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Substituting Nonlinear Paths into Scalar Formulas

Substitute nonlinear coordinate formulas into scalar functions and identify exactly which parameter values make the resulting expression defined.

Beginner 10 min read

What You'll Learn

  • Replace each scalar-function input with its coordinate formula along the path.
  • Write the resulting scalar function of the parameter.
  • Determine parameter restrictions from the original formula after substitution.
  • Combine restrictions to give the complete set of allowed parameter values.
  • Check substituted expressions at particular parameter values and catch false exclusions.

Introduction

A scalar function of several variables takes an ordered list of coordinates as input and returns a number. A path specifies those coordinates in terms of a parameter. Substituting the path into the function therefore produces a scalar expression in that parameter, wherever the formula is defined.

When the coordinate formulas are nonlinear, the substitution may involve powers, trigonometric functions, or other expressions in the parameter. It is important to replace each coordinate in its correct position and then check the resulting expression for parameter values at which it is not defined.

Substituting a Path into a Scalar Function

Suppose \(f\) is a scalar function of two variables and a path is given by \(\mathbf r(t)=(x(t),y(t))\). At a parameter value \(t\), the path gives the point \((x(t),y(t))\). If that point is an allowed input for \(f\), the value of \(f\) along the path is \(f(x(t),y(t))\).

Definition: The scalar function obtained by evaluating \(f\) along the path \(\mathbf r(t)=(x(t),y(t))\) is \(g(t)=f(x(t),y(t))\), for parameter values \(t\) such that \((x(t),y(t))\) is an allowed input for \(f\). Its allowed parameter values are exactly those for which the path point can be used in the original formula.

The substitution is made coordinate by coordinate. If, for example, \(f(x,y)=x^2+3y\) and the path is \(\mathbf r(t)=(t^2,\sin t)\), replace every \(x\) by \(t^2\) and every \(y\) by \(\sin t\). The order matters: the first coordinate of the path supplies \(x\), and the second supplies \(y\).

Worked Example: A Polynomial Formula Along a Nonlinear Path

Worked Example: A Polynomial Formula Along a Nonlinear Path

Let \(f(x,y)=x^2+3y\) and \(\mathbf r(t)=(t^2,\sin t)\). Find the scalar function along the path and evaluate it at \(t=\frac{\pi}{2}\).

Step 1: Substitute the coordinates. The path gives \(x(t)=t^2\) and \(y(t)=\sin t\). Therefore,

$$ g(t)=f(t^2,\sin t)=(t^2)^2+3\sin t=t^4+3\sin t. $$

Both \(t^4\) and \(\sin t\) are defined for every real \(t\), so the substituted expression is defined for every real parameter.

Step 2: Evaluate at the requested parameter. At \(t=\frac{\pi}{2}\), the path point is

$$ \mathbf r\left(\frac{\pi}{2}\right) = \left(\left(\frac{\pi}{2}\right)^2,\sin\frac{\pi}{2}\right) = \left(\frac{\pi^2}{4},1\right). $$

Substituting into the original formula gives

$$ f\left(\frac{\pi^2}{4},1\right) = \left(\frac{\pi^2}{4}\right)^2+3(1) = \frac{\pi^4}{16}+3. $$

Check: Evaluating the substituted expression gives \(g\left(\frac{\pi}{2}\right)=\left(\frac{\pi}{2}\right)^4+3\sin\frac{\pi}{2}=\frac{\pi^4}{16}+3\), in agreement with direct evaluation at the path point.

Finding the Allowed Parameter Values

Substitution does not automatically make every parameter value valid. The original formula may require a denominator to be nonzero or the input to a square root to be nonnegative. After substituting the path coordinates, impose those requirements on \(t\).

The key is to check the formula after making the substitution. A restriction in the original coordinates may simplify along a particular path. Conversely, a path may reach points where the formula is not defined for some parameter values. The correct answer comes from the expression obtained along the path, together with the requirements of the original formula.

1
Write the coordinate formulas.
Identify which path expression supplies each input variable.
2
Substitute into the scalar formula.
Replace each variable everywhere it occurs, preserving its position.
3
Impose the formula’s requirements.
Translate any restrictions into conditions on the parameter.
4
Combine and report the conditions.
Keep only parameter values satisfying all requirements, and check any claimed exclusions in the substituted expression.

Worked Example: A Square Root and a Denominator

Worked Example: A Square Root and a Denominator

Let \(f(x,y)=\frac{\sqrt{4-x}}{y-1}\) and \(\mathbf r(t)=(t^2,2t^2)\). Find the expression along the path and all parameter values for which it is defined.

Step 1: Substitute both coordinates. Here \(x=t^2\) and \(y=2t^2\), so

$$ g(t)=f(t^2,2t^2) = \frac{\sqrt{4-t^2}}{2t^2-1}. $$

Step 2: Apply the square-root requirement. The square root is real when its radicand is nonnegative:

$$ 4-t^2\geq 0 \quad\Longleftrightarrow\quad t^2\leq 4 \quad\Longleftrightarrow\quad -2\leq t\leq 2. $$

Step 3: Apply the denominator requirement. The denominator cannot be zero, so

$$ 2t^2-1\ne 0 \quad\Longleftrightarrow\quad t^2\ne\frac12 \quad\Longleftrightarrow\quad t\ne-\frac{1}{\sqrt2} \quad\text{and}\quad t\ne\frac{1}{\sqrt2}. $$

Step 4: Combine the conditions. The parameter must lie in \([-2,2]\), with the two denominator zeros removed. Thus the expression is defined for

$$ t\in[-2,2]\setminus\left\{-\frac{1}{\sqrt2},\frac{1}{\sqrt2}\right\}. $$

Check at an allowed value: At \(t=0\), the path point is \((0,0)\), and the original formula gives \(f(0,0)=\frac{\sqrt4}{-1}=-2\). The substituted expression gives \(g(0)=\frac{\sqrt4}{-1}=-2\).

