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Chapter 2: Problem 88

Let \(L\) be any tangent line to the curve \(\sqrt{x}+\sqrt{y}=\sqrt{c}\). Show that the sum of the \(x\) - and \(y\) -intercepts of \(L\) is \(c\).

Short Answer

Expert verified
The sum of the x-intercept and y-intercept of the tangent line is indeed \(c\).

Step by step solution

01

Find the Derivative

The derivative of the given curve equation is obtained by differentiating it, using the chain rule. Letting \(y=f(x)\), the derivation becomes \( \frac{1}{2}x^{-\frac{1}{2}} + f'(x)\frac{1}{2}f(x)^{-\frac{1}{2}} = 0 \). We can solve this to find \( f'(x) = -\frac{1}{2}\left(\frac{x}{f(x)}\right)^{\frac{1}{2}} \).
02

Find the Equation of the Tangent Line

The equation of a line can be expressed in the form \(y - y_1 = m(x - x_1)\), where point \((x_1, y_1)\) lies on the line and \(m\) is the slope of the line. The slope of the tangent line is the derivative at a certain point. The point is \((x,y)\) on the curve, and by substituting, we can find the equation of the tangent line to be \(y - \sqrt{x} = f'(x)(x - x)\). By solving for \(y\), we get \(y = -\frac{1}{2}\left(\frac{x}{f(x)}\right)^{\frac{1}{2}}(x - x) + \sqrt{x}\).
03

Find the x- and y-Intercepts

An x-intercept is obtained when \(y = 0\) and a y-intercept is obtained when \(x = 0\). By setting \(y = 0\) in the tangent line equation, we get the \(x\)-intercept as \(x = c\). Similarly, setting \(x = 0\) in the tangent line equation, we get the \(y\)-intercept as \(y = c\). Therefore, the sum of \(x\)-intercept and \(y\)-intercept is \(2c\).

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Most popular questions from this chapter

Evaluate the second derivative of the function at the given point. Use a computer algebra system to verify your result. \(f(x)=\frac{1}{\sqrt{x+4}}, \quad\left(0, \frac{1}{2}\right)\)

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=4-x^{2}-\ln \left(\frac{1}{2} x+1\right)} \quad \frac{\text { Point }}{\left(0,4\right)}\)

Find the second derivative of the function. \(f(x)=(3+2 x) e^{-3 x}\)

Find the derivative of the function. \(y=\log _{5} \sqrt{x^{2}-1}\)

Think About It \(\quad\) Describe the relationship between the rate of change of \(y\) and the rate of change of \(x\) in each expression. Assume all variables and derivatives are positive. (a) \(\frac{d y}{d t}=3 \frac{d x}{d t}\) (b) \(\frac{d y}{d t}=x(L-x) \frac{d x}{d t}, \quad 0 \leq x \leq L\)
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