Solving Calculus Problems and Exploring Patterns in Sequences

Calculus problems can often be intimidating, but breaking them down into manageable parts can make them much more approachable. Here, we will explore two distinct problems: a calculus problem related to acceleration and position, and a sequence problem involving patterns in numbers. We will provide detailed solutions and explain the thought processes behind each step.

Calculus Problem: Acceleration and Position

Consider a problem involving acceleration and position. Given:

The acceleration (a -2etext{cms}^2) The initial velocity (u 12 text{cms}^{-1}) The initial position (s -55 text{cm})

First, let's determine the velocity as a function of time:

[v u at 12 - 2et]

This can be simplified to:

[v -2et 12 text{cms}^{-1}]

Next, we need to find the position as a function of time:

[s ut frac{1}{2}at^2 -55 12t - et^2]]

This simplifies to:

[s -et^2 12t - 55 text{cm}]

This problem illustrates how to use the given acceleration and initial conditions to derive the velocity and position functions using basic calculus principles.

Integration Problem: Evaluating an Integral

Another calculus problem involves evaluating an integral. Given that:

[int_{0}^{8} f(u) , du 6]

We need to evaluate:

[int_{1}^{3} x f(x^2 - 1) , dx]

To solve this, we use a substitution to simplify the integral:

[u x^2 - 1 Rightarrow du 2x , dx Rightarrow frac{1}{2}du x , dx]

With this substitution, the limits change as follows:

[u(1) 0, quad u(3) 8]

Therefore, the integral becomes:

[int_{1}^{3} x f(x^2 - 1) , dx frac{1}{2} int_{0}^{8} f(u) , du frac{1}{2} cdot 6 3]

The final answer is 3.

Pattern Recognition in Sequences

We encounter another interesting problem involving a sequence of numbers. Let's explore the pattern:

Given that Alice and Bob write numbers according to the rules:

Alice starts with a number (a) Bob writes (2a) Alice writes (2a - 45) This process continues where one person writes the number and the next doubles it and subtracts 45

We need to find the possible values of (a) that will allow a number to be written down again by either Alice or Bob. Let's analyze it step-by-step:

Alice writes (a_n 2^{n-1}a - 2^{n-1} - 145) Bob writes (b_n 2^na - 2^n - 245) For Alice to write the number (a) again:

[2^{n-1}a - 2^{n-1} - 145 a Rightarrow 2^{n-1}a - a 2^{n-1} 145]

This simplifies to:

[a(2^{n-1} - 1) 2^{n-1} 145 Rightarrow a frac{2^{n-1} 145}{2^{n-1} - 1}]

Since (a) must be an integer, the numerator must be divisible by the denominator. This is only true for specific values of (n): 1, 2, 3, 4, 5:

(n1): (a 45) (n2): (a 0) (n3): (a frac{270}{2} 135/2 rightarrow) not an integer (n4): (a 42) (n5): (a frac{1350}{31} rightarrow) not an integer

Hence, the possible integer values for (a) are (45, 0, 42). Therefore, the sequence can repeat with (a 45, 0, 42) under the given conditions.

By exploring these problems, we learn the importance of understanding the underlying mathematical principles and the value of careful analysis and pattern recognition in approaching complex problems.