Chapter 7 The Chain Rule, Related Rates, and Implicit Differentiation 7.1 For each of the following, find the derivative of y with respect to x. (a) y 6 + 3y − 2x − 7x3 = 0 √ (b) ey + 2xy = 3 (c) y = xcos x Detailed Solution: (a) d 6 (y + 3y − 2x − 7x3 ) dx dy dy 6y 5 +3 − 2 − 21x2 dx dx dy (6y 5 + 3) dx dy dx
=
d (0) dx
= 0 = 21x2 + 2 =
21x2 + 2 6y 5 + 3
(b) d y (e + 2xy) dx dy dy ey + (2y + 2x ) dx dx dy (ey + 2x) dx dy dx
v.2005.1 - September 12, 2005
=
d √ ( 3) dx
= 0 = −2y = −
ey
2y + 2x
1
Math 102 Problems
Chapter 7
(c) First take the logarithm of both sides: ln y = ln(xcos x ) = (cos x)(ln x), then differentiate both sides with respect to x. d d (ln y) = (cos x · ln x) dx dx 1 1 dy = − sin x · ln x + cos x · y dx x dy cos x = y − sin x · ln x + dx x cos x cos x = x − sin x · ln x + x
7.2 Consider the growth of a cell, assumed spherical in shape. Suppose that the radius of the cell increases at a constant rate per unit time. (Call the constant k, and assume that k > 0.) (a) At what rate would the volume, V , increase ? (b) At what rate would the surface area, S, increase ? (c) At what rate would the ratio of surface area to volume S/V change? Would this ratio increase or decrease as the cell grows? [Remark: note that the answers you give will be expressed in terms of the radius of the cell.] Detailed Solution: We are given the information that
(a) For the volume:
dr = k. dt
4 V = πr 3 3 dV 4π d(r 3 ) 4π 2 dr = = 3r = 4πr 2 k. dt 3 dt 3 dt
(b) For the surface area: S = 4πr 2 dS d dr = (4πr 2) = 4π(2r) = 8πrk. dt dt dt (c)
Thus
S 4πr 2 3 = = 3 V (4/3)πr r d 3 3 dr d S 3k = =− 2 =− 2. dt V dt r r dt r
The derivative is negative so that the ratio of surface area to volume is decreasing as the cell grows.
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2
Math 102 Problems
7.3
Chapter 7
Growth of a circular fungal colony
A fungal colony grows on a flat surface starting with a single spore. The shape of the colony edge is circular (with the initial site of the spore at the center of the circle.) Suppose the radius of the colony increases at a constant rate per unit time. (Call this constant C.) (a) At what rate does the area covered by the colony change ? (b) The biomass of the colony is proportional to the area it occupies (factor of proportionality α). At what rate does the biomass increase? Detailed Solution: The area of the colony is A = πr 2 and dr/dt = C. (a) dA/dt = π(2r)dr/dt = 2πrC. (b) M = αA so dM/dt = αdA/dt = α2πrC.
7.4
Limb development
During early development, the limb of a fetus increases in size, but has a constant proportion. Suppose that the limb is roughly a circular cylinder with radius r and length l in proportion l/r = C where C is a positive constant. It is noted that during the initial phase of growth, the radius increases at an approximately constant rate, i.e. that dr/dt = a. At what rate does the mass of the limb change during this time? [Note: assume that the density of the limb is 1 gm/cm3 and recall that the volume of a cylinder is V = Al where A is the base area (in this case of a circle) and l is length.] Detailed Solution: We have the volume of the cylinder V = πr 2 l and l = Cr so that V = πr 2 (Cr) = Cπr 3 . The mass of the limb is volume times density so M = 1V = Cπr 3 . Thus dM dr = Cπ(3r 2 ) = Cπ(3r 2 )a. dt dt
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3
Math 102 Problems
Chapter 7
7.5 A rectangular trough is 2 meter long, 0.5 meter across the top and 1 meter deep. At what rate must water be poured into the trough such that the depth of the water is increasing at 1 m/min when the depth of the water is 0.7 m? Detailed Solution: See Figure 7.1. Let x and y be the dimensions of the base of the trough, and z be the depth of the water in the trough. Then the volume of water in the trough V = xyz. x, y are constants and z is dz a function of time t. = 1. dt dz dV = xy = (2)(0.5)(1) = 1 m3 /min dt dt
x
z
y
Figure 7.1: Problem 7.5
7.6 Gas is being pumped into a spherical balloon at the rate of 3 cm3 /s. (a) How fast is the radius increasing when the radius is 15 cm? (b) Without using the result from (a), find the rate at which the surface area of the balloon is increasing when the radius is 15 cm. Detailed Solution: At time t the sphere has radius r, volume V = 34 πr 3 , and surface area S = 4πr 2 .
