Unit #5 - Implicit Diﬀerentiation, Related Rates

Some problems and solutions selected or adapted from Hughes-Hallett Calculus.

Implicit Diﬀerentiation

1. Consider the graph implied by the equation (c) Setting the y value in both equations equal to xy2 = 1. each other and solving to find the intersection point, the only solution is x = 0 and y = 0, so What is the equation of the line through ( 1 , 2) 4 the two normal lines will intersect at the origin, which is also tangent to the graph? (x, y) = (0, 0). Differentiating both sides with respect to x, Solving process: dy 3 3 y2 + x 2y = 0 (x − 4) + 3 = − (x − 4) − 3 dx 4 4 3 3 so x − 3 + 3 = − x + 3 − 3 4 4 dy −y2 3 3 = x = − x dx 2xy 4 4 6 x = 0 4 dy At the point ( 1 , 2), = −4, so we can use the x = 0 4 dx point/slope formula to obtain the tangent line y = −4(x − 1/4) + 2 Subbing back into either equation to find y, e.g. y = 3 (0 − 4) + 3 = −3 + 3 = 0. 2. Consider the circle defined by x2 + y2 = 25 4 3. Calculate the derivative of y with respect to x, (a) Find the equations of the tangent lines to given that the circle where x = 4. (b) Find the equations of the normal lines to x4y + 4xy4 = x + y this circle at the same points. (The normal line is perpendicular to the tangent line.) Let x4y + 4xy4 = x + y. Then (c) At what point do the two normal lines in- dy dy dy 4x3y + x4 + (4x) 4y3 + 4y4 = 1 + tersect? dx dx dx hence Differentiating both sides, dy 4yx3 + 4y4 − 1 = . dx 1 − x4 − 16xy3 d d x2 + y2 = (25) dx dx 4. Calculate the derivative of y with respect to x, dy given that 2x + 2y = 0 dx xey = 4xy + 5y4 dy −2x −x = = dx 2y y dy To solve for dx , we must think of y as a function of x (a) The points at x = 4, satisfying x2+y2 = 25, would and differentiate both sides of the equation, using the be y = 3 and y = −3. chain rule where appropriate: Using the point/slope formula for a line, and our dy dy dy dy ey + xey = 4y + 4x + 20y3 calculated , through (4, 3): y = −1.33333 ∗ dx dx dx dx (x − 4) + 3, and dy Now, we simplify and move the terms with a dx to the through (4, -3): y = 1.33333 ∗ (x − 4) + (−3) dy right, and keep the terms without a dx to the left: (b) If you have a line with slope m, the slope of a ey − 4y = (4x + 20y3 − xey) dy perpendicular line will be −1/m. dx Finally, we can solve for dy : 3 dx y Through (4, 3): y = 4 (x − 4) + 3, and dy e − 4y through (4, -3): y = 3 (x − 4) + (−3) = 4 dx 4x + 20y3 − xey

1 To simplify a little we multiply both sides through by 5. Use implicit differentiation to find the equation √ of the tangent line to the curve xy3 + xy = 14 2 x + y: at the point (7, 1). dy √ √ 1 − 4x2y x + y = 4xy2 x + y − 1 dx d To get the slope, we take the derivative of both √ dx dy (4xy2 x + y) − 1 = √ sides. dx 1 − 4x2y x + y d d xy3 + xy = (14) 7. Find all the x-coordinates of the points on the dx dx curve x2y2 + xy = 2 where the slope of the tan- dy dy (1)y3 + x 3y2 + (1)y + x = 0 gent line is −1. dx dx dy We need to find the derivative by implicit differen- dy dx Gathering terms with , and those without it, tiation. Differentiating with respect to x on both sides dx of the equation, dy 3xy2 + x + y3 + y = 0 dy dy dx 2xy2 + x2 2y + y + x = 0 dx dx dy y3 + y = − dx 3xy2 + x dy Here, we could solve for , but we actually know the dx dy slope we want this time: = −1, so let’s just sub At the point (x, y) = (7, 1), dx that in now: dy 13 + 1 = − 2xy2 + x2 (2y(−1)) + y + x(−1) = 0 dx 3(7)(1)2 + 7 2 2 −2 −1 or 2xy − 2x y + y − x = 0 = = 28 14 factoring first terms: 2xy(y − x) + (y − x) = 0 Factoring common y − x:(y − x)(2xy + 1) = 0 To build a line that goes through the point (7, 1), and −1 with slope , we can use the point-slope line formula. Meaning either y − x = 0 (so y = x), or (2xy + 1) = 0. 14 We obtain the formula for the tangent line: Subbing each possibility into the equation for the curve, x2y2 + xy = 2 −1 y = (x − 7) + 1 14 If y = x, x2(x2) + x(x) = 2 x4 + x2 = 2 6. Find dy/dx by implicit differentiation. x4 + x2 − 2 = 0 √ 2 2 x + y = 9 + x y (x2 + 2)(x2 − 1) = 0

