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Chapter 3: Problem 38

a. Find an equation of the line tangent to the given curve at a. b. Use a graphing utility to graph the curve and the tangent line on the same set of axes. $$y=\frac{e^{x}}{4}-x ; a=0$$

Short Answer

Expert verified
Answer: The equation of the tangent line is $$y = -\frac{3}{4}x + \frac{1}{4}$$.

Step by step solution

01

Find the derivative

To find the derivative of $$y = \frac{e^x}{4} - x$$, we'll derive each term separately. The derivative of $$\frac{e^x}{4}$$ is $$\frac{1}{4}e^x$$, and the derivative of $$-x$$ is $$-1$$. Therefore, putting these together, the derivative of the function is: $$y' = \frac{1}{4}e^x - 1$$
02

Evaluate the derivative at a = 0

Now that we have the derivative, let's evaluate it at the given point $$a = 0$$: $$y'(0) = \frac{1}{4}e^0 - 1 = \frac{1}{4}(1) - 1 = -\frac{3}{4}$$ The slope of the tangent line at the point $$a = 0$$ is $$-\frac{3}{4}$$.
03

Find the coordinates of the point

To find the coordinates of the point where $$a=0$$, we will plug $$x=0$$ into the original function: $$y = \frac{e^0}{4} - 0 = \frac{1}{4}$$ So the coordinates of the point are $$(0, \frac{1}{4})$$.
04

Use the point-slope formula to find the equation of the tangent line

Now that we have the slope ($$-\frac{3}{4}$$) and coordinates ($$(0, \frac{1}{4})$$) of the point at $$a = 0$$, we can use the point-slope formula to find the equation of the tangent line: $$y - y_{1} = m(x - x_{1})$$ Plugging in the values: $$y - \frac{1}{4} = -\frac{3}{4}(x - 0)$$ Simplifying, we get: $$y = -\frac{3}{4}x + \frac{1}{4}$$ This is the equation of the tangent line.
05

Graph the curve and tangent line

Finally, use a graphing utility to graph the original function $$y = \frac{e^x}{4} - x$$ and the tangent line $$y = -\frac{3}{4}x + \frac{1}{4}$$ on the same set of axes. The student should overlay the graphs to see the tangent line touching the curve at the point $$(0, \frac{1}{4})$$.

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Most popular questions from this chapter

An observer is \(20 \mathrm{m}\) above the ground floor of a large hotel atrium looking at a glass-enclosed elevator shaft that is \(20 \mathrm{m}\) horizontally from the observer (see figure). The angle of elevation of the elevator is the angle that the observer's line of sight makes with the horizontal (it may be positive or negative). Assuming that the elevator rises at a rate of \(5 \mathrm{m} / \mathrm{s}\), what is the rate of change of the angle of elevation when the elevator is \(10 \mathrm{m}\) above the ground? When the elevator is \(40 \mathrm{m}\) above the ground?

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