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4
Math 102 Problems
Chapter 7
(a) V
=
4 3 πr 3
dV dr = 4πr 2 · dt dt dV dr dt = dt 4πr 2 3 = 4π · (15)2 1 cm/s = 300π (b) S dS dt dS dt dS dt
= 4πr 2 dr dt 2 dV = · r dt 2 = · (3) 15 2 cm2 /s = 5 = 8πr ·
7.7 1 A point moves along the parabola y = x2 in such a way that at x = 2 the x-coordinate is increasing 4 at the rate of 5 cm/s. Find the rate of change of y at this instant. Detailed Solution: 1 y = x2 4 1 dx 1 dy = x· = (2) (5) = 5 cm/s dt 2 dt 2
7.8
Boyle’s Law
In chemistry, Boyle’s law describes the behaviour of an ideal gas: This law relates the volume occupied by the gas to the temperature and the pressure as follows: P V = nRT where n, R are positive constants. (a) Suppose that the pressure is kept fixed, by allowing the gas to expand as the temperature is increased. Relate the rate of change of volume to the rate of change of temperature.
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5
Math 102 Problems
Chapter 7
(b) Suppose that the temperature is held fixed and the pressure is decreased gradually. Relate the rate of change of the volume to the rate of change of pressure. Detailed Solution: (a) If P is constant, we can rewrite the relationship in the form V (t) =
nR T (t) = kT (t) P
where k = nR/P is a constant. Then dV dT nR dT =k = . dt dt P dt (b) In this case T is constant, so we have V (t) =
nRT C = . P (t) P (t)
where C = nRT is a constant. Then by the reciprocal law, dV C dP nRT dP =− 2 =− 2 dt P dt P dt
7.9
Spread of a population
In 1905 a Bohemian farmer accidentally allowed several muskrats to escape an enclosure. Their population grew and spread, occupying increasingly larger areas throughout Europe. In a classical paper in ecology, it was shown by the scientist Skellam (1951) that the square root of the occupied area increased at a constant rate, k. Determine the rate of change of the distance (from the site of release) that the muskrats had spread. For simplicity, you may assume that the expanding area of occupation is circular. Detailed Solution: According to Skellam, if we put f (t) = A(t)1/2 , where A(t) is the occupied area, then f 0 (t) = k. If we make the simplifying assumption that the areas of expansion are circular, we can use the fact that the area of a circle, where both A(t) and r(t) are functions of time t, is A(t) = πr(t)2 which is the same as saying that f (t) = A(t)1/2 = π 1/2 r(t)
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6
Math 102 Problems
Chapter 7
Differentiating both sides with respect to t gives f 0 (t) = π 1/2 r 0 (t), which we can rearrange to get r 0 (t) = π −1/2 f 0 (t) = π −1/2 k since we assumed that f 0 (t) = k. The distance is thus increasing at a constant rate of π −1/2 k.
