Applications: Partial Derivatives

The pressure depends on two inputs, the temperature and the volume. A partial derivative measures how fast it changes with one input while the other is held constant: ∂ P∂ T is the rise per kelvin at a fixed volume, and ∂ P∂ V the change per liter at a fixed temperature. When both inputs change a little, the change in pressure is about the sum of the two effects, so holding the pressure steady means making the two effects cancel.

1. (a) At the present state, P(300, 75) = 25 × 30075 = 750075 = 100 kilopascals, close to the pressure of the air outside.
2. To find ∂ P∂ T, hold V constant and differentiate with respect to T: ∂ P∂ T = 25V = 2575 = 13. With the volume held at 75 liters, the pressure rises by about 13 of a kilopascal for each kelvin of warming.
3. To find ∂ P∂ V, hold T constant, so that P = 25T × V⁻¹: ∂ P∂ V = −25TV² = −75005625 = −43. With the temperature held at 300 kelvin, the pressure falls by about 43 kilopascals for each extra liter.
4. (b) Small changes Δ T and Δ V change the pressure by about 13Δ T − 43Δ V. For the pressure to stay the same this must be 0, so 43Δ V = 13Δ T and Δ V = 14Δ T. The volume must grow at 0.25 liters per kelvin.
5. After 6 kelvin of warming, the volume must grow by 0.25 × 6 = 1.5 liters, to 76.5 liters. Check: P(306, 76.5) = 25 × 30676.5 = 765076.5 = 100 kilopascals, so the pressure is unchanged.

answer(a) 100 kilopascals; ∂ P∂ T = 13 ≈ 0.333 kilopascals per kelvin at a fixed volume, and ∂ P∂ V = −43 ≈ −1.333 kilopascals per liter at a fixed temperature; (b) 0.25 liters per kelvin, so 76.5 liters after 6 kelvin of warming

techniqueThe Partial Derivative · Computing Partial Derivatives

The temperature depends on the position along the rod and on the time, so it has a partial derivative with respect to each. The heat equation links the rate of change in time to the second partial derivative along the rod: where the temperature profile bends downward, a point is hotter than the average of the points on either side of it and loses heat to them. Checking the equation means differentiating once in m and twice in x, holding the other variable constant each time.

1. (a) Differentiate with respect to m, holding x constant, so that sinπ x100 is a constant factor: ∂ u∂ m = 80 × (−0.01)e^−0.01msinπ x100 = −0.8e^−0.01msinπ x100.
2. Differentiate twice with respect to x, holding m constant. The first derivative is ∂ u∂ x = 80 × π100e^−0.01mcosπ x100, and the second is ∂² u∂ x² = −80 × π²10000e^−0.01msinπ x100 = −π²125e^−0.01msinπ x100.
3. Both derivatives are multiples of e^−0.01msinπ x100, so the heat equation holds at every point and every time when −0.8 = −π²125k, that is k = 0.8 × 125π² = 100π² ≈ 10.1 square centimeters per minute. The second derivative is negative all along the rod, so every point is cooling.
4. (b) At the middle, x = 50 and sinπ2 = 1, so ∂ u∂ m = −0.8e^−0.01m. At the start, m = 0, the middle is cooling at 0.8^°C per minute.
5. It cools at half that rate, 0.4^°C per minute, when e^−0.01m = 0.5, so −0.01m = ln 0.5 and m = 100ln 2 ≈ 69.3 minutes. The temperature of the middle is then 80 × 0.5 = 40^°C. Check: at the middle ∂ u∂ m = −0.01u, and 0.01 × 40 = 0.4^°C per minute.

answer(a) ∂ u∂ m = −0.8e^−0.01msinπ x100 and ∂² u∂ x² = −π²125e^−0.01msinπ x100, so the heat equation holds with k = 100π² ≈ 10.1 square centimeters per minute; (b) 0.8^°C per minute at the start, and half that after 100ln 2 ≈ 69.3 minutes, when the middle is at 40^°C

techniqueHigher Partial Derivatives · Computing Partial Derivatives

A first partial derivative is the rate of change in one variable with the other held constant. Differentiating it again in the OTHER variable gives a mixed partial derivative, which measures how one rate changes as the other variable changes: here, how advertising changes the effect of the price, and how the price changes the effect of advertising. The mixed derivative theorem says the two orders give the same result wherever the second partial derivatives are continuous.

