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a. Find the tangent to the curve \(y=2 \tan (\pi x / 4)\) at \(x=1\) b. Slopes on a tangent curve What is the smallest value the slope of the curve can ever have on the interval \(\quad-2

Short Answer

Expert verified
a. The tangent line is \(y = \pi x - \pi + 2\). b. The smallest slope is \(\frac{\pi}{2}\).

Step by step solution

01

Find the y-coordinate at x=1

Plug the value of \(x = 1\) into the function \(y=2 \, \tan \left( \frac{\pi x}{4} \right)\) to find the \(y\)-value. This gives us: \(y = 2 \, \tan \left( \frac{\pi \cdot 1}{4} \right) = 2 \, \tan \left( \frac{\pi}{4} \right) = 2 \cdot 1 = 2\). So, the point on the curve is \((1, 2)\).
02

Find the derivative of the function

Differentiate the function \(y = 2 \, \tan \left( \frac{\pi x}{4} \right)\) with respect to \(x\). Using the chain rule:\[\frac{dy}{dx} = 2 \cdot \sec^2\left(\frac{\pi x}{4}\right) \cdot \frac{d}{dx}\left(\frac{\pi x}{4}\right)\]. Simplifying this gives us \[\frac{dy}{dx} = \frac{\pi}{2} \cdot \sec^2\left(\frac{\pi x}{4}\right)\].
03

Calculate the slope at x=1

Substitute \(x = 1\) into the derivative to find the slope at this point: \[\frac{dy}{dx} = \frac{\pi}{2} \cdot \sec^2\left(\frac{\pi \cdot 1}{4}\right) = \frac{\pi}{2} \cdot \sec^2\left(\frac{\pi}{4}\right)\]. Since \(\sec\left(\frac{\pi}{4}\right) = \sqrt{2}\), this becomes \(\frac{\pi}{2} \cdot 2 = \pi\). Thus, the slope of the tangent is \(\pi\).
04

Write the equation of the tangent line

Using the point-slope form \(y - y_1 = m(x - x_1)\) with the point \((1, 2)\) and slope \(\pi\), the equation is \(y - 2 = \pi(x - 1)\). Expanding this gives the tangent line equation: \(y = \pi x - \pi + 2\).
05

Analyze the slope of the function over given interval

From Step 2, the slope is \(\frac{dy}{dx} = \frac{\pi}{2} \cdot \sec^2\left(\frac{\pi x}{4}\right)\). At its smallest, \(\sec^2(\theta)\) is 1 (when \(\theta=0\)); thus, the minimal slope over the real line is \(\frac{\pi}{2}\). However, since \(-2<x<2\), avoid where tangent is undefined (when \(\sec(\theta)\) is undefined), ensuring it remains smallest at \(x=0\).

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Key Concepts

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

Derivative
The derivative is a fundamental concept in calculus. It measures how a function changes as its input changes - in simpler terms, it's the rate of change or the slope at any given point on a curve. When you find the derivative of a function, you're finding a new function that gives the slope of the original function's graph at each point.
For example, in the exercise, we have the function \(y = 2 \tan \left(\frac{\pi x}{4}\right)\). To discover how fast \(y\) is changing at any point \(x\), we calculate its derivative. This involves taking the derivative of \(\tan\), which is \(\sec^2\), and applying necessary adjustments based on the chain rule. The result gives us the slope at each point \(x\).
  • The notation \(\frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\).
  • The derivative here simplifies to \(\frac{\pi}{2} \sec^2\left(\frac{\pi x}{4}\right)\).
The derivative tells us how \(y\) behaves concerning \(x\). It's like a magic tool to predict the curve's behavior without needing to graph it.
Slope of Tangent Line
Understanding the slope of a tangent line is crucial in calculus. A tangent line is a straight line that touches a curve at a single point without crossing it. The slope of this line tells you how steep the curve is at that exact point. For the exercise, the tangent line to the curve at \(x=1\) is a major focus.
To find this slope, you plug \(x=1\) into the derivative of the function. For the function \(y = 2 \tan \left(\frac{\pi x}{4}\right)\), we calculated the derivative to be \(\frac{\pi}{2} \sec^2\left(\frac{\pi x}{4}\right)\). When evaluating this at \(x=1\), it leads to the slope \(\pi\).
  • This slope means that for every unit increase in \(x\), \(y\) increases by \(\pi\) units at \(x=1\).
  • In essence, the slope quantifies the curve's sharpness at a particular point.
  • Understanding this helps in predicting how the curve behaves locally around that point.
The equation of the tangent line can then be formed using point-slope formula using this slope and the point of tangency.
Chain Rule
The chain rule is a powerful technique in calculus used to differentiate composite functions, which are functions within functions. It allows us to find the derivative of more complicated functions by breaking them down into simpler parts.
In our exercise, we utilized the chain rule to find the derivative of the function \(y = 2 \tan \left(\frac{\pi x}{4}\right)\). This function can be seen as the composition of the tangent function and the linear function \(\frac{\pi x}{4}\).
Applying the chain rule involves:
  • Firstly, taking the derivative of the outer function \(2 \tan(u)\), where \(u = \frac{\pi x}{4}\), resulting in \(2 \sec^2(u)\).
  • Secondly, multiplying by the derivative of the inner function \(\frac{d}{dx}\left(\frac{\pi x}{4}\right)\), which simplifies to \(\frac{\pi}{4}\).
After multiplying these derivatives, we arrive at \(\frac{\pi}{2} \sec^2\left(\frac{\pi x}{4}\right)\). The chain rule elegantly connects the dots between the different layers of our function, helping us understand how each part contributes to the overall rate of change.

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Most popular questions from this chapter

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