Variation · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

11 Variation · g10-11-01 · 5–10 min · TEACHER · ANSWERS

Which power controls the multiplier?

Does doubling x always double y?

Predict: For y=kxⁿ, predict y(2x)/y(x) for n=1,2,3.

Compare	Computed observation
Linear direct variation Constant k = 2; Power n = One; Positive input x = 1	Output multiplier: 2; identity multiplier 2ⁿ=2.
Square variation Constant k = 2; Power n = Two; Positive input x = 1	Output multiplier: 4; identity multiplier 2ⁿ=4.
Zero proportionality constant Constant k = 0; Power n = Three; Positive input x = 2	Output multiplier: undefined (0/0); identity multiplier 2ⁿ=8.

Conclusion: The fixed invariant is y/xⁿ=k. Doubling x multiplies y by 2ⁿ when a ratio is defined. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: Here x>0 and n∈{1,2,3}. If k=0, both outputs vanish and their ratio is undefined.

Justification: k(2x)ⁿ=2ⁿkxⁿ by exponent laws. Dividing requires y(x)≠0; the identity itself also holds at k=0.

Ask: Distinguish an output identity from a ratio that may be undefined.

Watch for: “Directly proportional to x²” means kx², not (kx)².

Extension: Use a noninteger positive scale t and derive the multiplier tⁿ.

Exit: y=3x². If x is tripled, what happens to y?

y becomes nine times as large.

Before: Powers and ratios · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

11 Variation · g10-11-02 · 5–10 min · TEACHER · ANSWERS

A fixed product instead of a fixed ratio

What remains fixed in inverse variation?

Predict: Predict the new output when x is multiplied by t in y=k/x.

Compare	Computed observation
Double a positive input Constant k = 4; Input x = 1; Input multiplier t = 2	The fixed product is 4; y(tx)=y(x)/2.
Negative input Constant k = 4; Input x = -2; Input multiplier t = 2	The fixed product is 4; y(tx)=y(x)/2.
Excluded zero input Constant k = 0; Input x = 0; Input multiplier t = 2	x=0 is excluded even when k=0.

Conclusion: The product xy=k is fixed; for nonzero t the new output is y/t. x=0 is excluded. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: x≠0 and t>0; even if k=0, the original expression k/x remains undefined at zero.

Justification: (tx)·[k/(tx)]=k and k/(tx)=(k/x)/t, under the nonzero-denominator conditions.

Ask: Ask what survives algebraic simplification when the input is zero.

Watch for: Multiplying x by two does not subtract a fixed amount from y.

Extension: Compare negative and positive x while retaining the same k.

Exit: y=12/x. Find y at x=3 and x=6.

4 and 2 respectively; both products xy equal 12.

Before: Division and domain restrictions · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

11 Variation · g10-11-03 · 5–10 min · TEACHER · ANSWERS

Change one factor or both?

In joint variation, do two doublings still give a doubling?

Predict: For y=kxz, compare doubling x alone with doubling both x and z.

Compare	Computed observation
Two nonzero factors Constant k = 2; Input x = 1; Input z = 2	Doubling one factor gives 2y; doubling both gives 4y, including zero cases.
A negative factor Constant k = 2; Input x = -1; Input z = 2	Doubling one factor gives 2y; doubling both gives 4y, including zero cases.
A zero factor Constant k = 2; Input x = 0; Input z = 2	Doubling one factor gives 2y; doubling both gives 4y, including zero cases.

Conclusion: One doubled factor doubles y; two doubled factors multiply y by four. Zero inputs remain legitimate but can make ratios undefined. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: k,x,z are real; y/(xz) is used only when xz≠0.

Justification: k(sx)(tz)=st(kxz) by associativity and commutativity. This identity does not divide by any potentially zero output.

Ask: Ask groups to alter just one factor before altering both.

Watch for: Joint variation requires specifying which variables are held fixed.

Extension: Let x scale by s and z by t; determine the combined multiplier.

Exit: y=2xz; initially x=3,z=4. What if x doubles and z halves?

y stays 24 because the two multipliers have product one.

Before: Products and direct variation · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

11 Variation · g10-11-04 · 5–10 min · TEACHER · ANSWERS

The fixed part changes the ratio

Does a straight graph always mean direct proportion?

Predict: Compare y=a+bx with y=bx while keeping the same rate b.

Compare	Computed observation
A fixed starting amount Fixed part a = 3; Rate b = 2; Input x = 2	Doubling difference: y(2x)−2y(x)=-3=−a
Direct variation boundary Fixed part a = 0; Rate b = 2; Input x = 2	Doubling difference: y(2x)−2y(x)=0=−a
Zero input Fixed part a = 3; Rate b = 2; Input x = 0	Doubling difference: y(2x)−2y(x)=-3=−a

Conclusion: A fixed part a gives y(0)=a. Equal x increments give equal output increments, but output ratios are not generally fixed. Finite displayed cases suggest a conjecture; the justification below states why it holds under the stated conditions.

Conditions: x is nonnegative; a and b are model parameters, not measured empirical facts.

Justification: y(2x)−2y(x)=−a, so doubling outputs holds for all x iff a=0. Also y(x+h)−y(x)=bh.

Ask: Ask students to identify the initial amount before comparing per-unit rates.

Watch for: Linearity with an intercept is not the same as direct proportionality through the origin.

Extension: From two input-output pairs, identify a and b and test a third pair.

Exit: A charge is 20+3x. Compare x=2 and x=4.

26 and 32; doubling x does not double the fixed 20.

Before: Affine functions and ratios · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.