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ISU MATH 165 - Math Problem

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Michael W.SI Math 1651. Find the derivative of f(x) =ax + bcx + dwhere a, b, c, d are constant realnumbers.Using quotient rule, f0(x) =a(cx + d) − c(ax + b)(cx + d)2=ad − bc(cx + d)22. Find the points on the graph of y = x3/2− x1/2at which the tangentline is parallel to the line y = 1 − 2x.We need the slope of any tangent line, which is given by the derivativedydx=32x1/2−12x−1/2=3x − 12√xTwo lines are parallel if and only if they have the same slope. Sincethe slope of the line y = 1 − 2x is -2, we are left to solvedydx=3x − 12√x= −2We rewrite to get 3x + 4√x − 1 = 0 which is a quadratic equationin√x. Therefore, using the quadratic formula, we must have√x =−2 ±√73. Since√x must be nonnegative, we are left with the solution√x =−2 +√73. Equivalently, x = −2 +√73!2. So there is onlyone point on the graph at which the tangent line is parallel to the liney = 1 − 2x.13. Find the x−coordinates of all points on the graph f (x) = x + 2 cos(x)in the interval [0, 2π] at which the tangent line is horizontal.The tangent line is horizontal when f0(x) = 0. So f0(x) = 1 −2 sin(x).When 0 = 1 − 2 sin(x),12= sin(x). By the unit circle, we know thatsin(x) =12when x =π6or x =5π6.4. Find the derivative of f (x) =x2−2x33.By the chain rule, we find that f0(x) = 3x2−2x32·2x +6x45. Finddydxif xy3+ x2y2+ 3x2− 6 = 1.By implicit differentiation,y3+ 3xy2dydx+2xy2+ 2x2ydydx+ 6x = 0.Thendydx= −6x + y3+ 2xy23xy2+ 2x2y6. Finddydxif x2tan(y) + y10sec(x) = 2x.So using implicit differentiation,2x tan(y) + x2sec2(y)dydx+ 10y9dydxsec(x) + y10sec(x) tan(x) = 2Some algebraic operations then give usdydx=2 − y10sec(x) tan(x) − 2x tan(y)x2sec2(y) + 10y9sec(x)27. Air is being pumped into a spherical balloon at a rate of 5 cm3/min.Determine the rate at which the radius of the balloon is increasingwhen the diameter of the balloon is 20 cm.Firstly, we know that the volume of the balloon is increasing at a rateof 5 cm3/min. This can be represented asdVdt= 5. We want to findthe rate at which the radius of the balloon is increasing, so we wantdrdt=? This is when r =12d = 10cm. Recall that the formula for thevolume of a sphere is V =43πr3. So if we differentiate this implicitlywe getdVdt=43π3r2drdt. By substituting in things that we know, wehave 5 =43π3(10)2drdt⇒drdt=180πcm/min.38. A 15 foot ladder is resting against the wall. The bottom is initially 10feet away from the wall and is being pushed towards the wall at a rateof14ft/sec. How fast is the top of the ladder moving up the wall 12seconds after we start pushing?We have defined the distance of the bottom of the latter from thewall to be x and the distance of the top of the ladder from the floorto be y. These are changing with time, so we should write x(t) andy(t), anddxdt,dydt. We know that the rate at which the bottom of theladder is moving towards the wall is given bydxdt= −14. We wantto find the rate at which the top of the ladder is moving away fromthe floor, ordydt. We need to find a relationship between x and y.Using the Pythagorean Theorem, x2+ y2= 152= 225. All that weneed to do at this point is to differentiate both sides with respect tot, remembering that x and y are really functions of t and so well needto do implicit differentiation. Doing this gives an equation that showsthe relationship between the derivatives. So 2xdxdt+ 2ydydt= 0 (*).Determining x and y is actually fairly simple. We know that initiallyand the end is being pushed in towards the wall at a rate of14ft/secand that we are interested in what has happened after 12 seconds. Weknow that distance=rate×time=14(12) = 3. So the end of the ladderhas been pushed 3 feet and so after 12 seconds we must take x = 7.To find y after 12 seconds all we need to do is reuse the PythagoreanTheorem with the values of x we just found, so y =√225 − x2=√176.Plugging this into (*) and solving fordydtis 2(7)(−14+ 2√176dydt= 0 ⇒dydt=74√176≈


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ISU MATH 165 - Math Problem

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