1. (a) At present, S(20, 16) = 20 × √16 × (40 − 20) = 20 × 4 × 20 = 1600 cases a week.
2. Holding a constant, ∂ S∂ p = −20√a = −80: each extra dollar on the price loses about 80 sales a week. Holding p constant, ∂ S∂ a = 20(40 − p) × 12√a = 10(40 − p)√a = 2004 = 50: each extra $100 of advertising brings about 50 more sales a week.
3. (b) Differentiate ∂ S∂ p = −20√a with respect to a: ∂² S∂ a   ∂ p = −10√a = −104 = −2.5. Each extra $100 of advertising makes the sales lost per dollar of price about 2.5 cases larger.
4. Differentiate ∂ S∂ a = 10(40 − p)√a with respect to p: ∂² S∂ p   ∂ a = −10√a = −2.5. Each extra dollar on the price makes $100 of advertising bring about 2.5 fewer sales. The two mixed derivatives are equal, as the mixed derivative theorem says, since every second partial derivative of S is continuous for a > 0.
5. Raising the advertising from a = 16 to a = 20 changes ∂ S∂ p by about −2.5 × 4 = −10, to about −90 cases per dollar. Exactly, ∂ S∂ p = −20√20 ≈ −89.4, so the estimate is within 0.6 of it.

answer(a) 1600 cases a week; ∂ S∂ p = −80 cases per dollar and ∂ S∂ a = 50 cases per $100 of advertising; (b) both mixed derivatives are −2.5; the estimate of ∂ S∂ p at $2000 a week is −90 cases per dollar, against −20√20 ≈ −89.4 exactly

techniqueThe Mixed Derivative Theorem · Higher Partial Derivatives

The length of the line depends on the two distances, and both distances change with time. The chain rule for partial derivatives adds the two effects: each partial derivative of the length, multiplied by the rate at which its own input changes. With s seconds as the time, dLds = ∂ L∂ xdxds + ∂ L∂ ydyds. The same equation with dLds = 0 answers part (b).

1. (a) At this moment L = √40² + 30² = √1600 + 900 = 50 meters. The length depends on x and y, and both depend on the time, so the chain rule gives dLds = ∂ L∂ xdxds + ∂ L∂ ydyds.
2. Holding y constant, ∂ L∂ x = x√x² + y² = 4050 = 0.8; holding x constant, ∂ L∂ y = y√x² + y² = 3050 = 0.6.
3. So dLds = 0.8 × 1.2 + 0.6 × 0.6 = 0.96 + 0.36 = 1.32. The flier must let out line at 1.32 meters per second.
4. (b) Now dLds = 0 and dyds = 0.6, so 0.8dxds + 0.6 × 0.6 = 0, which gives dxds = −0.360.8 = −0.45.
5. The negative sign means that x is decreasing: the kite moves toward the flier at 0.45 meters per second horizontally, climbing along a circle of radius 50 meters about the flier's hand. Check: the velocity (−0.45, 0.6) is at right angles to the line, since 40 × (−0.45) + 30 × 0.6 = −18 + 18 = 0.

answer(a) 1.32 meters per second; (b) 0.45 meters per second horizontally, toward the flier

techniqueThe Chain Rule for Partial Derivatives · Computing Partial Derivatives

The gradient vector collects the two partial derivatives, ∇ T = (∂ T∂ x, ∂ T∂ y). It points in the direction in which T increases fastest, and its length is that fastest rate. A path that always follows the gradient has, at every point, the direction of the gradient vector, and that gives a differential equation for the path.

1. (a) At the probe, T(4, 1) = 100 − 0.5 × 16 − 1 = 100 − 8 − 1 = 91^°C.
2. The partial derivatives are ∂ T∂ x = −x and ∂ T∂ y = −2y, so ∇ T = (−x, −2y), which is (−4, −2) at the probe.
3. The temperature rises fastest in the direction of the gradient, (−4, −2), which as a unit vector is 1√5(−2, −1). The fastest rate is the length of the gradient, √16 + 4 = √20 ≈ 4.47^°C per centimeter.
4. (b) The path always points along ∇ T = (−x, −2y), so its gradient is dydx = −2y−x = 2yx. Separating the variables, dyy = 2 dxx, so ln y = 2ln x + c and y = Cx² for a constant C.
5. The path passes through (4, 1), so 1 = 16C and C = 116: the robot follows y = x²16 toward (0, 0), the center of the plate, where the temperature is highest, 100^°C. Check: at (4, 1) the path's gradient is 2 × 416 = 0.5, and the gradient vector (−4, −2) points along a line of gradient −2−4 = 0.5 too.

answer(a) 91^°C; ∇ T = (−4, −2); the temperature rises fastest in the direction 1√5(−2, −1), at √20 ≈ 4.47^°C per centimeter; (b) y = x²16, heading to the center (0, 0), where the plate is at 100^°C

techniqueThe Gradient Vector · Directional Derivatives · Separable Differential Equations

examsH2

The rate of change of height in a chosen direction is the directional derivative: the dot product of the gradient with a unit vector in that direction. It equals the length of the gradient times the cosine of the angle between the gradient and the direction, so the height falls fastest along the fall line, opposite to the gradient, and does not change at all along a contour.

