Problem 18
Question
Use implicit differentiation to find the tangent line to the given curve at the given point \(P_{0}\). \(7 x^{1 / 2}-x v^{1 / 2}=3 \quad P_{0}=(9,4)\)
Step-by-Step Solution
Verified Answer
The tangent line is \(v = -15x + 139\) at \((9,4)\).
1Step 1: Differentiate Both Sides with Respect to x
The given equation is \(7x^{1/2} - xv^{1/2} = 3\). We differentiate both sides with respect to \(x\). Remember to use the product rule when differentiating a product of functions. Differentiate \(7x^{1/2}\) to get \(\frac{7}{2}x^{-1/2}\). For the second term, apply the product rule: \(u = x\) and \(v = v^{1/2}\), where \( v = v(x) \). This gives \(u'v + uv' = 1 \cdot v^{1/2} + x \cdot \frac{1}{2}v^{-1/2}\frac{dv}{dx}\). Differentiating both sides results in the equation: \[\frac{7}{2}x^{-1/2} - (v^{1/2} + \frac{1}{2}x v^{-1/2} \frac{dv}{dx}) = 0\].
2Step 2: Solve for dv/dx
This simplifies to: \[\frac{7}{2}x^{-1/2} - v^{1/2} - \frac{1}{2}x v^{-1/2} \frac{dv}{dx} = 0\]. Rearranging to solve for \(\frac{dv}{dx}\), we have: \[\frac{1}{2}x v^{-1/2} \frac{dv}{dx} = \frac{7}{2}x^{-1/2} - v^{1/2}\]. Finally, divide both sides by \(\frac{1}{2}x v^{-1/2}\) to isolate \(\frac{dv}{dx}\): \[\frac{dv}{dx} = \frac{(7x^{-1/2} - 2v^{1/2})}{x^{-1/2}}\].
3Step 3: Evaluate dv/dx at the Point (9,4)
Substitute \(x = 9\) and \(v = 4\) into the expression for \(\frac{dv}{dx}\). Remember \(v^{1/2} = 2\). Thus, the expression becomes: \[\frac{dv}{dx} = \frac{(7(9)^{-1/2} - 2 \times 2)}{(9)^{-1/2}}\]. Simplify this to: \[\frac{dv}{dx} = \frac{(\frac{7}{3} - 4)}{\frac{1}{3}} = 7 - 12 = -5\]. Therefore, \(\frac{dv}{dx} = -15\) at the point (9,4).
4Step 4: Write the Equation of the Tangent Line
The slope of the tangent line at \((9,4)\) is \(-15\). Use the point-slope form of the equation for a line: \(y - y_1 = m(x - x_1)\) where \((x_1, y_1) = (9,4)\) and \(m = -15\). Substitute these values: \(v - 4 = -15(x - 9)\). Simplify to find the equation of the tangent line: \[v = -15x + 139\].
Key Concepts
Tangent LineDerivativesProduct Rule
Tangent Line
A tangent line to a curve at a given point represents a straight line that just "touches" the curve at that point, without crossing it.
This concept is vital because it gives us an approximate idea of the curve's slope or steepness at a specific location.
When we deal with complex curves, such as the one given in the exercise, it's often necessary to use implicit differentiation to find the slope.In this exercise, determining the tangent line helped us find that at the point \((9, 4)\), the line's slope is \(-15\).
The process of finding this slope involves differentiating the implicit function and solving for the derivative, denoted as \(\frac{dv}{dx}\).
Once we have the slope, we can plug it into the point-slope equation, \(v - y_1 = m(x - x_1)\). This equation gives us the tangent's formula: \(v = -15x + 139\) at \((9,4)\).
Tangent lines are useful because they simplify complex curves into linear forms. You can use them to analyze behavior around points on the curve, which is highly useful in calculus.
This concept is vital because it gives us an approximate idea of the curve's slope or steepness at a specific location.
When we deal with complex curves, such as the one given in the exercise, it's often necessary to use implicit differentiation to find the slope.In this exercise, determining the tangent line helped us find that at the point \((9, 4)\), the line's slope is \(-15\).