7.10 A spherical piece of ice melts so that its surface area decreases at a rate of 1 cm2 /min. Find the rate that the diameter decreases when the diameter is 5 cm. Detailed Solution: Let r = r(t) be the radius at time t and D(t) = 2r(t) the diameter. Then the surface area is S = 4πr 2 = πD 2 and so
But
dS dt
dS dD = 2πD . dt dt = −1 and therefore
When D = 5 we have
7.11
dD 1 =− . dt 2πD dD 1 =− . dt 10π
A Convex lens
A particular convex lens has a focal length of f = 10 cm. The distance p between an object and the lens, the distance q between its image and the lens and the focal length f are related by the equation: 1 1 1 = + . f p q If an object is 30 cm away from the lens and moving away at 4 cm/sec, how fast is its image moving and in which direction? Detailed Solution: We can use the relationship 1 1 1 = + 10 p q to solve for q:
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1 1 1 p − 10 = − = q 10 p 10p
7
Math 102 Problems
Chapter 7
10p . p − 10 (The focal length of the lens is constant, but the other two quantities are changing and depend on t.) Differentiating each side with respect to t: q=
10p0 (t)(p − 10) − p0 (t)(10p) 100p0(t) dq = = − . dt (p − 10)2 (p − 10)2
We now use the fact that p(t) = 30, p0 (t) = −4, at the given instant. Then dq = 400/202 = 1. dt Therefore the image is moving toward the lens at 1 cm/sec.
7.12
A conical cup
Water is leaking out of a small hole at the tip of a conical paper cup at the rate of 1 cm3 /min. The cup has height 8 cm and radius 6 cm, and is initially full up to the top. Find the rate of change of the height of water in the cup when the cup just begins to leak. [Remark: the volume of a cone is V = (π/3)r 2h.] Detailed Solution: See Figure 7.2. Let r be the radius of the top surface of the water and h the height of the water in the conical cup at a given time. Then the volume of water is given by V = (π/3)r 2h and we are given that cm3 dV = −1 . dt min when r = 6 cm and h = 8 cm. First we need an expression that relates r to h. We want to find dh dt In the cone, the proportions of base radius to height are 6 to 8. The water always forms a conical shape and so, by similar triangles, r 6 6h = ⇒r= . h 8 8 Substituting this into the volume equation, we get: π3h3 π 36h2 h= . V = 3 64 16 To solve the problem, we now differentiate the volume: dV 3π d(h3 ) 3π 2 dh = = 3h . dt 16 dt 16 dt : and isolate dh dt dh 16 1 dV = . dt 9π h2 dt Substituting in our values for dV and h gives: dt 16 −1 dh = (−1) = ≈ 0.00884 cm/min. dt 576π 36π
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8
Math 102 Problems
Chapter 7 6
r 8 water
h
Figure 7.2: Conical Cup for problem 7.12 solution
7.13
Conical tank
1 m3 /min, and at 10 the same time is being pumped in the top at a constant rate of k m3 /min. The tank has height 6 m and the radius at the top is 2 m. Determine the constant k if the water level is rising at the 1 rate of m/min when the height of the water is 2 m. Recall that the volume of a cone of radius r 5 and height h is 1 V = πr 2 h. 3 Water is leaking out of the bottom of an inverted conical tank at the rate of
Detailed Solution: If V (t) is the volume of water in the tank at time t, then 1 dV =k− . dt 10 But we also know that (by similar triangles) h/r = 6/2 = 3 so r = h/3 and 1 πh3 V = πr 2 h = , 3 27 where r is the radius of the water and h is the height. Therefore
when h = 2 and
dh dt
dV πh2 dh 4π = × = dt 9 dt 45 1 = 5 . Plugging this into our original expression relating k=
7.14
dV dt
and k, it follows that
1 4π + ≈ 0.379. 10 45
The gravel pile
Gravel is being dumped from a conveyor belt at the rate of 30 f t3 /min in such a way that the gravel forms a conical pile whose base diameter and height are always equal. How fast is the height of the pile increasing when the height is 10 f t? (Hint: the volume of a cone of radius r and height 1 h is V = πr 2 h.) 3
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9
Math 102 Problems
Chapter 7
Detailed Solution: Let V (t) denote the volume of gravel at time t. Then πh3 1 V = πr 2 h = 3 12 since d = h, or simply r = h/2. Therefore πh2 dh dV = . dt 4 dt But we know that
