a cone is given by $V = \frac{1}{3}\pi r^2h$. We can relate r and h using similar triangles: $\frac{r}{h} = \frac{6}{12} = \frac{1}{2}$, so $r = \frac{1}{2}h$.
- Substituting into the volume equation: $V = \frac{1}{3}\pi (\frac{1}{2}h)^2h = \frac{1}{12}\pi h^3$
- Differentiating with respect to time: $\frac{dV}{dt} = \frac{1}{4}\pi h^2 \frac{dh}{dt}$
- Substituting known values: $10 = \frac{1}{4}\pi (4)^2 \frac{dh}{dt}$
- Solving for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{10}{\pi 4} = \frac{5}{2\pi} \approx 0.796$ meters per minute
Students often forget to use the chain rule when differentiating composite functions in related rates problems. Always remember that when you have a function of a function (like $r(h(t))$), you need to apply the chain rule.
Optimisation Problems
Optimisation problems involve finding the maximum or minimum value of a function within a given domain. These problems often require finding critical points using differentiation and determining whether they represent maxima, minima, or neither.
Steps to Solve Optimisation Problems
- Identify the quantity to be optimised and express it as a function of one variable.
- Determine the domain of the function based on the problem constraints.
- Find the critical points by setting the derivative equal to zero and solving.
- Evaluate the function at the critical points and at the endpoints of the domain.
- Compare the values to determine the optimal solution.
A farmer wants to fence off a rectangular field adjacent to a straight river. He has 100 meters of fencing and wants to maximise the area of the field. The side along the river doesn't need fencing. What dimensions will give the maximum area?
- Let x be the width of the field and y be the length. The area is A = xy.
- The perimeter constraint gives us: $x + 2y = 100$. Solving for y: $y = 50 - \frac{x}{2}$
- Substituting into the area function: $A(x) = x(50 - \frac{x}{2}) = 50x - \frac{x^2}{2}$
- Finding the critical point: $\frac{dA}{dx} = 50 - x$ Set this equal to zero: $50 - x = 0$ $\\x = 50$
- The domain is $0 \leq x \leq 100$. Evaluating at x = 0, 50, and 100:
A(0) = 0
A(50) = 1250
A(100) = 0 - The maximum area occurs when x = 50 meters and y = 25 meters.
In optimisation problems, always check the endpoints of the domain, even if they don't satisfy $\frac{dy}{dx} = 0$. Sometimes the optimal solution occurs at an endpoint.
Solving optimization problems involving multiple variables is difficult, therefore we would like to write it using one single variable as done above.
Applications and Importance
The techniques of implicit differentiation, related rates, and optimisation are crucial in various fields:
- Physics: Related rates are used to analyze motion and changing physical systems.
- Engineering: Optimisation is essential in designing efficient systems and structures.
- Economics: These methods are applied in cost minimization and profit maximization problems.
- Computer Science: Optimisation algorithms often use these calculus techniques at their core.
The ability to solve these types of problems demonstrates a deep understanding of calculus and its applications, which is highly valued in many academic and professional fields.
Common Challenges and Strategies
Students often face challenges when dealing with these advanced topics:
- Identifying the correct variables: In related rates and optimisation problems, clearly defining variables and their relationships is crucial.
- Setting up equations: Translating word problems into mathematical equations can be difficult. Practice and careful reading of the problem statement are key.
- Applying the chain rule: Many students struggle with when and how to apply the chain rule in implicit differentiation and related rates problems.
- Interpreting results: Understanding what the mathematical solution means in the context of the original problem is essential.
When solving these problems, always sketch a diagram if possible. Visual representation can greatly aid in understanding the problem and setting up the correct equations.
By mastering these techniques, students gain powerful tools for analyzing complex relationships and solving real-world problems, preparing them for advanced studies in mathematics and related fields.
