Math 221 – Exam II Tuesday Mar 23 – 5:30-7:00 PMAnswersI. (25 points.) Finddydx. Note: The book sometimes writes Dxy fordydx.(a) y = (x2− x + 1)−7Answer: Let u = x2− x + 1. Then y = (x2− x + 1)−7= u−7sodydx=ddu(u−7) ·dudx= (−7u−8) · (2x − 1)= −7(x2− x + 1)−8· (2x − 1)(b) y = sin4(x2+ 3x)Answer: Let u = x2+ 3x, then sin4(x2+ 3x) = sin4(u). Let v = sin(u), thensin4(u) = v4. It follows thatdydx=d(v4)dv·dvdu·dudx= (4v3) · (cos(u)) · (2x + 3)= (4 sin3(u))·(cos(u))·(2x+ 3)= 4 sin3(x2+ 3x) · cos(x2+ 3x) · (2x + 3)(c) y =µsin(x)cos(2x)¶3Answer: Let u =sin(x)cos(2x), thenµsin(x)cos(2x)¶3= u3. It follows thatdydx=d(u3)du·dudx= (3u2) ·µcos(x) cos(2x) − sin(x)(−2 sin(2x))cos2(2x)¶= 3µsin(x)cos(2x)¶2·µcos(x) cos(2x) + 2 sin(x) sin(2x)cos2(2x)¶1II. (20 points.) Writedydxin terms of x and y.(a) 4x3+ 7xy2= 2y3Answer:ddx[4x3+ 7xy2] =ddx[2y3]12x2+ (7y2+ 7x(2ydydx)) = 6y2dydx12x2+ 7y2+ 14xydydx= 6y2dydx12x2+ 7y2= (6y2− 14xy)dydx12x2+ 7y26y2− 14xy=dydx(b) xy + sin(xy) = 1Answer:ddx[xy + sin(xy)] =ddx[1](y + xdydx) + (cos(xy)(y + xdydx)) = 0(x + x cos(xy))dydx= −(y + y cos(xy))dydx= −y(1 + cos(xy))x(1 + cos(xy))dydx= −yxIII. (25 points.) Two particles move along a coordinate line. At the endof t seconds their directed distances from the origin, in feet, are given bys1= 4t − 3t2and s2= t2− 2t, respectively.(a) When do they have the same velocity?Answer: The velocities are given respectively by v1= 4 − 6t and v2= 2t − 2.Then v1= v2when 4 − 6t = 2t − 2, hence 8t = 6, so t=3/4 .(b) When do they have the same speed?Answer: Based on the previous problem, we know that the particles have thesame velocity at t = 3/4, so this is one answer. The other case where the speeds2can be equal is if the velocities have the opposite sign but the same magnitude (i.e.v1= −v2). Then 4 − 6t = 2 − 2t, so 2 = 4t and t = 1/2. The answers are thust = 1/2 and t = 3/4 .(c) When do they have the same position?Answer: The positions are the same when s1= s2. Then 4t − 3t2= t2− 2t, so4t2− 6t = 0 . Factoring, we have 2t(2t − 3) = 0, so t = 0 and t = 3/2 are thesolutions.IV. (25 points.) (a) Find the equation of the tangent line to the graph ofthe curvex2y2+ 4xy = 12yat the point (2, 1).Answer:ddx[x2y2+ 4xy] =ddx[12y](2xy2+ x2(2ydydx) + (4y + 4xdydx)) = 12dydx2xy2+ 4y = 12dydx− x2(2ydydx) − 4xdydx2xy2+ 4y = (12 − 4x − 2x2y)dydx2xy2+ 4y12 − 4x − 2x2y=dydxAt the point (2, 1), we havedydx=2(2)(1)2+ 4(1)12 − 4(2) − 2(2)2(1)=4 + 412 − 8 − 8=8−4= −2.Now we know the slope, so the equation of the tangent line is y − 1 = −2(x − 2),or y = −2x + 5 .(b) The equation in part (a) of this problem defines a differentiable functionf(x) implicitly for x near 2, i.e.x2f(x)2+ 4xf(x) = 12f(x).This function satisfies f(2) = 1. Find the linear function L(x) which bestapproximates f(x) for x near 2.Answer: This is the linear approximation L(x) = f(2) + f0(2)(x − 2). Since thegraph of the linear approximation is the tangent line, the answer is the same asthe answer to part (a), namely L(x) = −2x + 53V. (20 points.) Each edge of a variable cube is increasing at a rate of 1 inchper second. How fast is the volume of the cube increasing when an