dV = 30, dt
and therefore when h = 10 we get 4 dV 120 6 dh = = = ≈ 0.38 f t/min. dt πh2 dt 100π 5π
7.15
The sand pile
Sand is piled onto a conical pile at the rate of 10m3 /min. The sand keeps spilling to the base of the cone so that the shape always has the same proportions: that is, the height of the cone is equal to the radius of the base. Find the rate at which the height of the sandpile increases when the height is 5 m. Note: The volume of a cone with height h and radius r is V =
π 2 r h. 3
Detailed Solution: We would like to find h0 (t) at that point in time at which h(t) = 5m. (h0 (t) will be in m/min). We know that V 0 (t) = 10 for all t, and that V (t) =
π π r(t)2 h(t) = h(t)3 , 3 3
since r(t) = h(t) for all t. Using the chain rule, we get V 0 (t) = πh(t)2 h0 (t) and solving for h0 (t) we find h0 (t) =
V 0 (t) 10 = . 2 πh(t) πh(t)2
At the point in time when h(t) = 5, we therefore have h0 (5) =
v.2005.1 - September 12, 2005
2 10 = ≈ 0.1273 m/min. π25 5π
10
Math 102 Problems
Chapter 7
7.16 Water is flowing into a conical reservoir at a rate of 4 m3 /min. The reservoir is 3 m in radius and 12 m deep. (a) How fast is the radius of the water surface increasing when the depth of the water is 8 m? (b) In (a), how fast is the surface rising? Detailed Solution: (a) V = 13 πr 2 h. See Figure 7.3, the radius of the water surface is related to the water depth by r 3 1 4 = ⇒ h = 4r. So V = πr 2 (4r) = πr 3 . When h = 8, r = 2. h 12 3 3
3m 3m r
12 m
12 m h
h
Figure 7.3: Problem 7.16
dV dr = 4πr 2 dt dt dV dr dt = dt 4πr 2 4 = 4π(2)2 1 m/min = 4π (b) h = 4r, so
dh dr dh 1 1 =4 . =4 = m/min. dt dt dt 4π π
7.17 A ladder 10 meters long leans against a vertical wall. The foot of the ladder starts to slide away from the wall at a rate of 3 m/s. (a) Find the rate at which the top of the ladder is moving downward when its foot is 8 meters away from the wall.
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Math 102 Problems
Chapter 7
(b) In (a), find the rate of change of the slope of the ladder. Detailed Solution: Let x be the distance between the foot of the ladder and the foot of the wall and y be the distance between the foot of the wall and the top of the ladder. See Figure 7.4.
10 m y 3 m/s x
Figure 7.4: Problem 7.17
(a) x, y and the length of the ladder satisfies x2 + y 2 = 102 = 100. When x = 8, y = 6. Differentiate both sides of x2 + y 2 = 100 with respect to t:
2x
(b) Slope s =
√
100 − 82 =
dx dy + 2y = 0 dt dt −2x dx dy dt = dt 2y −x dx dt = y −8(3) = 6 = −4 m/s
y ds . Rate of change of slope is . x dt dy dt
· x − y · dx dt x2 (−4)(8) − 6(3) = 82 25 = − per sec 32
dS = dt
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12
Math 102 Problems
7.18
Chapter 7
Sliding ladder
A ladder 5 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate of 0.5 meter/min how fast is the top of the ladder sliding down the wall when the base of the ladder is 1 m away from the wall ? Detailed Solution: This is similar to problem 7.18. We relate the length of the ladder with its distance along the wall and the ground using Pythagoras: x2 + y 2 = 25. We differentiate this equation with respect to t: 2x
dy dy x dx dx + 2y = 0 =⇒ =− . dt dt dt y dt
But we have dx = 1/2 and x = 1. We can find y by plugging x = 1 into the equation x2 + y 2 = dt √ √ = − xy dx to get: 25 ⇒ 1 + y 2 = 25 ⇒ y = 24 = 2 6. We substitute these values into dy dt dt dy/dt =
−1 −x = √ . 2y 4 6
7.19 Ecologists are often interested in the relationship between the area of a region (A) and the number of different species S that can inhabit that region. Hopkins (1955) suggested a relationship of the form S = a ln(1 + bA) where a and b are positive constants. Find the rate of change of the number of species with respect to the area. Does this function have a maximum? Detailed Solution: Using the chain rule we find that ab dS = . dA 1 + bA This derivative is never zero, which means that there are no local maxima. The larger the area, the greater the number of species that would inhabit it according to this formula.