Check the boundary values: At \(t=2\) and \(t=-2\), the radicand is \(4-4=0\) and the denominator is \(2(4)-1=7\). Both values are allowed, and \(g(2)=g(-2)=0\). At \(t=\frac{1}{\sqrt2}\) or \(t=-\frac{1}{\sqrt2}\), the denominator is \(2(\frac12)-1=0\), so those values are correctly excluded.

Why the Domain Check Works

The parameter does not go directly into \(f\) as a single coordinate. It first determines a point on the path, and that point is then used as the input to \(f\). This gives a direct way to justify both the substitution and the parameter restrictions.

Result: If \(f\) is defined on a set \(D\) of coordinate points and \(\mathbf r(t)=(x(t),y(t))\), then \(f\) along the path is defined exactly for those \(t\) for which \(\mathbf r(t)\in D\).

This result follows from the meaning of evaluating a function: \(f\) can be evaluated at a point precisely when that point is an allowed input for \(f\). The path supplies the point \(\mathbf r(t)\), so the condition becomes \(\mathbf r(t)\in D\). Substituting its coordinates into the formula for \(f\) computes the value at that point.

1
Assume the path point is allowed.
If \(\mathbf r(t)\in D\), then \(f\) has a value at that input. Thus \(f(x(t),y(t))\) is defined.
2
Assume the path value is defined.
For \(f(x(t),y(t))\) to be an evaluation of \(f\), its input \((x(t),y(t))=\mathbf r(t)\) must be in \(D\).
3
Conclude equivalence.
The expression along the path is defined exactly when the path point is an allowed input. Therefore the permitted parameter set is \(\{t:\mathbf r(t)\in D\}\).

Worked Example: A Cancellation Along the Path

Worked Example: A Cancellation Along the Path

Let \(f(x,y)=\frac{1}{x+y}\) and \(\mathbf r(t)=(t^2,1-t^2)\). Are any parameter values excluded? Find the expression along the path.

Step 1: Substitute the coordinates into the denominator.

$$ x(t)+y(t)=t^2+(1-t^2)=t^2+1-t^2=1. $$

Step 2: Evaluate the formula along the path.

$$ g(t)=f(t^2,1-t^2) = \frac{1}{t^2+(1-t^2)} = \frac{1}{1} = 1. $$

Since the denominator is \(1\) for every real \(t\), it is never zero. Also, both coordinates \(t^2\) and \(1-t^2\) are real for every real \(t\). Therefore no parameter values are excluded: \(g(t)=1\) for every \(t\in\mathbb R\).

Check: At \(t=0\), the path point is \((0,1)\), and \(f(0,1)=1\). At \(t=1\), the path point is \((1,0)\), and \(f(1,0)=1\). Both agree with the substituted result.

A Common Mistake: Inventing a Restriction

A domain check must use the actual expression after substitution. In the preceding example, a mistaken expansion might claim

$$ t^2+(1-t^2)=1-t^4. $$

This is incorrect: the term \(-t^2\) does not become \(-t^4\). The correct expansion is

$$ t^2+(1-t^2)=t^2+1-t^2=1. $$

The incorrect expression \(1-t^4\) would falsely suggest that the denominator is zero at \(t=1\) and \(t=-1\). But direct substitution into the original denominator at either value gives \(x+y=1\), not zero. A quick check at \(t=1\) exposes the error: the path gives \((x,y)=(1,0)\), so \(x+y=1\).

Takeaway: Do not infer excluded parameter values from a coordinate or from an unverified simplification. Substitute carefully, simplify the full expression, and test any suspected exclusion in the original formula along the path.

Why Evaluating Along a Path Matters

Evaluating a scalar function along a path turns a formula in several coordinates into a one-parameter expression. This lets us ask how the scalar value behaves as the point moves along a specified curve. The immediate task is to obtain that expression correctly and know where it exists; later calculations can then use it without confusing the path’s coordinates with independent inputs.

For a path with more coordinates, the same method applies. If \(F(x,y,z)\) is a scalar formula and \(\mathbf r(t)=(x(t),y(t),z(t))\), substitute all three coordinates in their stated order to obtain \(F(x(t),y(t),z(t))\). Its allowed parameters are exactly those for which the resulting path point is an allowed input to \(F\).

Check Your Understanding

For each question, substitute the path coordinates carefully before deciding which parameter values are allowed.

  1. If \(f(x,y)=x+2y\) and \(\mathbf r(t)=(t^2,\sin t)\), what is the expression along the path?
  2. If \(f(x,y)=\frac{1}{x-y}\) and \(\mathbf r(t)=(t^2,t^2+3)\), are any parameter values excluded?
  3. If \(f(x,y)=\sqrt{y-x}\) and \(\mathbf r(t)=(t^2,4)\), what parameter values make the expression real?
  4. State the general condition on \(t\) for \(f\) to be defined along a path \(\mathbf r(t)\) when \(f\) is defined on a set \(D\).
  5. What should you do if your simplification suggests a parameter is excluded by a denominator?

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