Math 113 HW #7 Solutions

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1 Math 3 HW #7 Solutions 35 0 Given find /dx by implicit differentiation y 5 + x 2 y 3 = + ye x2 Answer: Differentiating both sides with respect to x yields 5y 4 dx + 2xy3 + x 2 3y 2 ) dx = dx ex2 + y2x)e x2 Re-arranging so that all terms containing /dx are on the left side gives Thus, 2 Given 5y 4 + 3x 2 y 2 e x2) dx = 2xyex2 2xy 3 dx = find /dx by implicit differentiation 2xyex2 2xy 3 5y 4 + 3x 2 y 2 e x2 + x = sinxy 2 ) Answer: Differentiating both sides with respect to x yields = cosxy 2 ) y 2 + x2y) ) dx Therefore, so we have that y 2 cosxy 2 ) = 2xy cosxy 2 ) dx, dx = y2 cosxy 2 ) 2xy cosxy 2 ) 30 Use implicit differentiation to find an equation of the tangent line to the curve at the point 0, 2) y 2 y 2 4) = x 2 x 2 5) Answer: Distributing the products, the given equation is equivalent to y 4 4y 2 = x 4 5 Differentiating with respect to x on both sides, we have that 4y 3 8y dx dx = 4x3 0x

2 Therefore, factoring out the /dx on the left and dividing both sides by 4y 3 8y yields dx = 4x3 0x 4y 3 8y Plugging in x = 0, y = 2, we see that, at the point 0, 2), the slope of the tangent line is 40) 3 00) 4 2) 3 8 2) = 0 Using the point-slope formula, then, the equation of the tangent line is y 2) = 0x 0) or, equivalently, y = 2 46 Find the derivative of the function Simplify where possible y = tan x Answer: Writing y as tan x ) /2 and using the Chain Rule and what we know about the derivative of tan x, dx = tan x ) /2 2 + x 2 = 2 + x 2 ) tan x 60 Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection Show that the two families of curves x 2 + y 2 = ax, x 2 + y 2 = by are orthogonal trajectories of each other, that is, every curve in one family is orthogonal to every curve in the other family Sketch both families of curves on the same axes Answer: Differentiating both sides of the equation for the first family, we see that meaning that 2x + 2y dx = a, dx = a 2x 2y gives the slopes of the tangent lines to the first family of curves Turning attention to the second family, differentiating yields ) meaning that 2x + 2y dx = b dx, 2y b) dx = 2x, 2

3 so dx = 2x 2y b gives the slopes of the tangent lines to the second family of curves At a point of intersection x 0, y 0 ), the two curves are orthogonal if and only if the expressions ) and 2) are negative reciprocals, ie if and only if 2) a 2x 0 2y 0 = 2y 0 b 2x 0 Cross-multiplying, this holds if and only if which is equivalent to 2ax 0 + 4x 2 0 = 4y by 0 4x y 2 0) = 2ax 0 + 2by 0 = 2ax 0 + by 0 ) But, since ax 0 = x y2 0 and by 0 = x y2 0, this is true, so the two families of curves are indeed orthogonal Figure : Blue curves: x 2 + y 2 = ax; red curves: x 2 + y 2 = by 36 6 Differentiate the function y = ln x 3

4 26 Let Answer: Using the Chain Rule and the fact that d dx ln x) = x, we see that Find y and y Answer: Using the Chain Rule, y = sec x + tan x dx = ln x) 2 x = x ln x) 2 y = lnsec x + tan x) sec x tan x + sec 2 x ) = sec x tan x + sec2 x sec x + tan x Factoring out sec x from both terms in the numerator, we see that Therefore, y = sec xsec x + tan x) sec x + tan x y = sec x tan x = sec x 40 Use logarithmic differentiation to find the derivative of the function y = 4 x 2 + x 2 Answer: First, take the natural log of both sides: x ln y = ln 2 + x 2 4 x 2 = ln + x 2 ) /4 = 4 ln We can further simplify this, using the rules of logarithms, as ln y = 4 Now, differentiating both sides yields or, after simplifying, y dx = 4 Multiplying both sides by y gives lnx 2 + ) lnx 2 ) ) x 2 + 2x x 2 2x y dx = 2x 4 x 2 + 2x ) dx = 4 y Finally, plugging in y = 4 x 2 + x 2 yields dx = 4 x x 2 2x x 2 + 2x x 2 + 2x ) x 2 ) + ), 2x ) 4

5 42 Use logarithmic differentiation to find the derivative of the function 39 y = x cos x Answer: Taking the natural log of both sides, Therefore, differentiating yields Multiplying both sides by y, or, after plugging in y = x cos x, ln y = ln x cos x ) = cos x ln x y dx = sin x ln x + cos x x = cos x sin x ln x x cos x ) dx = y x sin x ln x dx = xcos x cos x x sin x ln x ) 2 If a snowball melts so that its surface area decreases at a rate of cm 2 /min, find the rate at which the diameter decreases when the diameter is 0 cm Answer: We know that the surface area of a spherical) snowball is given by A = 4πr 2 and that the diameter is 2r Also, we know that da = and that, at the time of interest, the diameter is 0 cm, meaning that r = 5 Our goal is to determine the rate at which the diameter decreases, which will be 2 dr Differentiating the expression for A yields da = 4π2r)dr = 8πr dr Therefore, at the time of interest, we can plug in the known values for da and r: Therefore, = 8π5) dr so the diameter is decreasing at a rate of 20π dr = 40π, = 40π dr 006 cm/min 5

6 24 A trough is 0 ft long and its ends have the shapes of isosceles triangles that are 3 ft across at the top and have a height of ft If the trough is being filled with water at a rate of 2 ft 3 /min, how fast is the water level rising when the water is 6 inches deep? Answer: Let h denote the height of the water in the trough in feet), and let V denote the volume of the water in the trough Suppose w is the wih of the top of the water Since the trough is three times as wide as it is tall, w must be equal to 3h Hence, the volume of water in the trough is given by V = 2 wh 0 = 2 3h)h 0 = 5h2 We know that water is being added at a rate of 2 ft 3 /min, so interest, h = 05 We want to determine dh, so we differentiate the expression for V : Thus, = 52h)dh = 30hdh dh = 30h At the time of interest we can plug in the known values for and h: dh = ) = 2 5 = 4 5 = 2 Also, at the time of So, when the water is 6 inches deep, the water level is rising at a rate of 4 5 = 08 ft/min 32 When air expands adiabatically without gaining or losing heat), its pressure P and volume V are related by the equation P V 4 = C, where C is a constant Suppose that at a certain instant the volume is 400 cm 3 and the pressure is 80 kpa and is decreasing at a rate of 0 kpa/min At what rate is the volume increasing at this instant? Answer: We know that temperature and pressure are related by the equation We also know that, at a certain instant, P V 4 = C V = 400, P = 80, dp = 0 Our goal is to determine, so we differentiate the equation relating P and V : Solving for, we see that dp V P 4)V = 0 = dp V 4 dp 4P V 04 = 6 V 4P

7 Therefore, at the instant in question, = = = , so the volume is increasing at 357 cm 3 /min 7

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