d So x = −1, 1 are two solutions, with the corresponding Taking of both sides, dx y values, using y = x being y = −1 and 1 as well. d √ d The other possible case, 2xy + 1 = 0, leads to x + y = 9 + x2y2 dx dx 2 −1 −1 1 1 dy dy If y = −1/2x, x2 + x = 2 √ 1 + = 2xy2 + x22y 2x 2x 2 x + y dx dx 1 −1 or + = 2 Expanding left side: 4 2 1 1 dy 1 1 dy √ + √ = 2xy2 + x22y 2 x + y dx 2 x + y dx which is impossible, so the assumption that y = −1/2x must be impossible to use with this curve. dy Gathering terms with and without , 2 2 dx Therefore the only two points on the curve x y +xy = dy 2 which have slope = −1 are (x, y) = (1, 1) and dy 1 1 dx √ − 2x2y = 2xy2 − √ dx 2 x + y 2 x + y (−1, −1).

2 8. Where does the normal line to the ellipse

x2 − xy + y2 = 3

at the point (−1, 1) intersect the ellipse for the second time?

To obtain a normal (perpendicular) line, we ﬁnd a line perpendicular to the tangent line on the ellipse at the point (-1, 1). (Linear algebra students may have other ways to do this.) To get the slope of the tangent line, we use implicit diﬀerentiation, with respect to x: d d 9. The curve with equation 2y3 + y2 − y5 = x4 − x2 − xy + y2 = (3) dx dx 2x3 + x2 has been likened to a bouncing wagon dy dy (graph it to see why). Find the x-coordinates 2x − y + x + 2y = 0 dx dx of the points on this curve that have horizontal tangents. dy (−x + 2y) = −2x + y dx dy −2x + y d = Taking an implicit of both sides, dx (−x + 2y) dx d d 2y3 + y2 − y5 = x4 − 2x3 + x2 dx dx At the point on the ellipse (−1, 1), dy dy dy 6y2 + 2y − 5y4 = 4x3 − 6x2 + 2x dy −2(−1) + 1 dx dx dx = dx (−(−1) + 2(1)) = 3/3 = 1 dy Since we know we want points where dx = 0 (horizon- dy So the slope of the tangent line to the ellipse at (−1, 1) tal tangents), we’ll set dx = 0 immediately: is 1. 6y2(0) + 2y(0) − 5y4(0) = 4x3 − 6x2 + 2x The slope of the normal line will be perpendicular to 2 that, or −1/(1) = −1. factoring: 0 = 2x(2x − 3x + 1) 0 = 2x(2x − 1)(x − 1) The normal line, which also passes through (-1, 1), will therefore be

1 y = −1(x − (−1)) + 1 So x = 0, 2 and 1 are the points where the tangent to or y = −1(x + 1) + 1 the graph will be horizontal. or y = −x Note that the question only asked for the x coordinate of these points. As you can see on the graph below, all of these x coordinates correspond to multiple y values, To ﬁnd the intersections of this normal line with the el- but all of those points have horizontal tangents. lipse to ﬁnd a second crossing, we sub in this expression for y into the ellipse formula:

x2 − xy + y2 = 3 x2 − x(−x) + (−x)2 = 3 x2 + x2 + x2 = 3 x2 = 1x = ±1

Since x = −1 is the point we started at, x = +1 must be the other intersection. At that point, y = −x, so the point is (x, y) = (1, −1). Below is a graph of the scenario.

3 10. Use implicit differentiation to find an equation Substituting x = 1 back into the circle equation to of the tangent line to the curve find the matching y coordinates gives

y2(y2 − 4) = x2(x2 − 5) (1)2 + y2 − 2(1) − 4y = −1 y2 − 4y = 0 at the point (x, y) = (0, −2). y(y − 4) = 0 y = 0, 4

The points with horizontal tangents are (1,0) and (1,4). (b) The points with vertical tangents are those where dy the denominator of dx is zero (making the slope undeﬁned). From part (a), we have

dy −2x + 2 = dx 2y − 4 (The devil’s curve) Setting the denominator equal to zero gives