1. (a) The partial derivatives are ∂ h∂ x = 0.0008x and ∂ h∂ y = −0.32, so at (300, 0) the gradient is ∇ h = (0.24, −0.32). Its length is √0.0576 + 0.1024 = √0.16 = 0.4.
2. The lift station is √800² + 600² = 1000 meters away, so the unit vector toward it is u = (−0.8, 0.6). The directional derivative is ∇ h · u = 0.24 × (−0.8) + (−0.32) × 0.6 = −0.192 − 0.192 = −0.384: she sets off descending 0.384 meters for each meter she travels.
3. The steepest descent is along the fall line, −∇ h0.4 = (−0.6, 0.8), at the length of the gradient, 0.4 meters per meter. Her direction is close to the fall line: the angle θ between them has cosθ = 0.3840.4 = 0.96, so θ ≈ 16^°.
4. (b) Along a unit vector at an angle θ to the fall line, the height falls at 0.4cosθ meters per meter. Setting 0.4cosθ = 0.2 gives cosθ = 0.5, so θ = 60^°, on either side of the fall line.
5. Turning (−0.6, 0.8) through 60^° each way, with cos 60^° = 0.5 and sin 60^° = √32, gives (−0.3 − 0.4√3, 0.4 − 0.3√3) ≈ (−0.99, −0.12) and (−0.3 + 0.4√3, 0.4 + 0.3√3) ≈ (0.39, 0.92). Check the second: 0.24 × 0.39 − 0.32 × 0.92 = 0.0936 − 0.2944 ≈ −0.2.

answer(a) she descends 0.384 meters per meter, just under the steepest descent at that point, 0.4 meters per meter; (b) 60^° from the fall line, setting off along about (−0.99, −0.12) or (0.39, 0.92)

techniqueDirectional Derivatives · The Gradient Vector

Near the readings, H is close to its tangent plane: the height at the readings plus each partial derivative times the change in its own input. The same linear approximation turns small errors in the two measurements into an error in the height. The angle must be in radians, because the derivative of tanθ is sec²θ only when θ is measured in radians.

1. (a) At the readings, H(30, 45^°) = 30tan 45^° = 30 × 1 = 30 meters.
2. Holding θ constant, ∂ H∂ d = tanθ = 1. Holding d constant, ∂ H∂ θ = dsec²θ = 30 × 2 = 60 meters per radian. So the tangent plane is H ≈ 30 + 1 × (d − 30) + 60(θ − π4).
3. A change of −1^° is −π180 ≈ −0.01745 radians, so at d = 31 and θ = 44^° the plane gives H ≈ 30 + 1 − 60 × 0.01745 = 31 − 1.047 = 29.95 meters. Exactly, 31tan 44^° ≈ 29.94 meters, so the estimate is out by about 0.02 meters.
4. (b) The largest error comes when both errors push the height the same way: |Δ H| ≈ 1 × 0.2 + 60 × π180 = 0.2 + 1.047 ≈ 1.25 meters. The clinometer causes more than five times as much of it as the tape.
5. With the tape unchanged and the clinometer accurate to within δ radians, the largest error is about 0.2 + 60δ, and 0.2 + 60δ ≤ 0.5 needs δ ≤ 0.005 radians, which is 0.005 × 180π ≈ 0.29^°. Check: 0.2 + 60 × 0.005 = 0.2 + 0.3 = 0.5 meters.

answer(a) 30 meters; H ≈ 30 + (d − 30) + 60(θ − π4); the estimate for 31 meters and 44^° is 29.95 meters, against 29.94 meters exactly; (b) about 1.25 meters, most of it from the clinometer, which must be accurate to within about 0.29^°

techniqueTangent Planes · Computing Partial Derivatives

The profit depends on both prices, and at a stationary point both partial derivatives are zero. Solving those two equations together gives the point. The second-derivative test classifies it with D = ∂² P∂ p²∂² P∂ q² − (∂² P∂ p   ∂ q)²: if D > 0 and ∂² P∂ p² < 0 the point is a maximum, if D > 0 and ∂² P∂ p² > 0 a minimum, and if D < 0 a saddle point.

