Problem 14
Question
Find an equation for the tangent line to \(x^{2 / 3}+y^{2 / 3}=5\) at the point \((8,1)\).
Step-by-Step Solution
Verified Answer
The equation of the tangent line to the function \(x^{2 / 3}+y^{2 / 3}=5\) at the point \((8,1)\) is \(y = 3x - 23\).
1Step 1: Differentiate the given function implicitely
First, differentiate both sides of the function \(x^{2/3} + y^{2/3} = 5\) with respect to \(x\). Use the chain rule where needed (which states that the derivative of a composition of functions is the derivative of the outside function times the derivative of the inside function). This gives you \(\frac{2}{3}x^{-1/3} + \frac{2}{3}y^{-1/3} \frac{dy}{dx} = 0\).
2Step 2: Solve for \(\frac{dy}{dx}\) (slope of the tangent line)
Now, isolate \(\frac{dy}{dx}\) to solve for slope of the tangent line, this will give you \(\frac{dy}{dx} = - \frac{x^{-1/3}}{y^{-1/3}}\). Substitute the given point \((8, 1)\) into \(x\) and \(y\), which will provide the slope \(m\) of the tangent line.
3Step 3: Apply the point-slope form to find the equation of the tangent line
Finally, use the point-slope form of a line, \(y - y1 = m(x - x1)\), where \(x1 = 8\) and \(y1 = 1\) are the coordinates of the given point and \(m\) is the slope calculated in the previous step. Write the final equation of the tangent line.
Key Concepts
Implicit DifferentiationChain RulePoint-Slope Form
Implicit Differentiation
When dealing with equations involving two variables where one variable isn't isolated (like in our exercise), implicit differentiation is a crucial technique. It allows us to differentiate both sides of an equation with respect to a specific variable, often x, even if the equation is not explicitly solved for one variable. Differentiating implicitly comes in handy when equations have intricate combinations of x and y that are difficult to disentangle.
To perform implicit differentiation:
To perform implicit differentiation:
- Differentiate both sides of the equation with respect to x.
- Remember that when differentiating terms involving y, apply the chain rule by multiplying by \( \frac{dy}{dx} \) to account for y being a function of x.
Chain Rule
The chain rule is an essential tool in calculus, especially when it comes to differentiating composite functions. It tells us how to take the derivative of a function that is composed of one or more other functions.
The basic idea is:
The basic idea is:
- To differentiate a composite function, find the derivative of the outer function first.
- Then multiply this by the derivative of the inner function.
Point-Slope Form
Once we have the slope of a tangent line \(\frac{dy}{dx}\) from implicit differentiation, the next step is to write the equation of the tangent line. This is where the point-slope form of a line comes into play. It's particularly convenient when you know a point on the line and the slope.
The point-slope form is written as:
Remember, the point-slope form is an easy way to express a line, bridging the connection between the analytical slope calculated and the graphical representation of the tangent line.
The point-slope form is written as:
- \(y - y_1 = m(x - x_1)\)
Remember, the point-slope form is an easy way to express a line, bridging the connection between the analytical slope calculated and the graphical representation of the tangent line.
Other exercises in this chapter
Problem 13
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