This point happens to be at the bottom of the loop on 2y − 4 = 0 the y axis, so the slope there is zero. Therefore, the tangent line equation is simply: y = 2

y = −2 Substituting y = 2 back into the circle equation to find the matching x coordinates gives 11. Use implicit differentiation to find the (x, y) points where the circle defined by x2 + (2)2 − 2x − 4(2) = −1 x2 − 2x − 3 = 0 x2 + y2 − 2x − 4y = −1 (x − 3)(x + 1) = 0 has horizontal and vertical tangent lines. x = 3, − 1 (a) Find the points where the curve has a horizontal tangent line. This gives points on the circle with vertical tangents at (3,2), and (-1,2). (b) Find the points where the curve has a vertical tangent line. 12. The relation

dy x2 − 2xy + y2 + 6x − 10y + 29 = 0 (a) Use implicit differentiation, then set = 0. dx defines a parabola. d d x2 + y2 − 2x − 4y = (−1) (a) Find the points where the curve has a hor- dx dx izontal tangent line. dy dy 2x + 2y − 2 − 4 = 0 (b) Find the points where the curve has a ver- dx dx tical tangent line. dy (2y − 4) = −2x + 2 dx dy −2x + 2 dy = (a) Use implicit differentiation, then set = 0. dx 2y − 4 dx The only point with horizontal tangent is (2, 5). dy Setting now equal to zero gives dx (b) From the implicit derivative, get the equation in dy ... −2x + 2 the form = . 0 = dx ... 2y − 4 The point with vertical tangent is the point where dy −2x + 2 = 0 the denominator of is zero, or (1, 6). dx x = 1

4 13. The graph of the equation Solving for y0 gives

2 2 2x + y x + xy + y = 9 y0 = − . x + 2y is a slanted ellipse illustrated in this ﬁgure: Setting y0 = 0 and solving for x gives y 2x + y = 0 =⇒ x = − . 2 Substituting in the original equation gives

y2 y2 − + y2 = 9. 4 2 Thus 3y2 = 9 4 or y2 = 12. Think of y as a function of x. Hence the horizontal tangent lines have the equations (a) Differentiating implicitly, find a formula for √ the slopes of this shape. (Your answer will y = ± 12. depend on x and y.) The plus sign gives the upper tangent line, the minus (b) The ellipse has two horizontal tangents. sign the lower. By symmetry the vertical tangent lines The upper one has the equation character- have the equations ized by y0 = 0. To find the vertical tangent √ use symmetry, or think of x as a function x = ± 12. of y, differentiate implicitly, solve for x0 and then set x0 = 0. The rightmost vertical tangent line with the equation √ d Taking of both sides of the relation x = 12 dx x2 + xy + y2 = 9 touches the ellipse in the point where √ gives x 12 √ 2x + y + xy0 + 2yy0 = 0. y = − = − = − 3. 2 2

Related Rates

dV dh 14. Gravel is being dumped from a conveyor belt at r = h/2. We have , and want . a rate of 30 cubic feet per minute. It forms a dt dt pile in the shape of a right circular cone whose 1 Always true: V = πr2h base diameter and height are always equal. How 3 fast is the height of the pile increasing when the π h2 but since r = h/2 for this cone, V = h pile is 17 feet high? 3 4 Recall that the volume of a right circular cone dV π dh Taking d of both sides: = 3h2 with height h and radius of the base r is given dt dt 12 dt 1 by V = πr2h. dh 4 dV 3 so = dt πh2 dt Let V (t) = volume of the conical pile, and h and r be the height and bottom radius of the cone respectively. dV 3 The cone has a base diameter and h that are equal, so Subbing in dt = 30 ft /min, and h = 17,

