Section 7.4 Solutions

AP Calculus AB Section 7.4-Worksheet 2 d/dx (y) = (4y)/y int y ^ 2 * y = integrate 4x dx y * 2/2 + c_{1} = 2x ^ 2 + c_{2} lceil7 2 2 = 2x ^ 2 +c rceil y ^ 2 = 4x ^ 2 + c d/dx (y) = (g * x ^ 2)/(2y) 2y * dy = 9x ^ 2 * dx 2 + c_{1} = 3x ^ 3 + c_{0} y ^ 2 = 3x ^ 3 + c varphi^ 2 = exists (varphi) ^ 3 + c k = - 3 + c m = c 5.) d/dx (y) = 10x ^ 4 * y ^ 2 (dy)/(dx) = ( Box x^ prime prime y^ -2 y ^ a * dy = 10x ^ a * dx (y ^ - 1)/- 1 + c_{1} = 2x ^ 5 + c_{2} - 1/y = 2x ^ 5 + c (3,0) 0 ^ 2 = 4 * (x) ^ 2 + c 0 = 36 + c C - 36 = c aligned y^ 2 =4x^ 2 -36 \\ y = plus/minus sqrt(4x ^ 2 - 36) = plus/minus 2 * sqrt(x ^ 2 + 9) \\ c-1.40 aligned y ^ 2 = 3x ^ 3 + 19 y = plus/minus sqrt(3x ^ 3 + 19) (-1,1) 1/y = - 2x ^ 5 + c 1 = - 2 * (- 1) ^ 5 + c I = 2 + C 1/y = - 2x ^ 5 - 1 y = 1/(- 2x ^ 7 - 1) 6.) The function / is differentiable for all real numbers. The point (3, 1/4) on the graph of y = f(x) and the slope at each point (x, y) on the graph is given by d/dx (y) = y ^ 2 * (6 - 2x) Name: Key Date: Block: 2.) d/dx (y) = 2pi * y ^ 2 (3-1) d/dx (y) = (2x)/(y ^ - 2) aligned int y^ -2 dy= int2xdx\\ (y ^ - 1)/- 1 + c_{1} =x^ 2 +c. aligned - i/gamma = x ^ 2 + c 1/y = - x ^ 2 + 6 1 y = -x^ 2 +8 1/(- x ^ 2 + 8) 1 = - (x) + c y = - 1 + c 8 = c (1,1) d/dx (y) = sqrt(x/y) d/dx (y) = (sqrt(x))/(sqrt(7)) d/dx (y) = (x ^ (1 / a))/(y ^ (1 / a)) integrate y ^ (1/2) * 2 dy = x ^ (1/2) * dx + y ^ (3/2) = x ^ (3/2) + c I ^ (2/2) = I ^ (3/2) + C i = 1 + c 0 = C y ^ (3/2) = x ^ (3/2) + 0 (y ^ (3 / x)) ^ (2 / y) = (x ^ (2 / x)) ^ (2/3) y = x AP Free-Response Practice: a.) Draw the slope field for the given differential equation. 3 1 2 0 4 4 2 43 d_{7} / d * x D 0 16 2 8 2 24 -2 2 -8 3 -18 b.) Find (d ^ 2 * y)/(d * x ^ z) evaluate it at the point (3, 1/4) (dy)/(4x) = y ^ 2 * (6 - 2x) (d ^ 2 * y)/(d * x ^ 2) = y ^ 2 * (- 2) + (6 - 2x) - 2y * d/dx (y) = - 2y ^ 2 + (12y - 4xy)(6y ^ 2 - 2x * y ^ 2) -2( 1 4 )^ 2 + (12(t) - 4(z)(1/4)) (6(/)^ 2 -2(3)( 1 4 )^ 2 L(1/(l*psi)) - 1/9 - 1/8 Find y = f(x) by solving the differential d/dx (y) = y ^ 2 * (6 - 2x) with the initial condition (3, 1/4) d/dx (y) = (6 - 2x)/(y ^ - 2) y ^ - 2 * 4y = (6 - 2x) * dx y_{1} ^ - 1 + c_{1} = 6x - x ^ 2 + C_{1} - 1/y = 6x - x ^ 2 + c r/(y') - 6x + x ^ 2 + 6 1/(1/eta) = - xi * (3) + (3) ^ 2 + C 1/y = - 6x + x ^ 2 + 13 y = 1/(x ^ 2 - 6x + 1xi) 7.) For the differential equation d/dx (y) = (3 - x)/y 4 = - 18 + 9 + c dot 4 = - 9 + c a.) 13 = C Draw the slope field for the given differential equation 0 ۳/۲ 3 und de 1 1 2 4 -3 2 t 1 1/2 -1- 0 -2 -34 २ 221 -1/2 2 -4 4400 スー 2 1/2 hor Let y = f(x) be the particular solution to the given differential equation for 1 < x < 5 such that the line y = - 2i tangent to the graph of I(x) Find the x-coordinate of the point of tangency, and determine whether / has a local maximumlocal minimum, or neither at this point. Justify your answer. Se + d/dx (y) = 0 d/dx (y) = (3x)/y (3 ) 62 2/1 + 6(1) - (3) ^ 2 + 6 eta = it - eta + c (2 - x)/y = (by)/1 3 - x \approx c x = 3 (y ^ 2)/2 + C_{1} + J_{2} - (x ^ 2)/3 + C_{2} d_{1} * [[y_{11} ^ 2], [z ^ 2]] y^ prime prime = 6x -x^ prime prime +c y = root(- x ^ 22, m) + 6x - 5 ftias & local as indicated by the slope field where it is dec Switching to inc. c.) Let y = g(x) the particular solution to the given differential equation for - z < x < overline B , with initial condition Find g(6) = - 4 y = g(x) d/dx (y) = (3 - x)/y y ^ 2 = 6x - x ^ 2 + 6 (- 4) ^ 22 = 6(6) - (6) ^ 2 + 2 l_{6} = 36 - y_{6} + c y ^ 2 = 6x - x ^ 2 + 16x y = plus/minus sqrt(- x ^ 2 + 6x + 16) integrate y dy = 7 ^ 3 * (3 - x) * dx (y ^ 2)/2 + c_{1} < 3x - (x ^ 2)/2 + c_{2} A