Problem 31 - Single Variable Calculus

Question

Find an equation of the tangent line to the curve \(e^{y}+x y=e\) at \((0,1)\).

Step-by-Step Solution

Answer

The equation of the tangent line to the curve \(e^y + xy = e\) at the point (0, 1) is \(y = -\frac{1}{e}x + 1\).

Step 1: Implicit differentiation

Given the equation: \[e^y + xy = e\] Let's differentiate both sides of the equation with respect to x using implicit differentiation: \[\frac{d}{dx}\left(e^y + xy\right) = \frac{d}{dx}\left(e\right)\] Recall the chain rule: \(\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x)\) Apply chain rule for \(e^y\) term, and product rule for \(xy\) term: \[\frac{d}{dx}e^y + \left(y + x\frac{dy}{dx}\right) = 0\] Solve for \(\frac{dy}{dx}\): \[\frac{dy}{dx}\left(e^y + x\right) = -y\] \[\frac{dy}{dx} = -\frac{y}{e^y + x}\]

Step 2: Find the slope of the tangent line at (0, 1)

Now, we need to find the value of the derivative at the point (0, 1), which will give us the slope of the tangent line. Plug x = 0 and y = 1 into the derivative equation: \[\frac{dy}{dx} = -\frac{1}{e^1 + 0}\] \[\frac{dy}{dx} = -\frac{1}{e}\] So, the slope of the tangent line is \(-\frac{1}{e}\).

Step 3: Use the point-slope form to find the equation of the tangent line

We know the point is (0, 1) and the slope is \(-\frac{1}{e}\). Using the point-slope form of the line equation, which is: \(y - y_1 = m (x - x_1)\) Plug in the values: \(y - 1 = -\frac{1}{e}(x - 0)\) Simplify the equation: \(y = -\frac{1}{e}x + 1\) This is the equation of the tangent line to the curve \(e^y + xy = e\) at the point (0, 1): \[\boxed{y = -\frac{1}{e}x + 1}\]

Key Concepts

Implicit Differentiation

Implicit differentiation is a powerful technique in calculus used to find the derivative of an equation where the variables cannot be easily separated. Unlike explicit functions, where we can solve for one variable in terms of the other (e.g., y = f(x)), implicit functions involve expressions where the variables are mixed (e.g., F(x, y) = 0).

When dealing with implicit functions, such as e^y + xy = e, we differentiate both sides of the equation with respect to x. This might involve treating y as a function of x (y(x)), even if we don't have the explicit form. As a result, when we differentiate y, we need to use the chain rule and multiply by dy/dx, which accounts for the fact that y itself changes as x changes.

Differentiation of the given function requires application of various differentiation rules, including the chain rule for e^y and the product rule for the xy term, resulting in a new equation where dy/dx represents the slope of the tangent line at any point along the curve.

Tangent Line Equation

The equation of a tangent line to a curve at a specific point is a linear approximation of the curve near that point. In other words, it's the straight line that 'just touches' the curve without crossing it at that point. The slope of the tangent line at a point is exactly the value of the derivative of the function at that point.

To find this equation, we need two pieces of information: the slope of the curve at the point of tangency and the coordinates of the point itself. In our example, we found that the slope of the tangent line to the curve e^y + xy = e at the point (0, 1) is -1/e. Then, we employ the point-slope form, y - y₁ = m(x - x₁), where (x₁, y₁) is the point of tangency and m is the slope. This yields the final equation of the tangent line.

Chain Rule

The chain rule is a critical principle in calculus used for differentiating compositions of functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In mathematical terms, if we have a composite function f(g(x)), its derivative f'(x) is f'(g(x))g'(x).

Understanding the chain rule is essential for implicit differentiation, as you often deal with functions nested inside other functions. For example, differentiating e^y with respect to x calls for the chain rule since y is itself a function of x. This is why, in our initial differentiation step, we wrote the derivative as dy/dx multiplied by the derivative of e^y with respect to y, which is e^y once again.

Product Rule

The product rule is another fundamental theorem in calculus used when differentiating the product of two functions. If we have two differentiable functions u(x) and v(x), the product rule tells us that the derivative of their product u(x)v(x) is u'(x)v(x) + u(x)v'(x).

This rule is essential when differentiating expressions like xy, where both x and y are functions of some variable (in our case, x). When we applied the product rule to the xy term in our exercise, the result was the sum of y (derivative of x with respect to x is 1) and x(dy/dx) (since dx/dx is 1 and we multiply by the derivative of y, treating it as a function of x).