5 From this, the distance between the boats is D. We dh know dx and dy , and want dD . = 0.132 ft/min dt dt dt dt Always true: D2 = x2 + y2 15. When air expands adiabatically (without gain- d dD dx dy ing or losing heat), its pressure P and volume V Taking : 2D = 2x + 2y 1.4 dt dt dt dt are related by the equation PV = C where C dD 1 dx dy is a constant. Suppose that at a certain instant = x + y dt D dt dt the volume is 550 cubic centimeters, and the pressure is 91 kPa and is decreasing at a rate of 7 kPa/minute. At what rate in cubic centime- At noon, ship A is due west of ship B, so x = 40 and ters per minute is the volume increasing at this y = 0 nautical miles. instant? At the question time of 6 PM, each ship will have trav- (Pa stands for Pascal – it is equivalent to one eled 23 × 6 = 138 nautical miles,√ so x = 40 + 138 = 178 Newton/(meter squared); kPa is a kiloPascal or and y = 138. This gives D = 1782 + 1382 ≈ 225.23. 1000 Pascals. ) dx Also = 23 and dy = 23 knots (nautical miles per dt dt Let P and V be the pressure and volume respectively, hour). Subbing in these values into the derivative for- dP dV mula above gives and C be a constant. We know dt , and want dt . dD 1.4 = 32.269 knots Always true: PV = C dt d d Taking d : (PV 1.4) = (C) 17. The altitude of a triangle is increasing at a rate dt dt dt of 1 cm/min while the area of the triangle is dP dV V 1.4 + P 1.4V 0.4 = 0 increasing at a rate of 3 cm2/min. dt dt At what rate is the base of the triangle chang- dP dV V 1.4 + P 1.4V 0.4 = 0 ing when the altitude is 10 cm and the area is dt dt 105 cm2 ? dV − dP V 1.4 so = dt Let x be the base of the triangle, and y be the height, dt 1.4PV 0.4 so area = A = 1 xy. − dP 2 = dt V 1.4P Subbing in the known values gives y A dV = 30.22 cm3/min dt x 16. At noon, ship A is 40 nautical miles due west of ship B. Ship A is sailing west at 23 knots (Note that the triangle doesn’t have to be a right- and ship B is sailing north at 23 knots. How angled triangle as shown, since only the base and height fast (in knots) is the distance between the ships affect the area, not the internal angles.) changing at 6 PM? (Note: 1 knot is a speed of We are given dy and dA , and want dx . 1 nautical mile per hour.) dt dt dt 1 Always true: A = xy 2 The two boats can always be drawn in a triangular d dA 1 dx dy configuration. Let their east/west (horizontal on the Taking : = y + x graph) distance be x, and their north/south distance dt dt 2 dt dt be y. dx dx 1 dA dy solving for : = 2 − x dt dt y dt dt Boat B At the point where y = 10 cm and A = 105, we can find x = 2A/y = 21 cm. We are given dx = 1 cm/min b Boat B dt dA dx and dt = 3 cm/min. Subbing these values into dt , D y dx bb b = −1.5 cm/min x dt Boat A Boat A Reference Point

6 18. A street light is at the top of a 11 ft tall pole. A The tip of the shadow is moving at 11 ft/s (rate of woman 6 ft tall walks away from the pole with change of the distance from the shadow tip to the ﬁxed a speed of 5 ft/sec along a straight path. How point of the pole). fast is the tip of her shadow moving when she A related quantity, the size of the shadow (distance is 45 ft from the base of the pole? ds from the woman to the end of the shadow), or in dt the diagram, is growing at 6 ft/s. The question asks for the speed of the tip so 11 ft/s is Pole the correct answer here. 11 ft 19. A boat is pulled into a dock by a rope attached 6 ft to the bow of the boat and passing through a pulley on the dock that is 1 m higher than the bow of the boat.

x If the rope is pulled in at a rate of 1.2 m/s, how s = L − x fast is the boat approaching the dock when it is 9 m from the dock? L From the question, Rope • We know that the walking speed is 5 ft/sec, and Dock on the diagram above this represents the rate of dx Water change = 5 ft/sec. dt • We are asked for the speed of the tip of the We define variables for the two lengths that are chang- shadow. To measure this, we need to track the ing: distance between the tip of the shadow and a fixed dL point like the pole, so we need to find . • x = length of the rope, and dt We start by finding a relationship between x and L that • w = water-level distance between the boat and is always true. Based on the similar triangles 6:L − x the dock. and 11:L, L − x L x = 6 11 1 m Differentiating both sides with respect to t, d L − x d L w = dt 6 dt 11 In the problem, we are told that the rope is being pulled 1 dL 1 dx 1 dL − = in at 1.2 m/s, so 6 dt 6 dt 11 dt dx = −1.2. We want to find the speed of the boat dt dL dw Solving for , relative to the water, or . dt dt 1 1 dL 1 dx To get the relationship between those two rates, we − = 6 11 dt 6 dt start with an equation that is always true: 5 dL 1 dx = 2 2 2 66 dt 6 dt w + 1 = x dL 66 1 dx = Differentiating both sides with respect to time, dt 5 6 dt 11 dx d d = (w2 + 1) = x2 5 dt dt dt dx dw dx but the rate = 5, so 2w = 2x dt dt dt dL 11 = 5 = 11 ft/s dx dt 5 Note that we have , but we still need x and w. dt

