Questions with Solutions - Midterm II Exam | MATH 124, Exams of Analytical Geometry and Calculus

Download Questions with Solutions - Midterm II Exam | MATH 124 and more Exams Analytical Geometry and Calculus in PDF only on Docsity! Name Section MIDTERM II Math 124, Section C May 16, 2006 Problem Total Points Score 1 12 2 12 3 12 4 12 5 12 Total 60 6(Bonus) 2 - No book, notes or graphing calculators are allowed. You may use a scientific calculator. - Show all your work to get full credit. - Read instructions for each problem CAREFULLY. - Check your work! 1 1. (12pts) Find the following derivatives. You do not have to simplify. (a) [4pts] f(x) = tan( x 4 4√17x3+1) Answer. f(x)′ = sec2( x 4 4√17x3+1) · 4x3 4 √ 17x3+1− 51x6 4(17x3+1)3/4√ 17x3+1 Equivalently, f(x)′ = sec2( x 4 4√17x3+1) · ( 4x 3 4√17x3+1 − 51x 6 4(17x3+1)5/4 ) (b) [4pts] f(x) = xcos x Answer. f(x)′ == xcos x(sin x ln x + cos x x ) (c) [4pts] y = arccos(t) Answer. d 2y dt2 = − t√ (1−t2)3 2. (12pts) A (spherical) snowball is rolling down a snow covered hill in such a way that its radius is changing at the rate of 3 cm/min. Determine the rate of change of the volume of the snowball when the radius is 4 cm. Include units. (You may use the formula for the volume of a sphere of radius r: V = 4 3 πr3.) Answer. Differentiating the formula V = 4 3 πr3 we get dV dt = 4πr2 dr dt Plug in dr dt = 3, r = 4. Answer: dV dt = 192π cm3/min. 3. (12pts) Consider the curve given by the equation y2 = (x + 1)(x2 − 1/2) Use implicit differentiation to answer the following questions: [8pts] Find all values of x such that the tangent line to the curve at the point (x, y) is horizontal. How many such points are on the curve? Note: you do not have to compute the values of y. Answer. Differentiating implicitly, we get dy dx = 3x2 + 2x− 1 2 2y The tangent line is horizontal when 3x2 + 2x − 1 2 = 0. Solving for x using the quadratic formula, we get two solutions: x1 = −2 +√10 6 ≈ 0.194, x2 = −2− √ 10 6 ≈ −0.86 Plugging in back to the curve, we get (x1 + 1)(x 2 1 − 1/2) = −0.55 < 0. So, there is no y corresponding to x1, i.e. x1 does not give a point on the curve. For x2 = −0.86, we get (x2 + 1)(x22 − 1/2) = 0.033. This is positive, and we get TWO y-coordinates corresponding to x2. The points where the tangent line is horizontal are (−0.86,√0.033) and (−0.86,−√0.033). 2