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Chapter 3: Problem 78

Determine equations of the lines tangent to the graph of \(y=x \sqrt{5-x^{2}}\) at the points (1,2) and (-2,-2) Graph the function and the tangent lines.

Short Answer

Expert verified
Answer: The equations of the tangent lines are \(y = x + 1\) at the point (1, 2) and \(y = 3x - 4\) at the point (-2, -2).

Step by step solution

01

Find the derivative of the function

To find the slope of a tangent line to a function, we first need to find its derivative. Our given function is \(y = x\sqrt{5 - x^2}\). We can rewrite this as \(y = x(5 - x^2)^{1/2}\). Using the product rule and the chain rule, we derive the function with respect to \(x\): \(\frac{dy}{dx} = x\frac{1}{2}(5 - x^2)^{-1/2}(-2x) + (5 - x^2)^{1/2}\)
02

Find the slope (derivative) at the given points

Now we will evaluate the derivative at the given points (1, 2) and (-2, -2). At (1,2): \(\frac{dy}{dx} = 1\frac{1}{2}(5 - 1^2)^{-1/2}(-2(1)) + (5 - 1^2)^{1/2} = -1 + 2 = 1\) At (-2,-2): \(\frac{dy}{dx} = -2\frac{1}{2}(5 - (-2)^2)^{-1/2}(-2(-2)) + (5 - (-2)^2)^{1/2} = 4 - 1 = 3\) So, the slopes of the tangent lines at (1,2) and (-2,-2) are 1 and 3, respectively.
03

Find the equation of each tangent line using point-slope form

Now we will use the point-slope form of a line to find the equation of the tangent line at each given point: At (1,2) \(y - 2 = 1(x - 1) \Rightarrow y = x + 1\) At (-2,-2) \(y + 2 = 3(x + 2) \Rightarrow y = 3x - 4\) Our tangent lines are \(y = x + 1\) and \(y = 3x - 4\).
04

Graph the function and tangent lines

To complete the exercise, graph the given function \(y = x\sqrt{5 - x^2}\) along with the tangent lines \(y = x + 1\) and \(y = 3x - 4\). Verify visually that the tangent lines touch the function at the given points (1, 2) and (-2, -2). The function and tangent lines should intersect only at the given points (1, 2) and (-2, -2).

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