7.20
The burning candle
A candle is placed a distance l1 from a thin block of wood of height H. The block is a distance l2 from a wall as shown in Figure 7.5. The candle burns down so that the height of the flame, h1 decreases at the rate of 3 cm/hr. Find the rate at which the length of the shadow y cast by the block on the wall increases. (Note: your answer will be in terms of the constants l1 and l2 . Remark: This is a challenging problem.)
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13
Math 102 Problems
Chapter 7
y H
h1 l1
h
1
l2
Figure 7.5: Figure for Problem 7.20 Detailed Solution: Let h1 (t) be the height of the candle flame at time t, let y(t) be the length of the shadow at time t, and let L = l1 + l2 . Then L is constant. Also we will define h2 = H − h1 . Note that h1 = h1 (t) is changing with time and so is h2 . Looking at the geometry shown in Figure 7.5 above the dotted line, we see similar triangles having side lengths L, (y − h1 ), and h2 , l1 , Thus, by similar triangles, h2 H − h1 y − h1 = = . L l1 l1 The height of the shadow is therefore y(t) = L
H − h1 (t) h1 (t) + l1 L
Differentiating with respect to time: dy 1 dh1 1 dh1 . =L − + dt l1 dt L dt Simplifying leads to 1 dy dh1 1 . = L − + dt dt l1 L Using dh1 /dt = −3 leads to dy 1 1 . = −3(l1 + l2 ) − + dt l1 (l1 + l2 ) After simplifying we get dy l1 + l2 =3 −1 . dt l1 dy l2 =3 dt l1
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14
Math 102 Problems
Chapter 7
7.21 Use implicit differentiation to show that the derivative of the function y = x1/3 is y 0 = (1/3)x−2/3 . First write the relationship in the form y 3 = x, and then find dy/dx. Detailed Solution: Rewrite the equation in the form y3 = x and then differentiate both sides with respect to x : 3y 2y 0 = 1. Solving for y 0 we get y0 =
7.22
1 1 = = (1/3)x−2/3 . 2 (3y ) (3x2/3 )
Generalizing the Power Law
(a) Use implicit differentiation to calculate the derivative of the function y = f (x) = xn/m where m and n are integers. (Hint: rewrite the equation in the form y m = xn first.) √ (b) Use your result to derive the formulas for the derivatives of the functions y = x and y = x−1/3 . Detailed Solution: (a) Since y = f (x) = xn/m , we have y m = xn . So d(y m)/dx = d(xn )/dx = nxn−1 . Using the chain rule, dy m dy m dy dy = = my m−1 dx dy dx dx Therefore, my m−1 (dy/dx) = nxn−1 so that y dy n xn−1 n = = xn−1 m . m−1 dx my m y Substituting the expression y = xn/m in this last result, we get dy n xn−1 xn/m n n/m−1 = = x dx m xn m
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15
Math 102 Problems
Chapter 7
(b) We have y=
√
x = x1/2 .
Using the results in (a) we get
√ dy/dx = (1/2)x1/2−1 = (1/2)x−1/2 = 1/(2 x.)
For y = x−1/3 we get dy/dx = (−1/3)x−1/3−1 = (−1/3)x−4/3 .
7.23 The equation of a circle with radius r and center at the origin is x2 + y 2 = r 2 (a) Use implicit differentiation to find the slope of a tangent line to the circle at some point (x, y). (b) Use this result to find √ the equations of the tangent lines of the circle at the points whose x coordinate is x = r/ 3. (c) Use the same result to show that the tangent line at any point on the circle is perpendicular to the radial line drawn from that point to the center of the circle Note: Two lines are perpendicular if their slopes are negative reciprocals. Detailed Solution: (a) We are asked to find the slope of the tangent line to a circle with radius r and center at the origin, whose equation is x2 + y 2 = r 2 . Taking the derivative with respect to x on both sides we get: d 2 d 2 (x + y 2) = (r ) dx dx 2x +
dy 2 =0 dx
2x + 2y
dy =0 dx
Thus we can isolate y and find that dy 2x x =− =− . dx 2y y
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Math 102 Problems
Chapter 7
(b) We now want to find p the equations of the tangent lines of the circle at the points whose x coordinate is x = r 1/3. The corresponding y coordinate is given by r 2 /3 + y 2 = r 2
i.e. y = ±r
p
2/3.