edge is 3inches long?Answer: The formula for the volume of a cube is V = l3, where l is the edgelength. We want to calculatedVdtwhen l = 3. Using the Chain Rule, we havedVdt=dVdl·dldt= 3l2·dldt.We are given thatdldt= 1 inch/second, and are interested in the calculation forl = 3. Substituting values, we havedVdt= 3(3in)2· 1in/s = 27 in3/sVI. (25 points.) Determine where the graph of the functiong(x) = 4x3− 3x2− 6x + 12is increasing, decreasing, concave up and concave down. Then sketch thegraph.Answer: To determine intervals of monotonicity we use the first derivativeg0(x) = 12x2− 6x − 6 = 6(2x2− x − 1) = 6(2x + 1)(x − 1).Then g0(x) = 0 when x = −1/2, 1. The intervals having these endpoints are(−∞, −1/2), (−1/2, 1), (1, ∞).One each of these intervals the derivative g0(x) cannot change sign: else therewould be more zeros. To find the sign of g0(x) on such an interval we evaluate atone point; it does not matter which.• g0(−1) > 0 so g0(x) > 0 on (−∞, −1/2) so g is increasing on (−∞, −1/2);• g0(0) < 0 so g0(x) > 0 on (−1/2, 1) so g is decreasing on (−1/2, 1);• g0(2) > 0 so g0(x) > 0 on (1, ∞) so g is increasing on (1, ∞).4To determine intervals of concavity, we use the second derivativeg00(x) = 24x − 6 = 6(4x − 1).Then g00(x) = 0 when x = 1/4. Thus• g00(x) > 0 for x > 1/4, so g is concave up on (1/4, ∞) , and• g00(x) < 0 for x < 1/4, so g is concave down on (−∞, 1/4)The limits at ±∞ are limx→−∞g(x) = −∞ and limx→∞g(x) = ∞. Here is the graph.-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-10-505101520x4*x^3-3*x^2-6*x+125VII. (20 points.) Use the graph of f(x) to determine the properties of f atthe points given. For each point, write “Yes” or “No” in the box to indicatewhether the point x = c has the given property. (Write “Yes” in the columnheaded f0(c) DNE if the derivative of f (x) does not exist at x = c, i.e. if c isa singular point of f(x).)-1 0 1 2 3-10-505Point f0(c) > 0 f00(c) > 0 f0(c) DNE f0(c) = 0 Inflection Pt. Local Maxc = −1 Yes No No No No Noc = 0 No No No Yes No Yesc = 1 Yes Yes No No No Noc = 2 No No Yes No No Yesc = 3 No No No Yes Yes No6VIII. (20 points.) A function f(x) is concave up on an interval a ≤ x ≤ band c is a point between a and b.(a) Write the equation for the secant line through the points (a, f (a)) and(b, f (b)) and then write an inequality which expresses the fact that the graphof y = f(x) lies below this line for a ≤ x ≤ b,Answer: The point (x, y ) lies on this line if and only if the slope from (a, f(a))to (x, y) is the same as the slope from (a, f(a)) to (b, f(b)), i.e.y − f(a)x − a=f(b) − f(a)b − aso the secant line is y = f(a) +µf(b) − f(a)b − a¶(x − a). Since the graph lies belowthe secant line, the inequalityf(x) ≤ f(a) +µf(b) − f(a)b − a¶(x − a)holds for a ≤ x ≤ b.(b) Write the equation for the tangent line at the point (c, f(c)) and thenwrite an inequality which expresses the fact that the graph of y = f(x) liesabove this line for a ≤ x ≤ b,Answer: The point (x, y) lies on this line if and only if the slope from (c, f(c))to (x, y) is the same as f0(c), i.e.y − f(c)x − c= f0(c)so the tangent line is y = f(c) + f0(c)(x − …