Question

Carry out the following steps. a. Use implicit differentiation to find \(\frac{d y}{d x}\). b. Find the slope of the curve at the given point. $$x \sqrt[3]{y}+y=10 ;(1,8)$$

Short answer

The slope of the curve at the point (1, 8) is \(-\frac{24}{13}\).

Step-by-step solution

Rewrite the equation

First, we rewrite the equation so we can apply implicit differentiation more easily. The equation given is: $$x \sqrt[3]{y}+y=10$$ We can rewrite it as: $$x y^{\frac{1}{3}}+y=10$$

Apply implicit differentiation

Now, we will differentiate both sides of the equation with respect to x. Remember, whenever you differentiate a term containing y, you must multiply it by \(\frac{d y}{d x}\) since y is a function of x. Applying implicit differentiation gives: $$\frac{d}{dx}(x y^{\frac{1}{3}})+\frac{d}{dx}(y)=\frac{d}{dx}(10)$$

Differentiate each term

Now, we will differentiate each term. We need to use the product rule for the first term and the chain rule for the second term. The product rule states that \((uv)' = u'v + uv'\), and the chain rule states that \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Differentiating each term, we get: $$(x y^{\frac{1}{3}})' = 1 \cdot y^{\frac{1}{3}} + x \cdot \frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx}$$ $$(y)' = \frac{dy}{dx}$$ $$(10)' = 0$$ Now, we substitute these results back into our earlier equation: $$y^{\frac{1}{3}} + x \cdot \frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx} + \frac{dy}{dx}=0$$

Solve for \(\frac{dy}{dx}\)

Now, we need to solve the equation for \(\frac{dy}{dx}\). We will isolate the term containing the derivative on the left side of the equation and move the other terms to the right side. So, we get: $$x \cdot \frac{1}{3}y^{-\frac{2}{3}}\frac{dy}{dx} + \frac{dy}{dx} = -y^{\frac{1}{3}}$$ Then, we can factor out the \(\frac{dy}{dx}\) term: $$\frac{dy}{dx}(x \cdot \frac{1}{3}y^{-\frac{2}{3}} +1) = -y^{\frac{1}{3}}$$ Finally, we can solve for \(\frac{dy}{dx}\): $$\frac{dy}{dx} = \frac{-y^{\frac{1}{3}}}{x \cdot \frac{1}{3}y^{-\frac{2}{3}}+1}$$

Find the slope at the given point

Now that we have the derivative, we can find the slope of the curve at the given point (1, 8). We substitute x = 1 and y = 8 into the derivative and evaluate the expression: $$\frac{dy}{dx} = \frac{-8^{\frac{1}{3}}}{1 \cdot \frac{1}{3}8^{-\frac{2}{3}}+1}$$ $$\frac{dy}{dx} = \frac{-2}{\frac{1}{3}\cdot \frac{1}{4}+1}$$ $$\frac{dy}{dx} = \frac{-2}{\frac{1}{12}+1}$$ $$\frac{dy}{dx} = \frac{-2}{\frac{13}{12}}$$ $$\frac{dy}{dx} = -\frac{24}{13}$$ Thus, the slope of the curve at the point (1, 8) is \(-\frac{24}{13}\).

Key concepts

Slope of a Curve

Understanding the concept of the 'slope of a curve' at a specific point helps students in Calculus visualize the steepness and direction of a line tangent to the curve. The slope is essentially the rate at which the y-value of a curve is changing with respect to its x-value. In graphical terms, it is the angle of the tangent line at a particular point on the curve.

To find this slope, we often differentiate the equation of the curve with respect to x, which gives us the derivative, or the slope function. This is what we mean when we talk about finding \(\frac{dy}{dx}\). To find the slope at a specific point, we simply plug the x and y coordinates of the point into the derivative. This is what we did in the exercise when we evaluated \(\frac{dy}{dx}\) at the point (1, 8), revealing that the slope of the curve at this point is \( -\frac{24}{13}\).

Product Rule

The 'product rule' is a powerful tool in differential calculus used when differentiating products of two functions. It's expressed as \( (uv)' = u'v + uv' \), where u and v are functions of x. This rule helps us break down complex expressions into simpler parts that are easier to differentiate.

For the given problem, using the product rule allowed us to differentiate \( x y^{\frac{1}{3}} \), treating \( x \) and \( y^{\frac{1}{3}} \) as separate functions. We differentiated x to get 1 and \( y^{\frac{1}{3}} \) using the chain rule, considering the dependency of y on x.

Chain Rule

The 'chain rule' is essential for differentiating composite functions: functions of functions. The rule can be summarized as \( (f(g(x)))' = f'(g(x)) \cdot g'(x) \). It allows us to differentiate the inner function and then multiply by the derivative of the outer function.

In our exercise, the chain rule helped differentiate \( y^{\frac{1}{3}} \) with respect to x, since y is itself a function of x. This rule often works in tandem with the product rule to differentiate more complex expressions, making it an indispensable technique in calculus.

Differentiation Techniques

There are numerous 'differentiation techniques' to tackle different kinds of problems. From the basic power rule to more complex methods like implicit differentiation, each technique has its place. Implicit differentiation was used in this exercise because y was a function of x, but not isolated on one side of the equation. We applied the product and chain rules as part of our differentiation technique to find the equation for the slope function \(\frac{dy}{dx}\).

Combining these techniques allows for the differentiation of almost any function, whether it's polynomial, rational, trigonometric, exponential, or logarithmic. Practice in recognizing when and how to apply these techniques is crucial for mastery in Calculus.

Calculus

The field of 'calculus' encompasses various concepts like limits, derivatives, integrals, and infinite series. It is a profound tool for analyzing changes and motion, and is extensively used across science, engineering, economics, and more. The exercise we went through encapsulates several key aspects of differential calculus, which focuses on the concept of change, slopes, and tangents.

Through calculus, we can solve for rates of change and understand the behavior of functions, as seen in finding the slope of a curve. Slope calculations form the basis for understanding motion and change, highlighting the endless applications of calculus in real-world problems and theoretical analysis alike.