University:
Montgomery College
Course:
MATH 181: Calculus I
Academic year:

2019
Views:

227

Pages:

8
Author:

fynegafn8ps

Name: Key (10 points) (1) Let y = f(x) be implicitly defined as x ^ (cos y) = y ^ (sin x) Compute y' in terms of X, and y. (Hint: Use Natural Logarithms) Cosy * ln(x) = Sin * x' * ln(y) - sin y * y' * ln(x) + cos y * 1/x = cos x * ln(y) + sin x * 1/y * y' y' * [- sin y * ln(x) - (sin x)/y] = cos x * ln(y) - (cos y)/x (y^ prime =& O 0 xln y- Giy x - delta n yln x- Gin x y )= (x * cos xbnx)/(- (y ^ (cos x) - sin x)/2) = (y(x * sin y - sin y))/(x(- y * sin x - sin x)) (2) A man 6 ft tall walks at a rate of 1.5 ft/sec away from a lamppost that is 20 feet high. At what rate is the length of his shadow changing when he is 50 feet away from the lamppost. (10 points) 6/x = 20 2 x+y (dy)/(dt) =1.5ft/sec d/dt (x) = 6 d/dt (y) = 1.5ft / i * ec 6x + 6y = 20x 6y = 14x H lom.. X = 6/14 * Y dx dt = b 14 ( dy dt ) Rightarrow (dx)/(dt) = 6/14 * (1.5) = 9/14 =0.643 ft sec 1 (3) A particle starts at the origin and moves along the parabola y= 2x ^ 2 : such that its distance from the origin increases at 7 meters per second. How fast is the particle's x-coordinate changing as it passes through the point (-2,8)? (10 points) ( 0 , ) D= sqrt((x + 0) ^ 2 + (x ^ 2 + 0) ^ 2) Rightarrow D= sqrt x^ 2 +4x^ 4 8 D = sqrt(8 ^ 2 + (- 2) ^ 2) D ^ 2 = x ^ 2 + 4x ^ 4 -2 2D * (dD)/(dt) =2 x dx dt +Ibx^ eta dx dt Rightarrow2 D dD dt = dx dt [2 x + i6x^ 3 ] (dx)/(dt) = (2D * (dD)/(dt))/(2x + lb * x ^ 2) = 2( sqrt 68 )(7j 2(-2)+lb(-z)^ 2 = approx - 0.875m /sec River (4) A rancher intends to fence off three sides of a rectangular region along a straight stretch of river (no fence along the river). The enclosed area is to be 3600 square meters. X a) Find an equation for the total length of fencing material, F(x) where x represents the lengths of the sides that are perpendicular to the river, (5 points) 2x + y = F(x) F(x) = 2x + 3600/x and (metens) xy = 36oo = iy = 3600/x b) What is the least amount of fencing material that is needed? (5 points) F(x) = 2x + 3b * 0x ^ - 1 F' * (x) = i - 3600/(x ^ 2) = 0 2x² 2x² = 3600X x² = 1800 2 then; F(42.43)= 169.71 meters is needed for fencing material X=42.43 meters 2 (5) Find the most general antiderivative of the following functions. f(x) = 12x ^ (3/4) + 6x ^ (1/3) - 5 (3 points) matrix (x)=& 1 4 +1 \\ 12 4+1 x^ 1 4 +1 + 6 * x/3 + 5x + C = 48/7 * x ^ (7/4) + 9/2 * x ^ (4/3) - 5x + C b) f(x) = (5 - 2/(sqrt(x)))(5 + 2/(sqrt(x))) = 25 - 4/x (2 points) F(x) = 25x - ln|x| + c c) f(t) = (4t ^ 8 + 2t ^ 2)/(sqrt(t)) * (sqrt(t))/2 =2t + ^ 8 t^ 2 (3 points) F(t) = (2t)/9 ^ 9 + (t ^ 3)/3 + C F(x)= pi^ 8) f(x)= pi^ * +ex \\ F(x) = (Q * x ^ 2)/2 + C . (2 points) e) f(t) = e ^ t + sec(t) * tan t + 2 (5 points) F(t) = rho ^ t + sec(t) + 2t + c 1) f(t) = 10/(t ^ 2 + 1) + 9/(sqrt(1 - t ^ 2)) F(t) = 10arctan(t) + 9arcsin(t) + c (5 points)