Problem 2
Question
For small positive values of \(z, z<\tan z\) (Equation 7.5\(),\) but 'just barely so' a. Compute $$ \tan z-z \quad \text { for } \quad z=0.01 $$ for \(\quad z=0.001 \quad\) and for \(\quad z=0.0001\). b. Compute $$ \frac{\tan z}{z} $$ for \(\quad z=0.01\) for \(\quad z=0.001 \quad\) and for \(\quad z=0.0001\). c. Note that the slope of the tangent to \(y=\tan t\) at (0,0) is $$ [\tan t]_{t=0}^{\prime}=\lim _{h \rightarrow 0} \frac{\tan (0+h)-\tan 0}{h}=\lim _{h \rightarrow 0} \frac{\tan h}{h} $$ What is your best estimate of $$ [\tan t]_{t=0}^{\prime} ? $$
Step-by-Step Solution
Verified Answer
The best estimate of \( [\tan t]_{t=0}^{\prime} \) is 1.
1Step 1: Compute \( \tan z - z \) for \( z = 0.01 \)
First, compute \( \tan z \) where \( z = 0.01 \). We find that \( \tan(0.01) \approx 0.0100003333 \). Now, calculate \( \tan z - z = 0.0100003333 - 0.01 = 0.0000003333 \).
2Step 2: Compute \( \tan z - z \) for \( z = 0.001 \)
Now, compute \( \tan z \) for \( z = 0.001 \), so \( \tan(0.001) \approx 0.001000000167 \). Thus, \( \tan z - z = 0.001000000167 - 0.001 = 0.000000000167 \).
3Step 3: Compute \( \tan z - z \) for \( z = 0.0001 \)
Next, compute \( \tan z \) for \( z = 0.0001 \), resulting in \( \tan(0.0001) \approx 0.00010000000033 \). Therefore, \( \tan z - z = 0.00010000000033 - 0.0001 = 0.00000000000033 \).
4Step 4: Compute \( \frac{\tan z}{z} \) for \( z = 0.01 \)
Calculate \( \frac{\tan(0.01)}{0.01} = \frac{0.0100003333}{0.01} = 1.00003333 \).
5Step 5: Compute \( \frac{\tan z}{z} \) for \( z = 0.001 \)
Calculate \( \frac{\tan(0.001)}{0.001} = \frac{0.001000000167}{0.001} = 1.000000167 \).
6Step 6: Compute \( \frac{\tan z}{z} \) for \( z = 0.0001 \)
Calculate \( \frac{\tan(0.0001)}{0.0001} = \frac{0.00010000000033}{0.0001} = 1.0000000033 \).
7Step 7: Estimate the Slope of \( y = \tan t \) at \( t = 0 \)
Using the results from steps 4-6, as \( z \) approaches 0, \( \frac{\tan z}{z} \) approaches 1. Therefore, the best estimate is \( [\tan t]_{t=0}^{\prime} = 1 \).
Key Concepts
Tangent FunctionLimit EstimationDerivative at a PointApproximations
Tangent Function
When you think about the tangent function, it's often represented as \[ \tan(z) = \frac{\sin(z)}{\cos(z)} \] for small inputs, especially in calculus-related problems. The tangent function is particularly interesting because of how it behaves as the input, or angle \( z \), approaches smaller and smaller numbers. We are interested in its derivative, which gives a precise perspective on how this function behaves near certain points.
For the exercise at hand, we explored how \( \tan(z) \) is just slightly larger than \( z \) itself for small values like 0.01, 0.001, and 0.0001. This close comparison is intrinsic to understanding the concept of limits and derivatives in calculus, as we need to understand what happens to functions as their inputs approach certain special points, like zero.
For the exercise at hand, we explored how \( \tan(z) \) is just slightly larger than \( z \) itself for small values like 0.01, 0.001, and 0.0001. This close comparison is intrinsic to understanding the concept of limits and derivatives in calculus, as we need to understand what happens to functions as their inputs approach certain special points, like zero.
