x 2 y 2 +3xy ] = d dx dx [10y] dy dx = 2xy2 +3y

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Bridget Booth
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1 MA7 - Calculus I for thelife Sciences Final Exam Solutions Spring -May-. Consider the function defined implicitly near (,) byx y +xy =y. (a) [7 points] Use implicit differentiation to find the derivative at (, ). d [ x y +xy ] = d dx dx [y] xy +yx dy dx +y+xdy dx =dy dx yx dy dx +xdy dx dy dx = xy y Substituting x= andy =, weget dy dx = xy +y yx +x dy dx = 4+ (,) 8+6 dy dx = 7 (,) 4 (b) [ points] Determine the equation of the tangent line at (, ). The equation of the tangentline is y = 7 4 (x )+.. Let f (x) bethe function defined byf(x)=e x for x. (a) [6 points] Write an equation for the line tangent to the graph of f (x) at the point where x=. f (x)= d dx e x =e x d dx x= x e x, So the slope of the tangent line at x = is f () = e. The equation for the tangentline is y = e (x )+e.

d [ x y +xy ] = d dx dx [y] xy +yx dy dx +y+xdy dx =dy dx yx dy dx +xdy dx dy dx = xy y Substituting x= andy =, weget dy dx = xy +y yx +x dy dx = 4+ (,) 8+6 dy dx = 7 (,) 4 (b) [ points] Determine

2 -May- Final Exam Solutions (b) [4 points] Find the point where this tangent line crosses the x-axis. The line crosses the x-axiswhen y =. e (x )+e= e (x )= e x = x= The tangentline crosses the x-axisat (,).. Let f be thefunction defined by f (x)=x x x for x. (a) [ points] Find where the graph of f crosses the x-axis. Show your work. Set f (x)=and solve x x x= x(x x )= x(x+)(x )= The solutions are x =, x =, x =, but we are restricted to [,], so the only two places where f crosses the x-axiswill bex=and x=. (b) [4 points] Find the intervals on which f is increasing. Find the derivative and the critical points. Now solve Only x = + 7 f (x)=x x x x = x= ± 8 6 = ± 7 lies in our interval. We know that f (x) is decreasing if f (x) < which happens when < x < + 7. Also, f (x) is increasing if f (x)> which happenswhen + 7 <x<. (c) [4 points] Find the absolute maximum and absolute minimum value of f. Justify your answer. f (x)iscontinuousontheclosedinterval[,], soitattainsitsmaximumand minimum. There are three points to check: x =, x = + 7, and x =. Spring MATH 7-4

Set f (x)=and solve x x x= x(x x )= x(x+)(x )= The solutions are x =, x =, x =, but we are restricted to [,], so the only two places where f crosses the x-axiswill bex=and x=.

3 -May- Final Exam Solutions FromtheFirstDerivative Testwecanseethattheabsoluteminimumoccurs at x = + 7. Evaluating f atand, we have that f ()= and f () =, so the absolute maximum occurs at x=. Themaximumvalueoff on[,]isandtheminimumvalueis [points] Aprojectile is fired toward the sea from acliff feet above the water. If the launch angle is 45 and the muzzle velocity is k, then the height h of the projectile above thewater when it is x feet downrangefrom the launch site is See the figure. h(x)= k x +x+ feet. Determine the maximum values of h and the distance downrange where this maximum occurs if k =8 feet per second. h cliff x sea level h(x)= k x +x+ h (x)= 64 k x+ Set equal to zero andsolve 64 k x+= x= k 64 Since h (x) <, this will be the absolute maximum. Thus, the downrange distance is x= k =,5 feet. The height at that distance ish(5)=6,45 feet 64 Spring MATH 7-4

If the launch angle is 45 and the muzzle velocity is k, then the height h of the projectile above thewater when it is x feet downrangefrom the launch site is See the figure. h(x)= k x +x+ feet.

4 -May- Final Exam Solutions 5. Find the requested quantities in each of the following. (a) [5 points] Let +cos(x ) if x, f (x)= ax if <x<, x x+b if x. where a and b are constants. Find the values of a and b which make f (x) is continuous on [, ]. We need lim f (x)= lim x x +f (x). This gives the equation +cos()=a a=5 Likewise, we need lim f (x)= lim x x +f (x). This gives a = ()+b 8=4 4+b b=8 These values for a and b will make f (x)continuous. (b) [5 points] Evaluate lim n 6n n+7 (n+). 6. Find the following limits 6n n+7 6n n+7 lim n (n+) = lim n n +6n+9 n 6 n = lim n n + 7 n n n n +6 n n + 9 n = 6 =6 (a) [4 points] lim x sin(x) xcos(x). Use l Hospital s Rule since both the numerator and denominator go to. lim x sin(x) xcos(x) =lim x = = cos(x) cos(x) xsin(x) 4 Spring MATH 7-4

