Problem 10
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. $$\sqrt{x}-2 \sqrt{y}=0 ;(4,1)$$
Step-by-Step Solution
Verified Answer
Answer: The slope of the curve at the point (4,1) is \(\frac{1}{4}\).
1Step 1: Rewrite the Function
Rewrite the given function with a rational exponent to make it easier to differentiate:
$$x^{\frac{1}{2}} - 2y^{\frac{1}{2}}=0$$
2Step 2: Implicit Differentiation
Differentiate both sides of the equation with respect to x, treating y as an implicit function of x:
$$\frac{d}{dx}(x^{\frac{1}{2}}) - \frac{d}{dx}(2y^{\frac{1}{2}})=\frac{d}{dx}(0)$$
Using the power rule and chain rule, we get:
$$\frac{1}{2}x^{-\frac{1}{2}} - 2\cdot\frac{1}{2}y^{-\frac{1}{2}}\cdot\frac{dy}{dx}=0$$
3Step 3: Solve for \(\frac{dy}{dx}\)
Simplify the equation and solve for \(\frac{dy}{dx}\):
$$\frac{1}{2\sqrt{x}} - \frac{1}{\sqrt{y}}\cdot\frac{dy}{dx}=0$$
Next, isolate \(\frac{dy}{dx}\):
$$\frac{dy}{dx}=\frac{1}{\sqrt{y}}\cdot\frac{1}{2\sqrt{x}}$$
4Step 4: Substitute the Given Point
Substitute the given point \((4,1)\) into the derivative equation to find the slope of the curve at that point:
$$\frac{dy}{dx}\Big|_{(4,1)}=\frac{1}{\sqrt{1}}\cdot\frac{1}{2\sqrt{4}}$$
5Step 5: Calculate the Slope
Simplify the expression:
$$\frac{dy}{dx}\Big|_{(4,1)}=\frac{1}{1}\cdot\frac{1}{2\cdot 2}=\frac{1}{4}$$
So, the slope of the curve at the point \((4,1)\) is \(\frac{1}{4}\).
Key Concepts
Chain RuleSlope of a CurvePower Rule
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. To understand the chain rule, imagine that you have two functions nested within each other, much like a set of Russian dolls. If you want to take the derivative of this combined function, you can't just differentiate each part separately \textemdash you need a rule that links the derivatives of the inner and outer functions together.
Mathematically, if we have a function h(x) that can be written as g(f(x)), where g and f are differentiable functions, then the derivative of h with respect to x (h'(x)) is given by the product of the derivative of g with respect to f (g'(f(x))) and the derivative of f with respect to x (f'(x)). This is written as:
Mathematically, if we have a function h(x) that can be written as g(f(x)), where g and f are differentiable functions, then the derivative of h with respect to x (h'(x)) is given by the product of the derivative of g with respect to f (g'(f(x))) and the derivative of f with respect to x (f'(x)). This is written as:
\( h'(x) = g'(f(x)) \times f'(x) \)
In the context of our problem, when deriving \( y^{1/2} \), we treated y as a function of x and applied the chain rule. This resulted in multiplying the derivative of \( y^{1/2} \) with respect to y by \( \frac{dy}{dx} \), acknowledging that y is itself varying with x.Slope of a Curve
The slope of a curve at any given point is the rate at which the y-coordinate changes with respect to the x-coordinate as you move along the curve. To find the slope, we calculate the derivative of the function describing the curve. This derivative, represented by \( \frac{dy}{dx} \), is also known as the instantaneous rate of change and is crucial in understanding the behavior of the curve at a particular point.
The slope is tangent to the curve at the point of interest and can be interpreted as the steepness of the curve at that place. In our example, once we determined the expression for \( \frac{dy}{dx} \), we evaluated it at the specific point (4,1) to find the slope at that exact location. The computed slope of \( \frac{1}{4} \) tells us how steep the curve is at the point (4,1). If the slope were higher, the curve would be steeper, and conversely, a smaller slope would indicate a more gradual incline or decline at that point.
The slope is tangent to the curve at the point of interest and can be interpreted as the steepness of the curve at that place. In our example, once we determined the expression for \( \frac{dy}{dx} \), we evaluated it at the specific point (4,1) to find the slope at that exact location. The computed slope of \( \frac{1}{4} \) tells us how steep the curve is at the point (4,1). If the slope were higher, the curve would be steeper, and conversely, a smaller slope would indicate a more gradual incline or decline at that point.
Power Rule
The power rule is one of the most straightforward and frequently used differentiation rules in calculus. It provides a way to differentiate functions of the form \( x^n \) where n is any real number. According to the power rule, the derivative of \( x^n \) with respect to x is \( nx^{n-1} \). This simple rule is immensely powerful because it can be applied to various functions to find their derivatives quickly and efficiently.
In our exercise, the power rule was employed when taking the derivative of \( x^{1/2} \) and yielded \( \frac{1}{2}x^{-1/2} \) as a result. For the y term, even though it involved an implicit differentiation, the power rule was still applied initially to get \( \frac{1}{2}y^{-1/2} \) before invoking the chain rule because of y's dependency on x. Understanding the power rule is essential for any student wishing to master the techniques of calculus, as it simplifies the process of finding derivatives for a whole host of functions.
In our exercise, the power rule was employed when taking the derivative of \( x^{1/2} \) and yielded \( \frac{1}{2}x^{-1/2} \) as a result. For the y term, even though it involved an implicit differentiation, the power rule was still applied initially to get \( \frac{1}{2}y^{-1/2} \) before invoking the chain rule because of y's dependency on x. Understanding the power rule is essential for any student wishing to master the techniques of calculus, as it simplifies the process of finding derivatives for a whole host of functions.
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