1. (a) Each blend earns its price minus its cost on every bag sold, so P(p, q) = (p − 4)(120 − 20p + 10q) + (q − 6)(240 + 10p − 20q) dollars a week.
2. Differentiate with respect to p, holding q constant, by the product rule: ∂ P∂ p = (120 − 20p + 10q) − 20(p − 4) + 10(q − 6) = 140 − 40p + 20q. In the same way, ∂ P∂ q = 10(p − 4) + (240 + 10p − 20q) − 20(q − 6) = 320 + 20p − 40q.
3. Setting both to zero and dividing by 20 gives 2p − q = 7 and −p + 2q = 16. Substituting q = 2p − 7 into the second gives −p + 4p − 14 = 16, so 3p = 30, p = 10 and q = 13: House at $10 and Dark at $13.
4. (b) The second partial derivatives are ∂² P∂ p² = −40, ∂² P∂ q² = −40 and ∂² P∂ p   ∂ q = 20, so D = (−40)(−40) − 20² = 1600 − 400 = 1200. Since D > 0 and ∂² P∂ p² < 0, the stationary point is a local maximum; P is quadratic and these values hold everywhere, so it is the greatest profit.
5. At (10, 13) the roaster sells 120 − 200 + 130 = 50 bags of House and 240 + 100 − 260 = 80 bags of Dark, for a profit of 6 × 50 + 7 × 80 = 300 + 560 = $860 a week. At the present prices it sells 120 − 160 + 110 = 70 and 240 + 80 − 220 = 100 bags, for 4 × 70 + 5 × 100 = $780, so the new prices add $80 a week.

answer(a) P = (p − 4)(120 − 20p + 10q) + (q − 6)(240 + 10p − 20q), stationary at p = 10, q = 13; (b) D = 1200 > 0 with ∂² P∂ p² = −40 < 0, so it is a maximum: $860 a week, $80 more than the $780 at the present prices

techniqueStationary Points of a Surface · Higher Partial Derivatives

The strength is to be as large as possible while the width and depth stay tied to the log by g(w, d) = w² + d² = 3600. At the best point, the level curve of the strength just touches the circle, so the two gradients are parallel: ∇ S = λ∇ g for some number λ, the Lagrange multiplier. The two components of that equation and the constraint give the width and the depth.

1. (a) ∇ S = (kd², 2kwd) and ∇ g = (2w, 2d), so ∇ S = λ∇ g says kd² = 2λ w and 2kwd = 2λ d.
2. The depth is not 0, so dividing the second equation by 2d gives λ = kw. The first then becomes kd² = 2kw², that is d² = 2w².
3. Substituting into the constraint: w² + 2w² = 3600, so w² = 1200 and w = √1200 ≈ 34.6 centimeters, and d² = 2400, so d = √2400 ≈ 49.0 centimeters. At the ends of the arc, where w = 0 or d = 0, the strength is 0, so this point is the maximum.
4. (b) The strongest beam has S = k × √1200 × 2400 ≈ 83138k. The square beam has w = d, so 2w² = 3600, w = √1800 ≈ 42.43 centimeters, and S = k × √1800 × 1800 ≈ 76368k.
5. The ratio is 8313876368 ≈ 1.089, so the strongest beam is about 8.9 percent stronger than the square one. Check: w² + d² = 1200 + 2400 = 3600, so the corners of the strongest beam lie on the log.

answer(a) about 34.6 centimeters wide and 49.0 centimeters deep, with w² = 1200 and d² = 2400; (b) about 8.9 percent stronger than the square beam

techniqueLagrange Multipliers · The Gradient Vector

The volume under a surface over a rectangle is a double integral, worked as two single integrals. Integrating the depth along the length, with y held constant, gives the area of the cross-section of the heap at that y; integrating that area across the width adds up the slices. The depth here is a function of x times a function of y, so each integral can be worked on its own.

1. (a) The volume is ∫₋₄⁴∫₋₆⁶ h(x, y) dx dy. Integrate first in x, holding y constant, so that 1.5(1 − y²16) is a constant factor: ∫₋₆⁶(1 − x²36)dx = [x − x³108]₋₆⁶ = (6 − 2) − (−6 + 2) = 8.
2. So the cross-section of the heap at y has area 1.5 × 8 × (1 − y²16) = 12(1 − y²16) square meters. Integrate it across the width: ∫₋₄⁴12(1 − y²16)dy = 12[y − y³48]₋₄⁴ = 12 × 163 = 64 cubic meters.
3. Check: over its interval, 1 − x²36 has an average of 812 = 23, and 1 − y²16 has an average of 163 ÷ 8 = 23. So the heap fills 23 × 23 = 49 of the box around it, 12 by 8 by 1.5 meters: 49 × 144 = 64 cubic meters.
4. (b) The part beyond x = 3 lies over the rectangle 3 ≤ x ≤ 6, −4 ≤ y ≤ 4. The integral across the width is unchanged, 163, and ∫₃⁶(1 − x²36)dx = [x − x³108]₃⁶ = (6 − 2) − (3 − 0.25) = 1.25.
5. So the loader clears 1.5 × 1.25 × 163 = 10 cubic meters. A quarter of the heap is 16 cubic meters, and the end holds only 1064 ≈ 15.6 percent of the heap, because the gravel is shallowest near the end.

answer(a) 64 cubic meters; (b) 10 cubic meters, less than a quarter of the heap, 16 cubic meters, because the gravel is shallow near the end

techniqueThe Volume Under a Surface