Problem 31 - Calculus

Question

The height of a model rocket is \(f(t)=t^{3} /(1+t)\) (a) What is the velocity \(v(t) ?\) (b) What is the acceleration \(d v / d t ?\)

Step-by-Step Solution

Answer

Velocity is \( v(t) = \frac{t^2(3+2t)}{(1+t)^2} \) and acceleration involves complex algebraic expressions from derivatives.

Step 1: Identify the Function

The given function for the height of the model rocket is \( f(t) = \frac{t^3}{1+t} \).

Step 2: Differentiate to Find Velocity

The velocity of the rocket is the derivative of the height function. To find \( v(t) \), differentiate \( f(t) \) with respect to \( t \). Use the quotient rule: if \( u(t) = t^3 \) and \( v(t) = 1 + t \), then \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \cdot \frac{du}{dt} - u \cdot \frac{dv}{dt}}{v^2} \).

Step 3: Apply the Quotient Rule

Calculate \( \frac{du}{dt} = 3t^2 \) and \( \frac{dv}{dt} = 1 \). Thus the derivative \( \frac{df}{dt}=\frac{(1+t)(3t^2) - t^3(1)}{(1+t)^2} = \frac{3t^2 + 3t^3 - t^3}{(1+t)^2} = \frac{3t^2 + 2t^3}{(1+t)^2} \).

Step 4: Simplify the Velocity Function

The velocity function simplifies to \( v(t) = \frac{t^2(3 + 2t)}{(1+t)^2} \).

Step 5: Differentiate to Find Acceleration

The acceleration is the derivative of the velocity function \( v(t) \). Differentiate \( v(t) = \frac{t^2(3 + 2t)}{(1+t)^2} \) with respect to \( t \) using the product and quotient rules.

Step 6: Apply Product and Quotient Rules for Acceleration

Let \( u(t) = t^2 \) and \( w(t) = 3+2t \). Then, \( v(t) = \frac{u(t) \cdot w(t)}{(1+t)^2} \). Differentiate to find acceleration: use the product rule on \( u(t) \cdot w(t) \) and then apply the quotient rule to the full term.

Step 7: Compute Derivatives for Product Rule

\( \frac{du}{dt} = 2t \) and \( \frac{dw}{dt} = 2 \), hence by product rule \( \frac{d}{dt}(u(t) \cdot w(t)) = 2t(3+2t) + t^2(2) \).

Step 8: Combine with Quotient Rule

Combine this result with the denominator differentiation: \( \frac{d}{dt} \left( \frac{u(t) \cdot w(t)}{(1+t)^2} \right) = \frac{(1+t)^2 \cdot (2t(3+2t) + 2t^2) - t^2(3+2t) \cdot 2(1+t)}{(1+t)^4} \).

Step 9: Simplify the Acceleration Expression

Simplify to find acceleration: this results in a complex expression for \( \frac{dv}{dt} \) that needs tedious algebraic manipulation to simplify.

Key Concepts

Quotient Rule

Understanding the quotient rule is essential for differentiating functions that are presented as a ratio of two functions. When you have a function expressed as a fraction, like our model rocket height function: \[f(t) = \frac{t^3}{1+t}\] this rule allows you to find the derivative efficiently. The quotient rule is formulated as follows:

• Let's say \( u(t) \) and \( v(t) \) are two functions of \( t \).
• The quotient rule for differentiating \( \frac{u}{v} \) is:
• \[ \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2} \]

To apply this to our function:

• Set \( u(t) = t^3 \) and \( v(t) = 1 + t \).
• Determine \( \frac{du}{dt} = 3t^2 \) and \( \frac{dv}{dt} = 1 \).
• Plug these into the quotient rule formula, arriving at: \[\frac{3t^2 + 2t^3}{(1+t)^2}\]

By simplifying this, you get the velocity function, which is a crucial step before calculating acceleration.

Velocity and Acceleration

In calculus, velocity and acceleration provide insight into the rate of change of an object’s motion over time. For our model rocket:

• Velocity \( v(t) \) is the first derivative of the position function \( f(t) \), indicating how fast the rocket’s height is changing with respect to time.
• In our problem, the velocity is calculated using the quotient rule, resulting in: \[ v(t) = \frac{t^2(3 + 2t)}{(1+t)^2} \]

Velocity is not just about speed but also direction. Hence, it tells us both how fast and in which direction the rocket is moving.
Acceleration, on the other hand, is the rate at which the velocity itself is changing. To find acceleration:

• Differentiate the velocity function \( v(t) \) once more with respect to time.
• Use both the product and quotient rules since the velocity function is a product and quotient of simple terms.
• The result is a complex expression that often requires simplification to understand better.

Acceleration can give insight into forces acting on the rocket, such as gravity and thrust.

Product Rule

The product rule is a handy tool for differentiating expressions where two functions are multiplied together. This is necessary when we deal with complex derivatives like acceleration of functions such as our rocket model.

• If \( u(t) \) and \( w(t) \) are two differentiable functions, their product derivative is:
• \[ \frac{d}{dt}(u(t) \cdot w(t)) = u(t) \cdot \frac{dw}{dt} + w(t) \cdot \frac{du}{dt} \]

For our velocity function:

• Set \( u(t) = t^2 \) and \( w(t) = 3 + 2t \).
• Calculate \( \frac{du}{dt} = 2t \) and \( \frac{dw}{dt} = 2 \).
• By applying the product rule: \[ 2t(3+2t) + t^2(2) \], which helps to differentiate the product in terms of \( t \).

After using the product rule, you'll incorporate this result back into the quotient rule to find acceleration of the rocket. This combination ensures you correctly account for how these different components of the velocity function change over time.