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Chapter 3: Problem 10

Use the method of implicit differentiation to calculate \(d y / d x\) at the point \(P_{0}\) \(x^{4}-y^{4}=-15 \quad P_{0}=(1,2)\)

Short Answer

Expert verified
The derivative \(dy/dx\) at \((1,2)\) is \(\frac{1}{8}.\)

Step by step solution

01

Differentiate Both Sides

Start by differentiating both sides of the equation with respect to \(x\). For the left side \(x^4 - y^4 = -15\), differentiate each term separately. The derivative of \(x^4\) is \(4x^3\), and for \(y^4\), use the chain rule: \(d(y^4)/dx = 4y^3 \times dy/dx\). The right side is constant so its derivative is 0.
02

Set Derivatives Equal

Set the derivatives from Step 1 equal to each other: \[4x^3 - 4y^3 \frac{dy}{dx} = 0\]. This equation results from differentiating the left side and recognizing the right side's derivative is zero.
03

Solve for dy/dx

To solve for \(dy/dx\), first move \(4y^3 \frac{dy}{dx}\) to the other side: \[4x^3 = 4y^3 \frac{dy}{dx}.\] Now, divide both sides by \(4y^3\) to isolate \(\frac{dy}{dx}\): \[\frac{dy}{dx} = \frac{x^3}{y^3}.\]
04

Evaluate at Point \(P_0 = (1,2)\)

Substitute \((x, y) = (1, 2)\) into the expression for \(\frac{dy}{dx}\). This gives: \(\frac{dy}{dx} = \frac{1^3}{2^3} = \frac{1}{8}.\) Thus, the slope of the tangent line at point \(P_0\) is \(\frac{1}{8}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The Chain Rule is a fundamental concept in calculus used to differentiate compositions of functions. When a function is nested inside another function, we apply the Chain Rule to find its derivative.

In our exercise, we have the term \(y^4\). To differentiate it implicitly, we view \(y\) as a function of \(x\), say \(y = g(x)\). So, differentiating \(y^4\) becomes an application of the Chain Rule:
  • First, differentiate \(y^4\) with respect to \(y\), which gives us \(4y^3\).
  • Next, multiply that by the derivative of \(y\) with respect to \(x\), denoted as \(dy/dx\).
This gives us \(4y^3 \cdot \frac{dy}{dx}\), integrating both the inner and outer functions.
Derivative
The derivative measures how a function changes as its input changes. It's often likened to the slope of a curve at any point. In this problem, we want to find \(\frac{dy}{dx}\), which represents how \(y\) changes with respect to \(x\).

Using implicit differentiation, we differentiate both sides of the equation \(x^4 - y^4 = -15\) concerning \(x\). This involves differentiating explicitly for \(x\) and implicitly for \(y\), as shown here:
  • Differentiate \(x^4\) as \(4x^3\).
  • For \(y^4\), employ the Chain Rule to get \(4y^3 \cdot \frac{dy}{dx}\).
On the right side, as \(-15\) is constant, its derivative is \(0\). The equation is then solved for \(\frac{dy}{dx}\), yielding the expression \(\frac{dy}{dx} = \frac{x^3}{y^3}\).
Slope of Tangent Line
The slope of the tangent line at a specific point on a curve represents how steeply the curve is ascending or descending at that point. The tangent line itself is a straight line that just "touches" the curve at one point. When trying to determine the slope at a particular point, we substitute the point's coordinates into the derivative we found.

In this problem, at the point \(P_0 = (1,2)\), we substitute \(x = 1\) and \(y = 2\) into \(\frac{dy}{dx} = \frac{x^3}{y^3}\) and calculate:
  • Plug in values: \(\frac{dy}{dx} = \frac{1^3}{2^3} = \frac{1}{8}\).
Therefore, the slope at this point is \(\frac{1}{8}\), indicating a gentle uphill gradient.
Equation of a Tangent Line
To write the equation of a tangent line at a specific point, we use the point-slope form of a linear equation: \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point of tangency.

We already found the slope of the tangent to be \(\frac{1}{8}\) at the point \((1,2)\). Therefore, the equation becomes:
  • Insert these into the formula: \(y - 2 = \frac{1}{8}(x - 1)\).
Simplifying this further provides the tangent line equation: \(y = \frac{1}{8}x + \frac{15}{8}\). This elicits how the tangent line fits with the graph of the original equation at that specific spot.

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Most popular questions from this chapter

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=1+y / x, y(2)=1 / 2, x_{1}=3 / 2 ; \text { Exact solution: } y(x)= \\\ x \ln (x)+x(1 / 4-\ln (2)) \end{array} $$

Calculate the linearization \(L(x)=f(c)+\) \(f^{\prime}(c), \cdot(x-c)\) for the given function \(f\) at the given value \(c\) $$ f(x)=\cos ^{2}(x), c=\pi / 4 $$

Differentiate the given expression with respect to \(x\). $$ \operatorname{sech} \sqrt{x} $$

Use a central difference quotient to approximate \(f^{\prime}(c)\) for the given \(f\) and \(c .\) Plot the function and the tangent line at \((c, f(c))\). $$ f(x)=\sinh ^{-1}\left(\log _{2}(x)\right), \quad c=2.5 $$

Differentiate the given expression with respect to \(x\). $$ \arctan (\sinh (x)) $$
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