Functions of two variables and partial derivatives
A function of two variables, such as output as a function of capital and labour, has a graph that is a surface over the plane. Each level curve is where it takes one fixed value, and a partial derivative is an ordinary derivative in one variable with the other held fixed: the marginal product of labour at a fixed level of capital, for example.
In this topic
- Taught
- Before the final, flavour C
- Questions
- 15
- Total marks
- 44
- Suggested time
- 67 min
- Difficulty
Introductory to Capstone
Show the 4 hardest
Key results
- Partial derivatives
When the limits exist; in practice, differentiate in one variable with the other held constant.
- Mixed partial derivatives
When and are continuous near the point (Clairaut’s theorem).
- Estimating a change
When and are continuous near and and are small.
- Slope of a level curve
Along the level curve , when and are continuous near the point and there.
Read the theoryUBC Math 100 §14.2 Functions of two variablesUBC Math 100 §14.3 Sketching surfaces in 3DUBC Math 100 §15.1 Partial derivativesUBC Math 100 §15.2 Higher order partial derivatives
Worked example
Let . Find .
Strategy
Treat x as a constant and differentiate in y. Both factors depend on y, so the product rule applies, and the exponential needs the chain rule.
Working
Product rule. The chain rule on the exponential gives the factor x, the y-derivative of the exponent xy.
Substitute x = 2 and y = 1, so xy = 2.
Collect the multiples of e².
Answer
The answer is 8 e squared, about 59.11.
Check
The relative rate f_y/f is 2y/(x + y²) + x = 2/3 + 2 = 8/3 at (2, 1), and 8/3 of f(2, 1) = 3e² is 8e².
Where this goes wrong
Differentiating eˣʸ in y as eˣʸ, which gives 5e². The exponent is xy, and its y-derivative, x, multiplies the exponential.
On the paper
Say which variable is held constant, differentiate, and substitute the point last.
Leans on
The chain rule for exponentials, from The chain rule and logarithms
Practice questions
5 of the 15, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.
- Introductory2 marks · 3 min
Let . Which statement gives and the domain of correctly?
Answer this questionStrategy
Evaluate by substituting both coordinates. The domain is every point (x, y) where the formula makes sense: here, where the number under the square root is not negative.
Working
x = 2 and y = −1; the square of −1 is +1.
A square root needs an input that is not negative; zero is allowed.
The points within distance 3 of the origin: the disc of radius 3, its edge included.
Answer
The answer is f(2, −1) = 2, and the domain is the disc x² + y² ≤ 9: every point on or inside the circle of radius 3 about the origin..
Check
As 4 + 1 + 4 = 9, the point (2, −1, 2) is on the sphere x² + y² + z² = 9, and f(3, 0) = 0 is defined, so the circle is included.
Where this goes wrong
Leaving the circle out of the domain. The square root of 0 is 0, so the points with x² + y² = 9 are in; only the points outside the circle are excluded.
Leans on
Domains of functions, from Functions, graphs and inverses
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Let . Which describes the level curve ? - Introductory2 marks · 3 min
Let . Which describes the level curve ?
Answer this questionStrategy
A level curve is the set of points where f takes one fixed value. Set f(x, y) = 5 and complete the square to recognise the curve.
Working
The level curve f = 5: every point where the surface is at height 5.
Complete the square, adding 4 to both sides.
A circle with centre (2, 0) and radius 3. Every level f = c with c > −4 is a circle about the same centre, so the surface is a bowl with its lowest point over (2, 0).
Answer
The answer is The circle with centre (2, 0) and radius 3..
Check
The points (5, 0), (−1, 0) and (2, 3), each 3 from (2, 0), all give f = 5, for instance 4 − 8 + 9 = 5 at (2, 3).
Where this goes wrong
Adding 4 to complete the square on the left but not on the right, which gives (x − 2)² + y² = 5 and the wrong radius √5. Completing the square changes the equation unless both sides get the 4.