p p Therefore, one of the points is (r 1/3, r 2/3). By part (a) the slope of the tangent line at p p the point (r 1/3, r 2/3) would be p p √ (dy/dx) = −x/y = −(r 1/3)/(r 2/3) = −1/ 2. We can now use the point-slope method of finding the equation of the tangent line: p y − r 2/3 1 p = −√ 2 x − r 1/3 After simplifying we find that √ √ 2 1 r r 1 1 2r + r √ = √ −x + √ + √ = √ (−x + r 3) y = −√ x − √ 2 3 3 2 3 3 2
p p We can similarly do the calculations for the tangent line that goes through the point (r 1/3, −r 2/3) √ √ and the result is y = (1/ 2)(x − r 3). (Note: this second result can also be obtained by considering the symmetry of the two cases about the x axis.) (c) The slope of a line connecting the origin, (0, 0), and a point on the circle, (x, y) is m = y/x. The slope of the tangent line at the point (x, y) on the circle is, by our previous result, −x/y. Since these slopes are negative reciprocals, the two lines are perpendicular.
7.24 The equation of a circle with radius 5 and center at (1, 1) is (x − 1)2 + (y − 1)2 = 25 (a) Find the slope of the tangent line to this curve at the point (4, 5). (b) Find the equation of the tangent line. Detailed Solution: (a) Using implicit differentiation we have 2(x − 1) + 2(y − 1)dy/dx = 0 =⇒ dy/dx = −(x − 1)/(y − 1). At (4, 5) this gives −3/4 for the slope.
Another way to compute the slope is to use the fact that for a circle the tangent line is perpendicular to the radial line (the line from the center (1, 1) to (4, 5)). This slope is (5 − 1)/(4 − 1) = 4/3 and therefore the slope of the tangent line must be −3/4.
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17
Math 102 Problems
Chapter 7
(b) The equation of the tangent line is −3 y−5 = , x−4 4
or y = −(3/4)x + 8.
7.25
Tangent to a hyperbola
The curve x2 − y 2 = 1 is a hyperbola. Use implicit differentiation to show that for large x and y values, the slope dy/dx of the curve is approximately 1. Detailed Solution: Differentiating both sides : d 2 d (x − y 2 ) = 1 =⇒ 2x − 2y(dy/dx) = 0. dx dx Thus dy/dx = x/y. But on the hyperbola, x2 − y 2 = 1 so that p x = ± 1 + y 2.
p When x and y are very large compared to 1, it is approximately true that x ≈ ± y 2 = ±y. When both x and y are positive, this means that x ≈ y. Thus the slope of the tangent line for large positive values of x and y is dy/dx = x/y ≈ x/x = 1.
7.26
An ellipse
Use implicit differentiation to find the points on the ellipse x2 y 2 + =1 4 9 at which the slope is -1/2. Detailed Solution: For the ellipse
x2 a2
+
y2 b2
= 1, implicit differentiation gives 2x 2y dy =0 + 2 a2 b dx
solving for the slope, we find that dy xb2 = − 2. dx ya
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Math 102 Problems
Chapter 7
For the above ellipse, a = 2, b = 3 so that dy 9x =− . dx 4y = −1/2. From the equation of the ellipse Thus, the points at which this slope is - 1/2 satisfy − 9x 4y r 9x2 . Plugging this into − 9x = −1/2 gives us: we have the relation y = 9 − 4y 4 −1 −1 −9x −9 q q = = =⇒ 2 2 2 12 x12 − 14 12 1 − x4 r 1 3 1 9 1 10 1 − = =⇒ 2 = + = 2 x 4 2 x 4 4 4 2 4 =⇒ x = ± √ x2 = 10 10 s 9 4 9 Then y = 9 − = ±√ . 4 10 10 However, there should only be 2 points on the ellipse where the slope is −1/2, namely in the 1st and 3rd quadrants. To verify that these are the correct solutions, we plug ( √210 , √910 ) and = −1/2, and get −1/2 = −1/2 in both cases. Therefore the (− √210 , − √910 ) into the equation − 9x 4y points on the ellipse at which the slope is −1/2 are ( √210 , √910 ) and (− √210 , − √910 ).