This gives a = ()+b 8=4 4+b b=8 These values for a and b will make f (x)continuous. (b) [5 points] Evaluate lim n 6n

5 -May- Final Exam Solutions This can also bedone as follows: lim x sin(x) xcos(x) =lim x sin(x) x = = cos(x) ln(xe x ) (b) [4 points] lim. x x Use l Hospital s Rule since both the numerator and denominator go to infinity. ln(xe x ) lim = lim x x x xe x(ex +xe x ) e x +xe x = lim x xe x = lim x x += x 4 (c) [5 points] lim x x x+6. l Hospital s Rule does not apply. lim x 7. Find the following general antiderivatives. (a) [4points] (x +x+x +x ) dx x 4 x x+6 = 8 = (x +x+x +x ) dx= x + x +ln x x +C. (b) [points] (sin(x) cos(x)+sec (x) ) dx (sin(x) cos(x)+sec (x) ) dx= cos(x) sin(x)+tan(x)+c. ( e x e x ) (c) [points] dx. ( e x e x ) dx= (e x e x )dx= (ex +e x )+C 5 Spring MATH 7-4

ln(xe x ) lim = lim x x x xe x(ex +xe x ) e x +xe x = lim x xe x = lim x x += x 4 (c) [5 points] lim x x x+6. l Hospital s Rule does not apply. lim x 7.

6 -May- Final Exam Solutions 8. Use the Fundamental Theorem of Calculus(part ) to evaluate the following derivatives. (a) [5points] If F(x)= x cos (u)e u du then F (x)equals By thefundamental Theorem of Calculus, Part I, we know that (b) [5points] If G(x)= F (x)= d dx x x cos (u)e u du =cos (x)e x. cos(u)sin(u)du then G (x) equals By the Fundamental Theorem of Calculus, Part I, and the Chain Rule we have G (x)= d dx x 9. Find the following definite integrals. (a) [5points] 4x x dx cos(u) sin(u) du =cos ( (x ) ) sin ( (x ) ) d dx (x ) =6xcos(6x )sin(6x ) 4x x dx= x ln(x) =(8 ln) ( ln) =6 ln 4.94 (They may use the calculator to compute this integral, but in that case the answer is either right or wrong nopartial credit.) 6 Spring MATH 7-4

cos(u)sin(u)du then G (x) equals By the Fundamental Theorem of Calculus, Part I, and the Chain Rule we have G (x)= d dx x 9. Find the following definite integrals.

7 -May- Final Exam Solutions (b) [5 points] The function h(x) is graphed below Compute h(x) dx. This is to be done geometrically. h(x)dx= 4 5 h(x)dx+ h(x)dx+ h(x)dx+ h(x)dx+ h(x) dx 4 5 =()()+ (+4)()+ (4+)()+()()+ (+5)(5) =6. Find the requested areas. (a) [ points] Set up (but do NOT evaluate) the integral that will give the area enclosedbythecurvesy=cos(x),y =+sin(x)andtheverticallinex= π. Π Π Π Area= π [(+sin(x)) cos(x)]dx (The correct answer is + π They do not have to evaluate the integral, but IF they do, they must do it correctly.) 7 Spring MATH 7-4

(a) [ points] Set up (but do NOT evaluate) the integral that will give the area enclosedbythecurvesy=cos(x),y =+sin(x)andtheverticallinex= π.

8 -May- Final Exam Solutions (b) [7 points] The region R is enclosed by y =x/4 and y = x in the first quadrant. Find the area of R First, find the points of intersection by setting x = 4 x. These are x = and x=6. The area is then Area R = 6 ( x ) x dx= (They may use the calculator to compute this integral, but they must show the integral that they are evaluating and not just the answer.) BONUS [ points each] Compute the derivatives of the following functions. You do not need to simplify. (a) f (x)= x + e x. (b) f (x)=x 5 sin(x). (c) f (x)=cos(x+5x ). f (x)= xex (x +)e x e 4x f (x)=5x 4 sin(x)+x 5 cos(x). f (x)= (+5x )sin(x+5x ). (d) f (x)=e. f (x)= (e) f (x)=(4x ) +(8x+9). f (x)=(4x ) (8x)+(8x+9) (8). 8 Spring MATH 7-4

(They may use the calculator to compute this integral, but they must show the integral that they are evaluating and not just the answer.

Calculus 1: Sample Questions, Final Exam, Solutions

Calculus 1: Sample Questions, Final Exam, Solutions Calculus : Sample Questions, Final Exam, Solutions. Short answer. Put your answer in the blank. NO PARTIAL CREDIT! (a) (b) (c) (d) (e) e 3 e Evaluate dx. Your answer should be in the x form of an integer.