MATH 1080 Test 2 -Version A-SOLUTIONS Fall a. (8 pts) Find the exact length of the curve on the given interval.

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1 MATH 8 Test -Version A-SOLUTIONS Fall 4. Consider the curve defined by y = ln( sec x), x. a. (8 pts) Find the exact length of the curve on the given interval. sec x tan x = = tan x sec x L = + tan x = sec x = sec x = ln sec(x) + tan(x) = ln( + ) ln() L = ln( ) b. ( pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about the x- axis on the given interval. R = y = ln sec x SA = Or ln( sec x) + tan x y = ln( sec x) x = arcsec( e ) y and = ln SA = y + ( e ) y ( e ) y c. ( pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about th- axis on the given interval. R = x SA = x + tan x or SA = ln arcsec + ( e ) y

2 MATH 8 Test -Version A-SOLUTIONS Fall 4. Consider the integral f (x). a. (4 pts) Using the Trapezoidal Rule with n = 4, set up the expression that gives the approximation of Δx = 4 = f x f f x 4 f f (x). + f (.5) + f + f (.5) + f ( ) + f (.5) + f + f (.5) + f ( ) b. (4 pts) Using Simpson s Rule with n = 4, set up the expression that gives the approximation of Δx = 4 = f x f f x f f (x). + 4 f (.5) + f + 4 f (.5) + f ( ) + 4 f (.5) + f + 4 f (.5) + f ( ) b. ( pts) Give an example of an integral, f (x), that is impossible to evaluate exactly. It is impossible to evaluate the integrals a e x or sin( x ) because we can t find an antiderivative. Or It is also impossible to evaluate the integral when the function is determined from collected data and there is no formula for the function.

3 MATH 8 Test -Version A-SOLUTIONS Fall 4 4. Let R be the region bounded by the function f ( x) = [, ]. a. (8 pts) Evaluate the area of R. x and x - axis on the interval This is an improper integral and must be evaluated using limits. A = = lim x t t t = lim sin t x t ( x) ( t ) = lim sin sin A = The improper integral converges to. Thus the area of the region is finite. b. (8 pts) Use the disk/washer method to evaluate the volume of the solid generated when R is revolved about the x - axis. t = lim t x x t = lim t + x + x = t lim( ln ln ) x + + x t = lim ln t + ln + t = t The improper integral is divergent. Therefore, the volume is not finite.

4 MATH 8 Test -Version A-SOLUTIONS Fall Consider the curve defined parametrically by x = t + sint, y = t cost for t. a. (5pts) Find a Cartesian equation of the tangent(s) to the curve at the given point when t = 4. = + sint and = + cost = + sint = + cost m tan = t= 4 = y 4 = x 4 + y = x When t = 4, x, y = 4 +, 4. b. ( pts) Find all points on the curve where the tangent line is horizontal. Provide both t and (x, y). You must justify your answer! This occurs when = and. = when + sint =, t = = t= The point on the curve where the tangent line is horizontal is when t =, ( x, y ) =,. c. ( pts) Find all points on the curve where the tangent line is vertical. Provide both t and (x, y). You must justify your answer! This occurs when = and. = when + cost =, t = = t= The point on the curve where the tangent line is vertical is when t =, x, y = (, +).

5 MATH 8 Test -Version A-SOLUTIONS Fall 4 5. Consider the curve defined parametrically by x = e t, y = te t. a. ( pts) Set up an integral (Do not simplify or evaluate!) that represents the length of the curve on the interval e t e. = et te t = = et te t = e t ( t) and e t = t e t = et L = e e ( e ) t + e t t b. (4 pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about the x- axis on the interval e t e. e SA = te t e t e + e t t c. (4 pts) Set up an integral (Do not simplify or evaluate!) that represents the area of the surface obtained by rotating the curve about th- axis on the interval e t e. e SA = e t e t e + e t t

6 MATH 8 Test -Version A-SOLUTIONS Fall 4 7. Consider the curve defined in polar coordinates by r = cosθ. a. ( pts) Sketch the graph of the curve. θ r b. (8 pts) Find the exact area of the region enclosed by the curve. The curve is traced out for θ = to θ =. Therefore, we have A = ( cos θ) dθ = ( cosθ cos θ) dθ + + cos θ = cos θ dθ + = θ sinθ θ sinθ A =

MATH 8 Test -Version A-SOLUTIONS Fall 4 7 8. Consider the two curves defined in polar coordinates by r = and r = sinθ for θ. a. (5 pts) Set up an integral(s) (Do not simplify or evaluate!

7 MATH 8 Test -Version A-SOLUTIONS Fall Consider the two curves defined in polar coordinates by r = and r = sinθ for θ. a. (5 pts) Set up an integral(s) (Do not simplify or evaluate!) that represents the area of the region that lies inside both curves. Shade this region in the graph below. Where do the curves intersect? sinθ = sinθ = θ =,5 θ =,5 5 A= 4 ( sin θ) dθ+ d θ 4 or A= 4 ( sin θ) dθ+ d θ 5 or A= 4 ( sin θ) dθ+ d ( sin ) d θ+ θ θ 5 4 or A= 4 ( sin θ) dθ+ dθ+ ( sin θ) dθ 5 5 or A= 4 ( sin ) d ( sin ) θ θ d θ θ 4 or A= 4 ( sin ) d ( sin ) θ θ d θ θ or A= 4 d ( sin ) θ d θ θ or A= 4 θ d ( sin θ) dθ b. (5 pts) Set up an integral(s) (Do not simplify or evaluate!) that represents the area of the region that lies inside of the rose but outside the circle. Shade this region in the graph below. 5 5 A = 4 ( sinθ ) dθ ( ) dθ 5 or A = 4 ( sinθ ) dθ 4 or A = 4 ( sinθ ) dθ or A = 4 ( sinθ ) dθ 5 ( sinθ ) dθ + ( ) dθ or any equivalent area from part (a)

8 MATH 8 Test -Version A-SOLUTIONS Fall Consider the conic described by the equation y 4x + y x = 9. a. (8 pts) Rewrite the equation in standard form and state the type of conic equation it represents. ( y + y) 4( x + 4x) = 9 + ( y + y +) 4( x + 4x + 4) = ( y +) 4 x + ( y +) 4 This is a hyperbola. = 4 ( x + ) = b. ( pts) Graph the conic from part a. Find and label the vertices and center of the curve. The center is the point (-,- ). The vertices are the points (-,) and (-,- ). c. ( pts) Using the standard form from part a., provide a parameterization using a trigonometric identity for the top half of the curve with < θ <. Using the identity tan θ += sec θ sec θ tan θ =, one parameterization that gives the the top half of the curve with < θ < is y + = secθ and x + = tanθ Then y = secθ x = tanθ < θ <.

There are some trigonometric identities given on the last page.

There are some trigonometric identities given on the last page. MA 114 Calculus II Fall 2015 Exam 4 December 15, 2015 Name: Section: Last 4 digits of student ID #: No books or notes may be used. Turn off all your electronic devices and do not wear ear-plugs during