7 We are asked for the water speed when w = 9, but 3π we also need x at that point. Using the pythagorean c2 = a2 + b2 − 2ab cos equation again, 4 2 2 Filling in a = 16 and cos 3π = − √1 , 9 + 1 = x 4 2 √ x = 82 −1 c2 = 162 + b2 − 2(16)b√ 2 32 Subbing in to our related rates equation, c2 = 256 + b2 + √ b 2 dw dx 2w = 2x dt dt db We know = 4 km/min (plane’s speed), and we want dw √ dt 2(9) = 2( 82)(−1.2) dc dt √ to know . dw ( 82) dt = (−1.2) dt 9 To get the relationship between those rates, we differ- = −1.207 m/s entiate both sides of the equation above. d d 32 The boat is approaching the dock (distance is decreas- c2 = 256 + b2 + √ b ing) at 1.207 m/s. dt dt 2 dc db 32 db 20. A plane flying with a constant speed of 4 2c = 2b + √ km/min passes over a ground radar station at dt dt 2 dt an altitude of 16 km and climbs at an angle of dc 1 db 16 db = b + √ 45 degrees. At what rate, in km/min, is the dt c dt 2 dt distance from the plane to the radar station increasing 1 minute later? dc Recall the cosine law, We are asked for after 1 minute, or when b = 4 dt c2 = a2 + b2 − 2ab cos(θ) km/min × 1 min = 4 km. We need to solve for c as well, and we can do that using the cosine law formula where θ is the angle between the sides of length above, a and b. 32 c2 = 256 + (4)2 + √ (4) Here is a sketch of the scenario. Note that both b and 2 c are increasing as the plane flies away. c = 19.0397 Plane b Subbing all the known values into the related rates equation,

dc 1 16 b t = (4)(4) + √ (4) ( ) dt 19.0397 2 = 3.2172 km/min

π The plane is moving away from the radar station at a rate of 3.2172 km/min. 4 3π 21. Water is leaking out of an inverted conical tank c(t) 4 at a rate of 12000.0 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. a = 16 km • The tank has height 8.0 meters and the diameter at the top is 5.0 meters. • The depth of the water is increasing at 28.0

b centimeters per minute when the height of Radar the water is 4.0 meters. The cosine law gives us the “always true” relationship, Find the rate at which water is being pumped 3π based on the constant angle of : into the tank, in cubic centimeters per minute. 4

8 dh The first challenge in this problem is not to get dis- Subbing in the known values, h = 400 cm and = 28 tracted by the given outflow rate. If you imagine wa- dt ter flowing both into and out of the tank, you will cm/min, get a net rate of change of the volume, defined by dV dV 25 2 = rate water in − rate water out, where V = = π · 3 · (400) · (28) dt dt 768 actual volume of water in the tank. The information ≈ 1, 374, 000 cm3/min about the water level rising is directly tied to the volume, so we focus on that first. The represents the net rate of change of water, taking The second point is that we should use consistent units into account both the inflow and the outflow. Since we throughout. That will be easier if we standardize on are told that the outflow is 12, 000 cm3/min, centimeters. Also note that the diameter at the top is 5 m (500 cm), so the radius will be 250 cm. 250 cm 1374000 = rate water in − 12000 | {z } | {z } net flow rate water out

So the rate at which water is being poured in is r rate in ≈ 1, 374, 000 + 12, 000 = 1, 386, 000 cm3/min

800 cm 22. A spherical snowball is melting in such a way h that its diameter is decreasing at rate of 0.4 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 17 cm? (Note the answer is a positive number).

The volume formula for spheres relates the two quanti- dh dV ties, volume and radius/diameter. Let r be the radius, We are given , and want to know , so we need a dt dt D be the diameter, and V be the volume. relationship between h and V that will always be true. With start with the formula for the volume of a cone:

1 2 4 Volume of water in the tank = πr h V = πr3 3 3 There’s an extra r in there though, which we need to but D = 2r or r = D/2, put in terms of h or V . The h/r and 800/250 triangles 3 4 D are similar to each other, so so V = π 3 2 r 250 π = V = D3 h 800 6 5 so r = h 16 Taking the time derivative of both sides, Subbing this into the volume formula gives d π (V ) = D3 1 5 2 dt 6 V = π h h dV π dD 3 16 = 3D2 25 dt 6 dt V = πh3 768 dD We can now diﬀerentiate both sides with respect to t: Subbing in the known values, = −0.4 cm/min, dt and D = 17 cm, d d 25 (V ) = πh3 dt dt 768 dV π 2 dV 25 dh = 3(17) (−0.4) = π 3h2 dt 6 dt 768 dt ≈ 182 cm3/min

