MATH 271 Test #4T Solutions

Transcription

1 MATH 7 Test #4T Solutions You do not need to evaluate the integrals in problems () (5); just set them up. () ( points) Consider the curve which is represented b the parametric equations x = + t + t, = + e t, where t is an real number. Find the equation of the line tangent to this curve at the point (, ). Solution: The slope of the tangent line to a curve is d/dx. Here, x and are given in terms of the parameter t, so d dx = d/dt dx/dt = et + t. To find the slope of the line tangent to the specific point (, ), we need to find the value of t which gives x(t) = and (t) =. Hence, t must satisf + t + t = and + e t =. Onl one value of t will work, namel t =. The slope of the tangent line at the e point (, ) is + () =. Now we have a point and the slope, so we can use the point-slope form of the equation of a line: = (x ), or = x +. Grading: +5 points for writing d/dx in terms of t, +5 points for finding t, +5 points for the slope of the tangent line, +5 points for the equation of the tangent line.

2 () The polar coordinate equation r = + cos θ traces out a cardioid, covering it exactl once, when θ ranges from to π. (a) (5 points) Sketch the graph of this cardioid. Solution:. This problem was graded on a 3 5 x basis. (b) (5 points) Set up an integral which finds the perimeter of the cardioid. Solution: The general integral for arc length is b (dx ) + dt t = θ, x = r cos θ, and = r sin θ. Writing x and in terms of θ, we get x = ( + cos θ) cos θ = cos θ + cos θ a and = ( + cos θ) sin θ = sin θ + sin θ cos θ, so dx/dθ = sin θ + cos θ( sin θ) and d/dθ = cos θ + sin θ( sin θ) + cos θ cos θ. ( ) d dt. Here dt You are told that when θ ranges from to π, the cardioid is traced out exactl once, so a = and b = π. Putting all of this together, the perimeter of the cardioid is π ( sin θ + cos θ( sin θ)) + (cos θ + sin θ( sin θ) + cos θ cos θ) dθ. No further simplification is required. Grading for common mistakes: points for writing (A + B) as A + B ; 3 points for finding the surface area (including a factor of π in the integrand); 3 points for using the formula ( ( x dx) + d) dt for arc length/perimeter; 5 points for the integral + ( sin θ) dθ.

3 (3) ( points) Set up an integral to find the surface area of the surface ou get b rotating the curve = + 4 x ( x ) around the x-axis. Solution: The integral for the surface area of a curve rotated around the x-axis is b a π + ( ) dx (when the curve is given as a function of x). Here, a and b are the upper and lower limits on x, respectivel, and so the surface area is which simplified is π( + 4 x ) = (4 x ) / ( x), + π( + 4 x ) ( ) (4 x ) / ( x) dx, + x dx. (You did not need to 4 x simplif the integral, however.) Grading for common mistakes: 3 points for the integral of + dx. (4) ( points) Set up integrals to find the centroid of the region bounded b the curves = 3x and = x. Solution: To find the centroid of this region, ou need to find three integrals: the area A = T B dx, the x-coordinate of the centroid, which is x = x( T A B ) dx, and the -coordinate of the centroid, which is ȳ = ( A T B ) dx, where T is the curve on top, and B is the curve on the bottom. 3

4 To determine which function is the top and which one is the bottom, and what the limits on x are, the region needs to be found. It turns out that 3x = x (i.e., the curves cross) at x = and x = 3, and 3x > x for all values of x in between. Hence T = 3x and B = x. Thus A = x = A T B dx = 3 x( T B ) dx = A ( ) 3x x dx 3 ( ) x 3x x dx ȳ = ( A T B) 3 ( ) dx = 3x A x dx. Grading: +6 points for the area formula and integral, +7 points for the formula for x; +7 points for ȳ. Grading for common mistakes: 5 points for not including limits; points for ȳ = ( T B ) dx. (5) ( points) Set up an integral to find the area of the region which is under the curve parameterized b the equations x = + s 3 = 4 + 3s s 4 (where s is a real number) and above the x-axis. Note that = onl when s = ±. Solution: Setting up the integral for area when the region is described b parametric equations can be trick; the hint was intended to let ou know that nothing weird was going on. For instance, the curve is above the x axis when s is between and. (Also, the function for x is strictl increasing, so the curve never x at the right crosses itself.) This lets ou know that ou can use the formula dx. Since x and are given in terms of s, the area is s:x at the right s:x at the left dx dt dt = s= s= (4 + 3s s 4 )(3s ) ds. x at the left Grading: +5 points (total) were given for no integral; +5 points (total) were given for an integral with no limits, or the surface area of the region obtained b revolving the curve around the x-axis; points if limits for x were given ( 9 7 dx); point was given for the limits ds. 4

5 (6) (Extra credit: 5 points each) Evaluate the following integrals: (a) x sin(x) dx Solution: Since the integrand is the product of two functions, integration b parts is worth a tr. When doing integration b parts, an easier integral is desired, so we let u = x and v = sin(x), so that u = x and v = cos(x), and hence x sin(x) dx = x cos(x) ( x cos(x) ) dx = x cos(x) + x cos(x) dx. The power of x has gone down, so this integral is simpler than the previous one. Now we do integration b parts again, with u = x and v = cos(x), so that u = and v = sin(x), so that x cos(x) + x cos(x) dx = x cos(x) + x sin(x) sin(x) This integral can be evaluated using the substitution = x. This gives a final answer of x cos(x) + x sin(x) + cos(x) + C. 4 dx. (b) + 4t + t + t dt Solution: The radical with a quadratic function underneath it suggests a trig substitution in the most general case. It also indicates that the substitution u = + t + t should be tried (which is a lot easier). If u = + t + t, then du/dt = + 4t, so dt = du + 4t. Then + 4t + 4t dt = + t + t u du + 4t = du = u = u + C = + t + t + C. u / du = u/ / + C 5