- Useful in calculating angles and slopes.
- Essential for trigonometric calculations and transformations.
- Behaves predictably near zero, revealing deep insights into calculus applications.
Limit Estimation
Limit estimation is a foundational concept in calculus that helps us understand the behavior of functions as they approach specific values or infinity. In our exercise, we computed limits to compare \( \tan(z) \) with \( z \) for very small values of \( z \). Calculating these small differences shows how the limit of the difference, \( \tan z - z \), approaches zero as \( z \) becomes very small.
The magic of limit estimation is in its power to precise predict function behavior, especially when the function's direct calculation becomes difficult. For example, as we observed in the solutions, the limits for \( \tan z / z \) approached a crucial value, guiding us to understanding slope and tangency behaviors.
The magic of limit estimation is in its power to precise predict function behavior, especially when the function's direct calculation becomes difficult. For example, as we observed in the solutions, the limits for \( \tan z / z \) approached a crucial value, guiding us to understanding slope and tangency behaviors.
- Essential for making finite comparisons at infinite scales.
- Helps assess function continuity and catastrophic changes.
- Vital in defining derivatives and integrals precisely.
Derivative at a Point
The derivative represents the slope of the tangent line to the curve of a function at a given point. For the tangent function, calculating its derivative as \( z \) approaches zero reveals its rate of change. Specifically, at \( t = 0 \), the derivative \([\tan t]_{t=0}^{\prime}\) illustrates how the tangent function rises or falls around this point.
Our analysis led us to see that as \( z \) approaches zero, \( \frac{\tan z}{z} \) gets closer to the number 1. Therefore, the rate at which \( \tan(z) \) changes precisely at zero is truly 1. This lacks ambiguity and solidifies the derivative concept as a means to predict future behaviors based on immediate, known conditions.
Our analysis led us to see that as \( z \) approaches zero, \( \frac{\tan z}{z} \) gets closer to the number 1. Therefore, the rate at which \( \tan(z) \) changes precisely at zero is truly 1. This lacks ambiguity and solidifies the derivative concept as a means to predict future behaviors based on immediate, known conditions.
- Provides instantaneous rate of change.
- Essential for understanding function behaviors and extremes.
- Forms the backbone of differential calculus contexts.
Approximations
In calculus, approximations are an incredible tool for simplifying complex mathematical problems. They allow us to evaluate and predict function behavior near known points, using simpler functions that are easy to understand and compute. In the problem, we glean these skills by seeing \( \tan(z) \approx z \) when \( z \) is very small.
By focusing on approximations, especially through trivial calculations of \( \tan z - z \) and \( \frac{\tan z}{z} \), we were able to derive valuable estimates for calculations that would otherwise be cumbersome. This ability to approximate helps in many scientific fields where exact answers are unnecessary or impossible.
By focusing on approximations, especially through trivial calculations of \( \tan z - z \) and \( \frac{\tan z}{z} \), we were able to derive valuable estimates for calculations that would otherwise be cumbersome. This ability to approximate helps in many scientific fields where exact answers are unnecessary or impossible.
- Enable solving complex problems with less computational power.
- Provide a basis for numerical simulations and models.
- Used to derive solutions quickly when exact ones are not needed.
Other exercises in this chapter
Problem 1
For small positive values of \(z, \sin z
View solution Problem 2
Compute the derivative of $$ y(t)=1+2 t-5 t^{7}+2 e^{3 t}-\ln 6 t+2 \sin t-3 \cos t $$
View solution Problem 3
Show that the suggested solutions solve the associated equations. a. \(y=\cos t\) \(y^{\prime \prime}+y=0\) b. \(y=\cos 2 t\) \(y^{\prime \prime}+4 y=0\) c. \(\
View solution Problem 3
Compute the derivatives of a. \(\quad y=2 \sin t \cos t\) b. \(y=\sin ^{2} t+\cos ^{2} t\) c. \(\quad y=\sec t=\frac{1}{\cos t}\) d. \(y=\cot t\) e. \(y=\ln \co
View solution