9 23. The gas law for an ideal gas at absolute tem- df −1 p (a) = 2 T/p perature T (in kelvins), pressure P (in atmo- dL 2L spheres), and volume V is PV = nRT , where n df 1 is the number of moles of the gas and R = .0821 (b) √ dT 4L T p is the gas constant. Suppose that, at a certain instant, P = 8 atm and is increasing at a rate of √ df − T 0.10 atm/min and V = 10 L and is decreasing at (c) = dρ 3/2 a rate of 0.15 L/min. Find the rate of change of 4Lp T with respect to time at that instant if n = 10 (d) If we shorten the string length L, we are making moles. df ∆L negative. Also the derivative is negative. We start with the equality between the original quan- dL tities, Using our linearity relationship, we can estimate df PV = nRT ∆f ≈ (∆L). dL We note that n and R are constants, while P , V and dT Looking at the signs of each element, sign ∆f = T are changing with time. We want to find . dt (−)(−) = (+). Taking the derivative of both sides with respect to time, Since ∆f will be positive, the frequency will in- d d crease and the pitch will be higher. (PV ) = (nRT ) dt dt dP dV dT (e) If we increase the string tension T , and the deriva- V + P = nR df dt dt dt tive dT is positive, the frequency will be higher, so dT 1 dP dV the pitch will be higher. = V + P dt nR dt dt (f) If we increase the string density ρ, and the deriva- Filling in the known values from the question, df tive dρ is negative, the frequency will be lower, so dT 1 the pitch will be lower. = ((+0.1)(10) + (8)(−0.15)) dt (10)(0.0821) dT ≈ −0.2436 degrees/min 25. A potter forms a piece of clay into a right circu- dt lar cylinder. As she rolls it, the height h of the 24. The frequency of vibrations of a vibrating violin cylinder increases and the radius r decreases. s Assume that no clay is lost in the process. Sup- 1 T string is given by f = , where L is the pose the height of the cylinder is increasing by 2L ρ 0.4 centimeters per second. What is the rate at length of the string, T is the tension, and ρ is which the radius is changing when the radius is its linear density. 3 centimeters and the height is 12 centimeters? Find the rate of change of the frequency with respect to: 2 (a) the length (when T and ρ are constant) The cylinder of clay has a volume of V = πr h. We are dh dr given = 0.4 cm/s, and are asked to find . We (b) the tension (when L and ρ are constant) dt dt (c) the linear density (when L and T are con- are also told that no clay is lost, so we can also state dV stant) that the volume is unchanging, or = 0. dt The pitch of a note is determined by the frequency f. (The higher the frequency, Starting with the known relationship V = πr2h, we dif- the higher the pitch.) Use the signs of the ferentiate both sides with respect to t to get the related derivatives in (a) through (c) to determine rates. what happens to the pitch of a note: d d (d) when the effective length of a string is de- (V ) = (π r2h ) creased by placing a finger on the string so dt dt |{z} a shorter portion of the string vibrates. prod. rule dV dr dh (e) when the tension is increased by turning a = π2r h + πr2 tuning peg. dt dt dt (f) when the linear density is increased by dV switching to another string. Subbing = 0, and solving for the rate we want, dt

10 dr dθ π , Using dt = 0.1 and θ = 6 yields dt dr dh dy 1 0 = π2r h + πr2 = 2 (0.1) ≈ 0.2666 mi/min. dt dt dt cos2 (π/6) dr dh 2r h = −r2 dt dt 28. A road perpendicular to a highway leads to a dr r2 dh farmhouse located 5 mile away. An car traveling = − dt dt 2rh on the highway passes through this intersection dr r dh at a speed of 55mph. = − dt dt 2h How fast is the distance between the car and the Subbing in the values we know, r = 3, h = 12 and farmhouse increasing when the car is 7 miles dh past the intersection of the highway and the = 0.4, dt road? dr (3)(0.4) = − dt 2(12) dr = −0.05 cm/s dt The radius of the cylinder of clay is decreasing at 0.05 cm/s at that moment. Highway 5 miles 26. The volume V of a right circular cylinder of ra- Farmhouse Road dius r and height h is V = πr2h. l s dV dr (a) How is related to if h is constant b dt dt and r varies with time? Car dV dh (b) How is related to if r is constant dt dt and h varies with time? dV dh dr Let l denote the distance between the car and the farm- (c) How is related to and if both h dt dt dt house and let s denote the distance past the intersec- and r vary with time? tion of the highway and the road. As seen in the diagram, there is a right-angled triangle made, satisfying (a)2 πrh(dr/dt) l2 = 52 + s2. (b) πr2(dh/dt) Taking the derivative of both sides of this equation with 2 (c) π(2rh(dr/dt) + r (dh/dt)) respect to t yields 27. A hot air balloon rising vertically is tracked d d by an observer located 2 miles from the lift-oﬀ l2 = (52 + s2) point. At a certain moment, the angle between dt dt dl ds the observer’s line-of-sight and the horizontal is 2l = 2s π , and it is changing at a rate of 0.1 rad/min. dt dt 6 dl s ds How fast is the balloon rising at this moment? and so = . dt l dt

