Week #5 - Related Rates, Linear Approximation, and Taylor Polynomials Section 4.6

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1 Week #5 - Related Rates, Linear Approximation, and Taylor Polynomials Section 4.6 From Calculus, Single Variable by Hughes-Hallett, Gleason, McCallum et. al. Copyright 005 by John Wiley & Sons, Inc. This material is used by permission of John Wiley & Sons, Inc. SUGGESTED PROBLEMS. The desired rate is. At the date given, t = 0, so the rate is = 6.4e 0.0 t (0.0) billions of people per year (0) = 6.4 (0.0) billions of people per year = million people per year P = 8R dr = 8( R ) = 8 R 5. (a) How fast is an idea of rate, and is represented by the rate dy. dy = 0.(.5)t0.5 = 0. t At t =, dy () = 0. = 0. cm / hour At t =, dy () = cm / hour (b) Since both y = 0.t.5 and dy = 0.t0.5 increase as t increases, the thickness y and rate dy will both be maximal when t =. 0. (a) The rate of change is with respect to time, treating r as a constant, is = 500e rt/00 r 00 = 5 re rt/00 (i) Initially, (0) = 5 re0 dollars per year

2 (ii) At t =, () = 5 rer/00 = 5 re 0.0r dollars per year (b) If r is a function of t, we need the chain rule to be careful about the derivative: [ ( )] d r(t) t = 500e r(t) t/00 00 = 5e r(t) t/00 [ r (t) t + r(t) ] Substituting in t =, r = 4 and r () = 0., gives = 5e 4 /00 [0. + 4] 4.96 Thus the price is increasing by about $5 per year at when t =.. We are given that V = 9 and constant, R = 5, and dr = 0.. Differentiating the original equation, di = V ( R ) dr = 9(5) (0.) = 0.07 amps per second 5. (a) We are told that the body temperature ops by o F in the first hour. Since the body temperature is T(0) = e 0 = 98.6 at the time of death, this means T() = 98.6 = We use this information to solve for k: T() = 96.6 = e k() 8.6 = 0.6e k = e k ( ) 8.6 Take ln of both sides: ln = ln(e k ) = k 0.6 k (b) Now that we have k, we can solve for t when dt = o /hour: dt = = 0.6e t ( ) (0.6)(0.0676) = e t ( ) Take ln of both sides: ln = ln(e t ) = t (0.6)(0.0676) t 0.75 hours So where t 0.75 hours, the body is cooling at o F/ hour.

3 (c) After 4 hours, the body will have cooled to a temperature of T(4) = e t = o Using the coroner s approximate formula ( o in the first hour, o for each hour afterwards), the body would have cooled by 5 o F in the first 4 hours. This would imply a temperature of = 7.6 o, which is very close to the value of 74 o from the more realistic exponential model.. Let the volume of clay be V. The clay is in the shape of a cylinder, so V = πr L. We know dl = 0. cm/sec and we want to know when r = cm and L = 5 cm. Differentiating both sides of the equation with V with respect to time t gives = πrl + πrdl. However, the amount of clay is unchanged, so = 0 and so Solving for gives rl = rdl, = r dl L. When the radius is cm and the length is 5 cm, and the length is increasing at 0. cm per second, the rate at which the radius is changing is = 0. = 0.0 cm/sec. 5 Thus, the radius is decreasing at 0.0 cm/sec. 5. First, we define the length along the shore as x: We want to calculate dx dθ. We can do this by starting with the relationship x = tan(θ) so x = tan(θ) If we differentiate both sides with respect to θ, we get the desired rate: dx dθ = sec (θ) or = cos (θ)

4 9. Let r be the radius of the rainop. Then its volume V = 4 πr cm and its surface area is S = 4πr cm. It is given that = S = 8πr. Furthermore, differentiating our formula for V gives so from the chain rule, = 4πr, = and thus = = 8πr 4πr =. Since is a constant with value, the radius is increasing at a constant rate of cm/sec. 0. The essential relationship between the volume of a cone and its height and radius is V = πr h We want to find / or dh/. The facts we know are / = 0. m /hr r and h are related through the triangle information about the angle with the vertical (π/6 radians, or 0 degrees). Drawing the triangle, we can compare the cone with the standard 0/60 triangle: h π/6 π/6 r π/ π/ The similar triangles imply that h/r = / =, or that h = r or r = h () Going back to the relationship between V and r and h, we can use () to substitute for r: 4

5 V = πh h V = 9 πh The problem is we know /, and we re looking for dh/, but this equation only involves V and h. To get to the derivatives, we can use the implicit differentiation approach, and simply differentiate both sides of the equation (with respect to t): This allows us to solve for dh/: d (V ) = d ( = 9 π ) 9 πh ( h dh ) = πh dh dh = πh and using the fact that / = 0., dh = 0. πh This tells you that the rate the height of the cone is growing at depends on the current height (or size of the cone). Furthermore, if the current cone is big (large h), the rate of growth will be small. This makes sense, since the sand is being added at a constant rate, and if the cone is already large, any new sand will be distributed thinly over the whole cone. By a similar approach, but using r instead of h, you can arrive at the rate of change of r: 5

6 Differentiating: V = πr ( r) V = πr = π r = πr Solving for /: = πr But / = 0., r = h/ : = 0. πh = 0. πh You could get the same result (more easily) by differentiating the relationship r = h/. 4. (a) We differentiate a (t) + b (t) = c with respect to t to find d (a (t) + b d (t)) = c, a(t) a (t) b(t)b (t) = 0 giving a(t) a (t) = b(t)b (t) (b) (i) If Angela likes Brian, then a(t) > 0, so b (t) < 0. This means that b(t) is decreasing, so Brian s affection decreases when Angela likes him. (ii) If Angela dislikes Brian, then a(t) < 0, so b (t) > 0. This means that b(t) is increasing, so Brian s affection increases when Angela dislikes him. (c) Substituting b (t) = a(t) into (a(t) a (t) = b(t) b (t) gives a(t) a (t) = b(t) b (t) = b(t)( a(t)), so a (t) = b(t) (i) If Brian likes Angela, then b(t) > 0, so a (t) > 0. This means that a(t) is increasing, so Angela s affection increases when Brian likes her. (ii) If Brian dislikes Angela, then b(t) < 0, so a (t) < 0. This means that a(t) is decreasing, so Angela s affection decreases when Brian dislikes her. (d) When t = 0, they both like each other. This means that Angela s affection increases, while Brian s decreases. 6