The process of finding this slope involves differentiating the implicit function and solving for the derivative, denoted as \(\frac{dv}{dx}\).
Once we have the slope, we can plug it into the point-slope equation, \(v - y_1 = m(x - x_1)\). This equation gives us the tangent's formula: \(v = -15x + 139\) at \((9,4)\).
Tangent lines are useful because they simplify complex curves into linear forms. You can use them to analyze behavior around points on the curve, which is highly useful in calculus.
Derivatives
Derivatives are at the heart of calculus and serve as a tool for measuring how a function changes as its input changes.
In basic terms, the derivative tells us the slope of a function at any given point.For the exercise, the derivative of the function given tells us how the function \(v\) changes with \(x\). When we find \(\frac{dv}{dx}\), we are essentially calculating how sensitive \(v\) is to changes in \(x\).
This is crucial because knowing this relationship allows us to predict how the function behaves within small ranges.Whenever we perform implicit differentiation, as in our exercise, we differentiate expressions that involve both \(x\) and another variable, such as \(v\).
By considering implicit differentiation, we can work with more complex equations that can't be easily rearranged into simple functions.
The process usually involves applying rules like the product rule and chain rule to systematically find each derivative.
In basic terms, the derivative tells us the slope of a function at any given point.For the exercise, the derivative of the function given tells us how the function \(v\) changes with \(x\). When we find \(\frac{dv}{dx}\), we are essentially calculating how sensitive \(v\) is to changes in \(x\).
This is crucial because knowing this relationship allows us to predict how the function behaves within small ranges.Whenever we perform implicit differentiation, as in our exercise, we differentiate expressions that involve both \(x\) and another variable, such as \(v\).
By considering implicit differentiation, we can work with more complex equations that can't be easily rearranged into simple functions.
The process usually involves applying rules like the product rule and chain rule to systematically find each derivative.
Product Rule
The product rule is an essential tool in calculus used to find the derivative of products of two functions.
When you have a function composed of two sub-functions being multiplied, you can't simply differentiate each separately.
Instead, you use the product rule, which states: \[(uv)' = u'v + uv'\], where \(u\) and \(v\) are functions of \(x\).In our exercise, implicit differentiation requires the product rule to differentiate the term \(xv^{1/2}\).
By identifying \(u = x\) and \(v = v^{1/2}\), we can apply the product rule to differentiate.
This breaks down to \(1 \cdot v^{1/2} + x \cdot \frac{1}{2}v^{-1/2} \frac{dv}{dx}\).
Using the product rule is fundamental when dealing with problems involving implicit functions, as it ensures you correctly account for interactions between different terms.
Mastering the product rule is key for solving a variety of calculus problems, especially those involving complex relationships between variables.
When you have a function composed of two sub-functions being multiplied, you can't simply differentiate each separately.
Instead, you use the product rule, which states: \[(uv)' = u'v + uv'\], where \(u\) and \(v\) are functions of \(x\).In our exercise, implicit differentiation requires the product rule to differentiate the term \(xv^{1/2}\).
By identifying \(u = x\) and \(v = v^{1/2}\), we can apply the product rule to differentiate.
This breaks down to \(1 \cdot v^{1/2} + x \cdot \frac{1}{2}v^{-1/2} \frac{dv}{dx}\).
Using the product rule is fundamental when dealing with problems involving implicit functions, as it ensures you correctly account for interactions between different terms.
Mastering the product rule is key for solving a variety of calculus problems, especially those involving complex relationships between variables.
Other exercises in this chapter
Problem 18
Calculate the value of the given inverse trigonometric function at the given point. $$ \arccos (\cos (4 \pi / 3)) $$
View solution Problem 18
Use the method of increments to estimate the value of \(f(x)\) at the given value of \(x\) using the known value \(f(c)\) $$ f(x)=x e^{x}, c=0, x=0.12 $$
View solution Problem 18
An expression for \(f(x)\) is given. Compute the first, second, and third derivatives of \(f(x)\) with respect to \(x\). \(x \sin (x)\)
View solution Problem 18
Calculate the derivative of the given expression with respect to \(x\). $$ x^{4} \sin \left(1 / x^{2}\right) $$
View solution