Math 226 Quiz IV Review

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1 Math 226 Quiz IV Review. A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 3 ft/s. How rapidly is the area enclosed by the ripple increasing at the end of 0 s? 3 ft/s is our dr, since it s the rate at which the radius is increasing. We want to find the rate at which the area is increasing, da, so we want a relationship between A and r. We can use the fact that A = πr 2, since the ripple is circular. Thus, we know A = πr 2 da = 2πr dr So now the last piece of the puzzle will be finding r. Now since we are concerned with what happens at the end of 0 seconds, we know that the radius will be 3 ft/s 0 s = 30 ft at t = 0 s. So, we have da = 2π(30)(3) da = 80π ft 2 /s 2. A spherical balloon is to be deflated so that its radius decreases at a constant rate of 5 cm/min. At what rate must air be removed when the radius is 9 cm? Here, we have dr = 5 cm/min, and we want to find relationship we have between r and V is: V = 4 3 πr3 when r = 9 cm. The = 4 ( 3 π 3r 2 dr ) = 4πr 2 dr = 4π (9) 2 ( 5) = 4860π cm 3 /min

2 2 3. A 7 ft ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of 5 ft/s, how fast will the top of the ladder be moving down the wall when it is 8 ft above the ground? Using the picture above, we are trying to find, when y = 8 ft. Since the ladder top is moving down the wall, toward the ground, it should be a negative number. Thus we need a relationship between x and y. Since there is a right triangle, we can use the pythagorean theorem, and then differentiate with respect to t. So, 2x dx x 2 + y 2 = y = 0 = x dx y Now, since we have that y = 8 ft, we can solve for x. Using the pythagorean theorem, we have x = = 5. So, = = 75 8 ft/s

3 4. Water is being poured into a conical water tank so that the height of the water is increasing at a constant rate of 3 ft/min. If the base of the conical tank has a radius of 2 ft, and the height of the tank is 28 ft, find the rate at which water is being poured into the tank, with respect to the height of the water, h. 3 Notice I accidentally drew the cone upside-down...also, I will give h on the quiz. We are trying to find, and the relationship we have between V, r, h, is that V = 3 πr2 h. In this instance, we want only r or h. Since we are given that dh = 3 ft/min, we would rather use h. Thus, using the similar triangle argument, for the ratio of the radius of the height of the water at any time is r h = 2 28 = 3 7 r = 3 7 h Now we can substitute this value for r and differentiate with respect to t. Sew, V = 3 π ( 3 7 h ) 2 h = 3π 49 h3 = 9π dh h2 49 = 9π ( h 2 ) (3) 49 = 27π 49 h2 And since we have no specific height h, we must leave our answer in terms of h.

4 4 5. Find the local linear approximation of f at x 0 =. a) f(x) = 2 + x We know that f(x) f(x 0 ) + f (x 0 ) (x x 0 ). Noting that f (x) = f(x) (x ) (2 + ) 2 f(x) 3 (x ) 9, we have (2 + x) 2 b) f(x) = (4 + x) 2 Here, we have that f (x) = 2(4 + x) () = 2(4 + x). Using the same formula where near x 0, we know f(x) f(x 0 ) + f (x 0 )(x x 0 ), we have f(x) (4 + ) 2 + 2(4 + )(x ) f(x) (x ) 6. Use an appropriate local linear approximation to estimate the value of the given quantity, and then find the error in your approximation. a) 65 Here, we should use the function f(x) = x, and we should use x 0 = 64, since we ll get a nice number there. So, note that f (x) = 2. Using the formula for approximating x f(x) near x 0 = 64, we have f(x) ( ) (x 64) 64 f(x) 8 + (x 64) 6 Now, since 65 is close to 64, we can conclude that f(65) (65 64) = 6 6 Now, to find our error, we have that E(x) = , which tells us we got a pretty good estimation!

5 5 b) (.98) 3 This time, we are going to use the function f(x) = x 3, where x 0 = 2. Note that f (x) = 3x 2, so we have f(x) (2) 3 + 3(2) 2 (x 2) f(x) 8 + 2(x 2) Now, we can conclude that f(.98) 8 + 2(.98 2) = 7.76 For the error, we have E(x) = (.98) = c) ln (.0) This time, our function will be f(x) = ln x, and we will use x 0 =. f (x) = x. Thus, f(x) ln() + (x ) f(x) x Therefore f(.0).0 =.0. Our error is E(x) = ln(.0).0 = Use the differential to approximate y when x changes as indicated. a) y = x x 2 ; from x = 2 to x =.96 + To find, note that dx = (x2 + ) 2x(x) (x 2 + ) 2 = x 2 (x 2 + ) 2 dx Here, we have that dx = x =.96 2 =.04. So, y = (2)2 ((2) 2 (.04) = ) 2

6 6 b) y = tan x; from x = 0 to x = 0.2 First, notice that dx = sec2 x = sec 2 xdx Now since dx = x = = 0.2. Sough, y = sec 2 (0)(0.2) = () 2 (0.2) = Find the its. If you use L Hopital s rule, make sure you note this, along with the type of indeterminate form. xe x a) x 0 e x Notice that we have an indeterminate form of 0, so we can use L Hopital s rule. Therefore, we can look 0 at x 0 xe x e x + xe x e x = x 0 e x = + 0 = Using L H b) tan x ln x x 0 + First, I want to note that I won t put a problem this ugly on the exam... Here, our indeterminate form is 0 ( ), thus we will have to do some algebraic manipulation before we use L Hopital s rule. We will rewrite tan x ln x as ln x / tan x = ln x (tan x), or as tan x. Either way, we will still end up with the same answer. So, we have (on next / ln x page)

7 7 tan x tan x ln x = x 0 + x 0 + (ln x) Using L H = sec 2 x (/ ln x) ( (ln x) 2 )(tan x) x 0 + (ln x) 2 = ( sec 2 x x 0 + ln x + tan x ) (ln x) 2 (ln x) 2 = ( sec 2 x x 0 + (ln x) 3 + tan x ) (ln x) 4 = = ( ) c) x2 x e x This time our indeterminate form is. So, first doing some algebraic manipulation, we have e x x2 = e x x 2 e x Now, we may use L Hopital s rule as follows: ( ) x2 x e x e x x 2 = x e x ( e x x 2 + 2xe x) Using L H = x e x (x 2 2x) = x = ( x(x 2)) x = ( 2) =

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