Let y be the height of the balloon (in miles) and θ the We were told in the question that the car is moving at angle between the line-of-sight and the horizontal. ds = 55 miles/hr. When the car is s = 7 miles past Drawing a sketch with the ground and the balloon dt above the ground making a right angle, we have tan θ = the intersection, we can solve for l using√ the Pythago-√ 2 2 y . ras relationship we started with, l = 5 + 7 = 74. 2 dl 7 Taking the time derivative of both sides, This gives = 55√ ≈ 44.7553 mph dt 74 d d y (tan θ) = dt dt 2 29. Assume that the radius r of a sphere is expand- dθ 1 dy ing at a rate of 6in./min. The volume of a sphere sec2 θ · = is V = 4 πr3. dt 2 dt 3 dy dθ Determine the rate at which the volume is = 2 sec2 θ. dt dt changing with respect to time when r = 11in.

11 As the radius is expanding at 6 inches per minute, Let y be the distance between the dot of light and the dr we know that dt = 6 in./min. Taking the derivative point of intersection of the wall and the line through with respect to t of the equation V = 4 πr3 yields the searchlight perpendicular to the wall. Let θ be the 3 dV 4 2 dr 2 dr current angle between the beam of light and the line. = π 3r = 4πr . Substituting r = 11 y dt 3 dt dt Using trigonometry, we have tan θ = . dV 10 and dr = 6 yields = 4π112(6) ≈ 9123.19 in./min. Differentiating both side with respect to t, dt dt d d 1 30. The radius of a circular oil slick expands at a tan(θ) = y rate of 7 m/min. dt dt 10 dθ 1 dy (a) How fast is the area of the oil slick increas- sec2 θ · = ing when the radius is 26 m? dt 10 dt (b) If the radius is 0 at time t = 0, how fast is We want to find the speed of the light spot along the dy the area increasing after 2 mins? wall, which is defined by , so we solve for that: dt dy dθ (a) Let r be the radius of the oil slick and A its area. = 10 sec2 θ dA dr dt dt Then A = πr2 and = 2πr . dt dt We can now focus in on the particular scenario in the dr Substituting r = 26 and dt = 7, we find question: dA = 2π (26) (7) ≈ 1143.54 m2/min. π dt • θ = and 6 (b) Our formula for the rate of change of area requires dθ us to know r, but here we only know the time t. • = 4π. From the question, 2 revolutions per dt Our job then is to find out the radius of the oil slick minute = 4π radians per minute. after 2 minutes. dy Since the slick’s radius is growing at 7 m/min, and Evaluating for now gives: started at r = 0 at 0 minutes, we can just multiply dt to find that after 2 minutes, dy 1 = 10 (4π) r = (7 m/min) × (2 min) = 14 m. dt cos2 (π/6) We can use that r = 14 to compute the rate of area = 53.33π ≈ 167.55 mi/min change: dA dr = 2πr = 2π (14) (7) ≈ 615.752m2/min. dy dt dt Converting to miles per hour gives ≈ 10053.09 mph. dt 31. A searchlight rotates at a rate of 2 revolutions (Yes, this looks crazy fast, but if you imagine yourself per minute. The beam hits a wall located 10 watching as the light house beam rotates, it actually miles away and produces a dot of light that can approach an infinite speed as the angle gets close moves horizontally along the wall. How fast (in to 90 degrees from the wall.) miles per hour) is this dot moving when the an- 32. A kite 100 ft above the ground moves horizon- gle θ between the beam and the line through tally at a speed of 8 ft/s. At what rate is the the searchlight perpendicular to the wall is π ? 6 angle between the string and the horizontal de- Note that dθ/dt = 2(2π) = 4π rad/minute. creasing when 200 ft of string have been let out?