7.27
Motion of a cell
In the study of cell motility, biologists often investigate a type of cell called a keratocyte, an epidermal cell that is found in the scales of fish. This flat, elliptical cell crawls on a flat surface, and is known to be important in healing wounds. The 2D outline of the cell can be approximated by the ellipse x2 /100 + y 2/25 = 1 where x and y are distances in µ (Note: 1 micron is 10−6 meters). When the motion of the cell is filmed, it is seen that points on the “leading edge” (top arc of the ellipse) move in a direction perpendicular to the edge. Determine the direction of motion of the point (xp , yp ) on the leading edge. Detailed Solution: The ellipse is x2 /100 + y 2/25 = 1 so that its major and minor axes are a = 10, b = 5. The slope of the ellipse (i.e. of the edge of the cell) at a given point is 25x x xb2 =− dy/dx = − 2 = − ya 100y 4y A direction perpendicular to this would be a line whose slope is the negative reciprocal of dy/dx, p and at the point (xp , yp ) this direction would be m = 4y . namely m = 4y x xp
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Math 102 Problems
7.28
Chapter 7
The Folium of Descartes:
A famous curve (see Figure 7.6) that was studied historically by many mathematicians (including Descartes) is x3 + y 3 = 3axy (1.5a,1.5a)
a
-a
Figure 7.6: The Folium of Descartes in Problem 7.28 You may assume that a is a positive constant. (a) Explain why this curve cannot be described by a function such as y = f (x) over the domain −∞ < x < ∞. (b) Use implicit differentiation to find the slope of this curve at a point (x, y). (c) Determine whether the curve has a horizontal tangent line anywhere, and if so, find the x coordinate of the points at which this occurs. (d) Does implicit differentiation allow you to find the slope of this curve at the point (0,0) ? Detailed Solution: The Folium of Descartes is the curve x3 + y 3 = 3axy where a is a positive constant. (a) This curve cannot be described by a function such as y = f (x) over the domain −∞ < x < ∞ because it is not possible to solve for y in the equation and to obtain a single valued function y = f (x). Further, from the picture of the curve shown on the homework sheet, we see that it cannot be represented by a function since several values of x correspond to more than one value of y, i.e. the curve fails the vertical line property. (b) We can use implicit differentiation to find the slope of this curve at a point (x, y). The equation of the curve is x3 + y 3 = 3axy so 3x2 + 3y 2
dy dy = 3ay + 3ax dx dx
3(y 2 − ax)
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dy = 3(ay − x2 ) dx
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Math 102 Problems
Chapter 7 dy (ay − x2 ) = 2 dx (y − ax)
(c) Horizontal tangent(s) to the curve occur when dy/dx = 0, i.e. when (ay − x2 ) = 0 or y = x2 /a and when the equation of the curve is also satisfied. Plugging y = x2 /a into the equation of the curve, we find that x3 x6 3 =⇒ x6 = 2x3 a3 x + 3 = 3a a a which means that either x = 0 (in which case y = 0 too) or else x = 21/3 a. (d) At the point (0, 0), our result by implicit differentiation is that the slope of the curve is 0/0 which is undefined. Thus we cannot use this method. We also note from the sketch of the curve that there are two branches which both go through the origin. Thus there is not one unique tangent line at this point, and the derivative is not defined there. (Note: It is possible to discuss the slope of the various branches that cross the origin, but for this we would have to investigate the so-called parametric form of this curve, something that will be discussed in Math 200.)