Leans on
Completing the square, from Functions, graphs and inverses
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Let . Which statement gives and the domain of correctly? - Standard3 marks · 4 min
A firm’s output is units, where is units of capital and is units of labour. It now uses and . Which statement correctly gives and interprets the marginal product of labour, , there?
Answer this questionStrategy
The marginal product of labour is the partial derivative of output with respect to labour, capital held fixed. Compute it at (16, 81), then read it as extra output per extra unit of labour.
Working
K to the power 1/4 is a constant factor; the power rule acts on L to the power 3/4.
Fourth roots: 2⁴ = 16 and 3⁴ = 81.
With capital held at 16, output rises at 6 units per unit of labour, so one more unit of labour adds about 6 units of output.
Answer
The answer is The marginal product of labour is 6: with capital held at 16, one more unit of labour adds about 6 units of output..
Check
Raising labour from 81 to 82 with capital at 16 lifts output from 648 to about 653.99, a gain of about 6 units.
Where this goes wrong
Confusing the marginal product with the average product. Q/L = 648/81 = 8 is output per unit of labour so far; ∂Q/∂L = 6 is what the next unit of labour adds.
Leans on
The power rule with fractional exponents, from Differentiation rules
- Challenging3 marks · 6 min
Let . Find , the derivative in of , and check it against .
Answer this questionStrategy
Differentiate in x first, then differentiate the result in y. Clairaut’s theorem says the other order gives the same answer, which makes a good check.
Working
Chain rule on log: the derivative of the inside over the inside, with y held fixed.
Quotient rule in y with x fixed: the top has y-derivative 1 and the bottom has y-derivative x.
Substitute x = 2 and y = 1.
The other order gives the same expression, as Clairaut’s theorem promises.
Answer
The answer is negative 3/49, about −0.0612, in either order.
Check
With x = 2, f_x = (4 + y)/(5 + 2y) slips from 5/7 ≈ 0.7143 to about 0.7137 as y goes from 1 to 1.01, a slope near −0.06.
Where this goes wrong
Differentiating f_x in x a second time, which gives f_xx(2, 1) = −11/49. The subscripts give the order: f_xy is f_x differentiated in y.
Leans on
Differentiating a logarithm, from The chain rule and logarithms
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Let . Find at . - Capstone5 marks · 10 min
A firm’s weekly output is units, where and are its weekly spending on capital and on labour, in thousands of dollars. It now spends and . Find the two marginal products, decide which way one thousand dollars of spending should be moved between the inputs to raise output, and estimate the resulting increase in output.
Give the estimated increase in weekly output, in units.
Answer this questionStrategy
K and L are both measured in thousands of dollars, so each marginal product is the extra output per extra thousand dollars on that input. Compare them to see where money does more, then estimate the move as the output lost on one input plus the output gained on the other.
Working
Current output, for reference.
Hold L fixed. 642/3 = 16 and 82/3 = 4.
Hold K fixed. 641/3 = 4 and 81/3 = 2.
A thousand dollars more on labour adds about 20 units; on capital, about 1.25. Move money from capital to labour.
Cutting capital by one thousand dollars loses about 1.25 units; adding it to labour gains about 20.
The exact change confirms the estimate.
Answer
The answer is Move the thousand dollars from capital to labour: output rises by about 18.75 units, or 75/4..
Check
Actually moving the thousand gives Q(63, 9) ≈ 258.2, about 18.2 more than the present 240, close to the estimate.
Where this goes wrong
Counting only the gain from labour, 20 units, and forgetting that the money leaves capital. Moving spending is a trade: the estimate is the marginal product gained minus the marginal product given up.
Leans on
Linear approximation, from Linear approximation
Related topics
Needs
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
30 questions
Leads to
Optimization in two variables and Lagrange multipliers
Critical points and the second derivative test, the extremes of a function on a closed region, and Lagrange multipliers.
11 questions