Math 226 Quiz #5 Sample Problems 1. A stone dropped into a still pond sends out a circular ripple whose radius increases at a constant rate of 3 ft/s. At time t = 0, the radius is 0 feet. How rapidly is the area enclosed by the ripple increasing at time t = 10 seconds? dr , since it’s the rate at which the radius is increasing. We want to find the rate at dt dA , so we want a relationship between A and r. We can use the fact which the area is increasing, dt 2 that A = πr , since the ripple is circular. Thus, we know 3 ft/s is our A = dA dt πr2 = 2πr dr dt So now the last piece of the puzzle will be finding r. Now since we are concerned with what happens at the end of 10 seconds, we know that the radius will be 3 ft/s ·10 s = 30 ft at t = 10 s. So, we have dA dt = 2π(30)(3) dA dt = 180π ft2 /s ≈ 565.5 ft2 /s 2. A spherical balloon is to be deflated so that its radius decreases at a constant rate of 15 cm/min. At what rate is air being removed when the radius is 9 cm? dr dV = −15 cm/min, and we want to find when r = 9 cm. The relationship we dt dt have between r and V is: Here, we have = 4 dV dt = 4 dV dt = 4πr2 = 4π (9) (−15) = −4860π cm3 /min V 3 3 πr3 π 3r 2 dr ! dt dr dt 2 3. A 17 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 is the top of the ladder moving down the wall when it is 8 ft above the ground? dy , when y = 8 ft. Thus we need a relationship dt between x and y. Since there is a right triangle, we can use the pythagorean theorem, and then differentiate with respect to t. So, Using the picture above, we are trying to find x2 + y2 2x dy dx + 2y dt dt dy dt = 172 = 0 = −x dx y dt Now, since we have that y = 8 ft, we can solve for x. Using the pythagorean theorem, we have √ x = 172 − 82 = 15. So, dy −15 75 = ·5=− ft/s dt 8 8 2 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 top of the conical tank has a radius of 12 ft, and the height of the tank is 28 ft, find the rate at which water is being poured into the tank, in terms of the height h. dV 1 , and the relationship we have between V, r, h, is that V = πr2 h. In dt 3 dh this instance, we want only r or h. Since we are given that = 3 ft/min, we would rather use h. dt Thus, using the similar triangle argument, for the ratio of the radius of the height of the water at any time is r 12 3 3 = = ⇒ r= h h 28 7 7 We are trying to find Now we can substitute this value for r and differentiate with respect to t. Sew, V = = dV dt = = dV dt = 1 3 π 3π 49 9π 49 3 7 2 h h h3 h2 9π 49 27π 49 dh dt h2 (3) h2 ft3 /min (Notice that our answer is in terms of h, as requested in the question) 3 5. Find the local linear approximation of f at x0 = 1. a) f (x) = 1 2+x Noting that f 0 (x) = −1 , we have (2 + x)2 f (x) ≈ = = = f (x0 ) + f 0 (x0 ) (x − x0 ) 1 2 + x0 1 2+1 1 − 3 1 9 − + 1 (2 + x 0 )2 (x − x0 ) −1 (x − 1) (2 + 1)2 (x − 1) b) f (x) = (x + 4)3 Noting that f 0 (x) = 3(x + 4)2 , we have f (x) ≈ (x0 + 4)3 + 3 (x0 + 4)2 (x − x0 ) = (1 + 4)3 + 3(1 + 4)2 (x − 1) = 125 + 75(x − 1) c) f (x) = tan−1 (x) Here, f 0 (x) = 1 1 + x2 . Hence, f (x) ≈ = = tan−1 (x0 ) + tan−1 (1) + π 4 + 1 2 1 1 + x20 1 1 + 12 (x − 1) 4 (x − x0 ) (x − 1) 6. Use an appropriate local linear approximation to estimate the value of the given quantity. √ a) 65 √ Here, we should use the function f (x) = x, and we should use x0 = 64, since we’ll get a nice number there. So, note that f 0 (x) = 1 2 √ . Using the formula for approximating f (x) near x0 = 64, x we have √ f (x) ≈ 64 + = 8+ 1 16 ! 1 √ 2 64 (x − 64) (x − 64) Now, since 65 is close to 64, we can conclude that f (65) ≈ 8 + 1 16 (65 − 64) = 129 16 b) 1.983 This time, we are going to use the function f (x) = x3 , where x0 = 2. Note that f 0 (x) = 3x2 , so we have Now, we can conclude that f (x) ≈ (2)3 + 3(2)2 (x − 2) f (x) ≈ 8 + 12(x − 2) f (1.98) ≈ 8 + 12(1.98 − 2) = 7.76 c) ln(1.01) This time, our function will be f (x) = ln x, and we will use x0 = 1. f 0 (x) = f (x) ≈ ln(1) + f (x) ≈ x−1 1 1 (x − 1) Therefore f (1.01) ≈ 1.01 − 1 = .01. 7. Find dy dx a) y = ex (x3 + 5) dy = ex x3 + 5 + 3x2 ex dx 5 1 x . Thus, √ b) y = sin−1 ( x) dy = q dx 1 ! 1 1 √ 2 2 √x = p 2 x(1 − x) 1− x c) y = ln (cos (ex )) dy 1 (− sin (ex )) (ex ) = −ex tan (ex ) = dx cos (ex ) d) y = p cos−1 (x2 )    dy 1 −1 2 −1/2  x 1  (2x) = − p  = cos x − p  2 − 1 dx 2 cos (x2 ) (1 − x4 ) 1 − (x2 ) e) y = etan(x) (1 + x2 ) dy = etan(x) sec2 (x) · 1 + x2 + 2x etan(x) = etan(x) 2x + sec2 (x) + x2 sec2 (x) dx f) y = 1 π tan−1 (2x) ! dy 1 1 2 (2) = = dx π 1 + (2x)2 π (1 + 4x2 ) 8. Find a point on the graph of y = e3x at which the tangent line passes through the origin. Here, we have that dy = 3e3x . So we want the (x, y) where y − 0 = 3e3x (x − 0), or y = 3xe3x . dx Now, we have y = e3x and y = 3xe3x , telling us that 3xe3x = e3x 3x = 0 e (3x − 1) = 0 = 1 3xe 3x −e 3x x 3 Notice that e3x is never equal to zero, so the point is when x = 1/3, which means y = e3(1/3) = e. So, 1 (x, y) = , e 3 6