Here is a diagram showing the arrangement of the el- Kite motion ements in the problem. It is a view on the scene from above. As the lighthouse rotates its beam, the angle θ b Kite will change. Light spot b L 100 ft

θ Light beam y b Person x

θ dx

b We note that the kite motion of 8 ft/s represents dt Light House dθ in this diagram. We are searching for , so we need 10 miles dt

12 a relationship between x and θ. The simplest is • R1 = 80 and R2 = 100, and 100 tan θ = dR dR x • 1 = +0.3 and 2 = 0.2. d dt dt Taking of both sides, dt d d 100 (tan θ) = dt dt x dR This covers all the values in our expression for , ex- dθ −100 dx dt 2 cept for R itself. To find that, we can use the original sec θ = 2 dt x dt equation to solve for overall resistance: dθ −100 dx = dt x2 sec2 θ dt dθ −100 cos2 θ dx 1 1 1 = = + dt x2 dt R 80 100 1 9 We can use the L value of 200 ft (200 ft of string) with = Pythagoras to solve for x = 173.21 ft, and θ ≈ 0.5236 R 400 dx 400 rad. Subbing those and = 8 into the related rates so R = ≈ 44.444 ohms dt 9 equation gives dθ −100 cos2(0.5236) dR = 8 Substituting in all these values into our expression dt 173.212 dt dθ gives = −0.02 radians/s dt 33. If two resistors with resistances R1 and R2 are dR 2 1 dR1 1 dR2 = R 2 + 2 connected in parallel, as in the figure, then the dt R1 dt R2 dt total resistance R, measured in Ohms (Ω), is 2 1 1 given by: = (44.444) 2 (0.3) + 2 (0.2) 1 1 1 80 100 = + dR R R1 R2 = 0.132 ohms/s dt If R1 and R2 are increasing at rates of .3 Ω/s and .2 Ω/s, respectively, how fast is R increasing when R1 = 80 Ω and R2 = 100 Ω? 34. A water trough is 10 m long and a cross-section has the shape of an isosceles trapezoid that is 30 R1 cm wide at the bottom, 80 cm wide at the top, and has height 50 cm. If the trough is being filled with water at a rate of 0.2 m3/min, how fast is the water level rising when the water is 30 cm deep? V R2

Diagrams are essential for this problem! There are several ways to solve this problem. The most straightforward is to take the relationship that is given, and diﬀerentiate with respect to t. d 1 d 1 1 = + dt R dt R1 R2 −1 dR −1 dR1 −1 dR2 2 = 2 + 2 R dt R1 dt R2 dt dR Solving for gives: dt dR 2 1 dR1 1 dR2 = R 2 + 2 dt R1 dt R2 dt In the problem, we are given the following values:

13 We can now diﬀerentiate both sides with respect to dV dh time to ﬁnd the relationship between and : dt dt

dV dh dh = 10h + 3 dt dt dt

dV Subbing in = 0.2 m3/min, and h = 0.3 m, dt

dh dh 0.2 = 10(0.3) + 3 dt dt When water is filling up the trough, we have a con- dh 0.2 = 6 nection between the filled depth h, and the volume V . dt dV dh dh 0.2 1 We are given that = 0.2m3/min, and we want = = m/min, or dt dt dt 6 30 when h = 30 cm. dh 1 10 = 100 × = ≈ 3.33 cm/min. We need an equation that relates h and V at any time dt 30 3 during the filling process. The depth of the trough is 10 m (not shown in the 35. A voltage V across a resistance R generates a diagrams), so current I = V/R. If a constant voltage of 4 V = 10 × (cross-section area up to height h) volts is put across a resistance that is increasing at a rate of 0.7 ohms per second when the The cross-section area in the diagram is made up of resistance is 7 ohms, at what rate is the current three elements: T , R and T . If we express everything 1 2 changing? in terms of meters (so 30 cm = 0.3 m), R = 0.3h

The areas of T1 and T2 can be determined using similar We know dR/dt = 0.7 when R = 5 and V = 9 and we triangles, as their height/widths are in a 50 cm/25 cm want to know dI/dt. Diﬀerentiating I = V/R with V ratio: h/width = 0.50/0.25, so constant gives width = 0.5h. 1 This means that (area of T1) = (area of T2) = hw = dI 1 dR 2 = V − , 1 1 2 h(0.5h) = h2. dt R dt 2 4 Bringing it all back into the volume expression, so substituting gives

V = 10 × (cross-section area up to height h) dI 1 = 4 − 2 · 0.7 ≈ −0.0571 amp/s . V = 10(T1 + R + T2) dt 7 1 1 V = 10 h2 + 0.3h + h2 4 4 (Note that we know the units are amp/s from the Leib- dI 1 2 nitz form of the derivative, . It is, of course, equally V = 10 h + 0.3h dt 2 correct to use the expression on the right-hand side to 2 V = 5h + 3h determine the units, V ohm , or, V/ohm*s.) ohm2 s