Problem 21 - The Calculus with Analytic Geometry

Question

\(y=x^{2}-5 x+1 ; x=s^{3}-2 s+1 ; s=\sqrt{t^{2}+1}\)

Step-by-Step Solution

Answer

y = ((t^2 + 1)^{3/2} - 2sqrt{t^2 + 1} + 1)^2 - 5((t^2 + 1)^{3/2} - 2sqrt{t^2 + 1} + 1) + 1

Step 1: Express 'x' in terms of 't'

First, express the variable 's' in terms of 't'. Given that \[ s = \sqrt{t^{2}+1} \]Next, substitute this expression into the formula for 'x' to express 'x' in terms of 't':\[ x = (\sqrt{t^{2}+1})^3 - 2(\sqrt{t^{2}+1}) + 1 \]Simplify this to get 'x' in terms of 't'.

Step 2: Simplify 'x' in terms of 't'

Simplifying the expression from Step 1:\[ x = (t^2 + 1)^{3/2} - 2\sqrt{t^2 + 1} + 1 \]Now, 'x' has been expressed in terms of 't'.

Step 3: Substitute 'x' into 'y' to express 'y' in terms of 't'

Now, take the expression for 'x' from Step 2 and substitute it into the formula for 'y':\[ y = ( (t^2 + 1)^{3/2} - 2\sqrt{t^2 + 1} + 1 )^2 - 5( (t^2 + 1)^{3/2} - 2\sqrt{t^2 + 1} + 1 ) + 1 \]Simplify this expression to get 'y' in terms of 't'.

Step 4: Simplify 'y' in terms of 't'

Simplify the expression obtained in Step 3. This involves expanding the square and combining like terms.\[ y = ( (t^2 + 1)^{3/2} - 2\sqrt{t^2 + 1} + 1 )^2 - 5( (t^2 + 1)^{3/2} - 2\sqrt{t^2 + 1} + 1 ) + 1 \]The result will be the simplified form of 'y' as a function of 't'.

Key Concepts

chain rule

The chain rule is a key formula in calculus used to find the derivative of composite functions. It states that if a variable depends on another variable, which in turn depends on a third variable, we can compute the derivative by multiplying the derivatives along the chain. For example, if we have three variables such that

• x depends on s
• s depends on t

then the derivative of y with respect to t can be found using the chain rule as follows: \[ \frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{ds} \cdot \frac{ds}{dt} \]
In our exercise, we need to differentiate the expressions by constantly applying the chain rule from the outer function to the innermost variable.

The approach involves working systematically through each layer. This offers a powerful way to handle complex dependencies.

composite functions

A composite function is a function formed by combining two or more functions. The output of one function becomes the input of another. In our exercise, we deal with several layers of composite functions:

• s in terms of t: \[ s = \sqrt{t^2 + 1} \]
• x in terms of s: \[ x = s^3 - 2s + 1 \]
• y in terms of x: \[ y = x^2 - 5x + 1 \]

To express y directly in terms of t, we substitute each function into the next, carefully combining the operations at each stage. Ensuring that every step maintains the integrity of the functions involved is crucial. This helps to handle the expressions correctly in each substitution stage.

simplification of expressions

Simplification is the process of making an expression easier to work with by combining like terms, factoring, and performing arithmetic operations. Let's see how this works in our exercise.
When simplifying x from s:
\[ x = (\sqrt{t^2 + 1})^3 - 2(\sqrt{t^2 + 1}) + 1 \]
It simplifies to:
\[ x = (t^2 + 1)^{3/2} - 2(t^2 + 1)^{1/2} + 1 \]
This makes it easier to substitute into y.
After substituting x into y, you will simplify:

• Expand all parentheses
• Combine like terms
• Reduce the expression

This step-by-step simplification is essential for a clearer and more manageable final expression.

function composition

Function composition involves creating a new function by applying one function to the result of another. In the given exercise, we have three levels of composition. We can denote this as follows:

• If f(s) = s^3 - 2s + 1
• And g(t) = \sqrt{t^2 + 1}

Then the function h(t) = f(g(t)) evaluates to:
\[ h(t) = \left( (\sqrt{t^2 + 1})^3 - 2\sqrt{t^2 + 1} + 1 \right) \]
We also need y expressed as another composition:
If p(x) = x^2 - 5x + 1, then
\[ y(h(t)) = p(h(t)) \]
This can be substituted as a nested function step-by-step. Helping students understand function composition is crucial, as it lays the groundwork for more advanced calculus topics.