7.29
Isotherms in the Van-der Waal’s equation:
In thermodynamics, the Van der Waal’s equation relates the mean pressure, p of a substance to its molar volume v at some temperature T as follows: a (p + 2 )(v − b) = RT v where a, b, R are constants. Chemists are interested in the curves described by this equation when the temperature is held fixed. (These curves are called isotherms). (a) Find the slope, dp/dv, of the isotherms at a given point (v, p). (b) Determine where points occur on the isotherms at which the slope is horizontal. Detailed Solution: The isotherms in the Van-der Waal’s equation are given by a (p + 2 )(v − b) = RT (∗) v where a, b, R are constants. In this case, since T is held fixed, it is also a constant. (a) We can find the slope using implicit differentiation. Using the product rule and implicit differentiation we get a dp a ( − 2 3 )(v − b) + (p + 2 )(1) = 0. dv v v Thus a a dp = (2 3 ) − (p + 2 )/(v − b). dv v v
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Math 102 Problems (b)
dp dv
Chapter 7
= 0 when (2
i.e. when
a 2a(v − b) − (pv 3 + av) a ) − (p + )/(v − b) = =0 v3 v2 v 3 (v − b) 2a(v − b) − pv 3 − av = 0
Solving for p we get 2a(v − b) − va a(v − 2b) = . 3 v v3 To find points on the isotherms at which the slope is 0, we need to make sure that the values a(v − 2b) into (*): of p, v satisfy the equation of the isotherms, (*). Therefore we plug p = v3 p=
av − 2ab + av (v−b) = RT =⇒ (2av−2ab)(v−b) = RT v 3 =⇒ 2av 2 −2avb−2avb+2ab2 = RT v 3 3 v This leads to a cubic equation in v, namely RT v 3 − 2av 2 + 4ahv − 2ab2 = 0. We will find out how to solve such equations using Newton’s method later on in this course.
7.30
The circle and parabola
A circle of radius 1 is made to fit inside the parabola y = x2 as shown in figure 7.7. Find the coordinates of the center of this circle, i.e. find the value of the unknown constant c. [Hint: Set up conditions on the points of intersection of the circle and the parabola which are labeled (a, b) in the figure. What must be true about the tangent lines at these points?] y
(0,c) (a,b)
x
Figure 7.7: Figure for Problem 7.30
Detailed Solution: The equation of the circle is (y − c)2 + x2 = 1. We now gather the needed conditions to solve for c.
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Math 102 Problems
Chapter 7
Condition 1: (a, b) is on the parabola so that b = a2 . Condition 2: (a, b) is on the circle so that (b − c)2 = 1 − a2 . Condition 3: The parabola and the circle share a tangent line at the point (a, b). This implies that the slope computed from dy/dx = 2x = 2a at this point matches the slope computed for the tangent to the circle. To compute that slope, observe that the slope of the radius vector joining (a, b) to (0, c) is (c − b)/(0 − a) = −(c − b)/a. The tangent is perpendicular to this radius vector, and thus has slope which is the negative reciprocal, i.e. a/(c − b). We now know that a/(c − b) = 2a (from equating slopes) which means that c − b = 1/2. Using √ 2 2 2 (b − c) = 1 − a now gives us 1 − a = 1/4 which implies that a = 3/2 and b = a2 = 3/4. Thus c = b + 1/2 = 5/4. The coordinate of the center of the circle is (0, 5/4).
7.31 Consider the curve whose equation is x3 + y 3 + 2xy = 4, y = 1 when x = 1. (a) Find the equation of the tangent line to the curve when x = 1. (b) Find y 00 at x = 1. (c) Is the graph of y = f (x) concave up or concave down near x = 1? Hint: Differentiate the equation x3 + y 3 + 2xy = 4 twice with respect to x. Detailed Solution: (a) Differentiating the equation x3 + y 3 + 2xy = 4 with respect to x, and not forgetting that y depends on x, gives 3x2 + 3y 2 y 0 + 2y + 2xy 0 = 0. Putting x = 1 and y = 1 and then solving for y 0 we get 3 + 3y 0 + 2 + 2y 0 = 0 =⇒ y 0 = −1. Thus the equation of the tangent line is y − 1 = −1(x − 1). (b) Differentiating the equation 3x2 + 3y 2y 0 + 2y + 2xy 0 = 0 with respect to x gives 6x + 6yy 0 + 3y 2y 00 + 2y 0 + 2y 0 + 2xy 00 = 0. 4 Now set x = 1, y = 1 and y 0 = −1 to get 6 − 6 + 3y 00 − 2 − 2 + 2y 00 = 0 =⇒ y 00 = . 5 (c) Since y 00(1) > 0 the graph is concave up near x = 1.
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