PRACTICE FINAL. Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 10cm.


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1 PRACTICE FINAL Problem 1. Find the dimensions of the isosceles triangle with largest area that can be inscribed in a circle of radius 1cm. Solution. Let x be the distance between the center of the circle and the base of the isosceles triangle. Then the base of the triangle has length x 2 and the height of the triangle is 1 + x. Therefore, area of the triangle is given by We first find the critical points of A(x): A(x) = 1 x 2 (1 + x) x [ 1, 1] A (x) = 1 x 2 x(1 + x) = 1 x2 1x x 2 1 x 2 1 x 2 Thus A (x) = when 1 1x 2x 2 =. The solution to this equation is given by 1 ± 9 = 1 or 5 4 For x = ±1, we have A(x) =. For x = 5, we get A (5) = which is the maximum area of the largest isosceles triangle that can be inscribed in the circle of radius 1 cm. For x = 5 we get that the base of the isosceles triangle has length 1 3. The other sides of the triangle have length equal to = 1 3. Thus the triangle is equilateral of side length 1 3 cm. Problem 2. Find area of the region bounded by the curves x = y 2 2, y = 1, y = ln(x) and y = 1. Solution. We compute the area by integrating relative to y. The its of integral are y = 1 and y 1. The left function is x = y 2 2. The right function is x = e y (so that y = ln(x)). Thus the definite integral we need to compute is ] 1 (e y y 2 + 2) dy = [e y y y = (e e 1 23 ) Problem 3. Evaluate the following its: (a) x 3e 2x 3 tan(2x) Solution. The it is an indeterminate form /. We apply the L Hospital rule 3e 2x 3 6e 2x = x tan(2x) x 2 sec 2 (2x) = 6 2 = 3 (b) (x3 7x 2 )e x
2 2 PRACTICE FINAL Solution. The it is an indeterminate form.. We apply L Hospital rule repeatedly to compute the it as follows: (x3 7x 2 )e x = = (c) x + 7x 2 6x 14 e x = e x = 6 e x = 1 + x 1 + sin(x) Solution. We start by rationalizing the numerator 1 + x 1 + sin(x) = x sin(x) The it x + x sin(x) ( 3x 2 14x e x ) x sin(x) can be computed by L Hospital rule as follows: x sin(x) 1 cos(x) sin(x) x + = x + 3x 2 = x + 6x = cos(x) = 1 x Thus the final answer is Problem 4. x x 1 + sin(x) (a) Evaluate the integral = x + (1 + x 1/4 ) 2 dx. ( x sin(x) = = 1 12 ) ( ) x sin(x) Solution. (1 + x 1/4 ) 2 = 1 + 2x 1/4 + x 1/2. An antiderivative of this function is Thus the definite integral is x + 2 x5/4 5/4 + x3/2 3/2 = x x5/ x3/2 (1 + x 1/4 ) 2 dx = (b) Find the derivative of g(x) = Solution. calculus: [ x x5/4 + 2 ] 1 3 x3/2 = = x 4 cos(t 2 ) dt. The derivative of g(x) can be computed using the fundamental theorem of ( ) d x 4 cos(t 2 ) dt = d ( u ) ( d(x cos(t 2 4 ) ) ) dt = cos(u 2 )4 = cos(x 8 )4 dx du dx In the intermediate step, we used u = x 4.
3 (c) If f(x) is a continuous function and PRACTICE FINAL 3 2 f(x) dx = 6, then compute π/2 f(2 sin(θ)) cos(θ) dθ. Solution. By the substitution rule, setting x = 2 sin(θ) gives dx = 2 cos(θ)dθ and we get π/2 f(2 sin(θ)) cos(θ) dθ = 1 f(x) dx = 3 2 Problem 5. Consider the function f(x) = ln(x 2 3x + 2). (a) Find the domain and range of f(x). Solution. f(x) = ln(x 2 3x + 2) = ln((x 1)(x 2)). Since ln is only defined for positive input values, the domain of f(x) is (, 1) (2, ). (b) Find the intervals of increase and decrease of f(x). Find the local max. and min. points of f(x). Solution. f (x) = 2x 3 x 2. Since the denominator is positive on the domain, we have 3x + 2 f (x) < for x < 3/2 f (x) > for x > 3/2. Thus f (x) = for x = 3/2. But this point is not in the domain of f(x), so f(x) has no critical points. (c) Find the intervals of concavity and inflection points of f(x). Solution. 2 f (x) = (x2 3x + 2)(2) (2x 3) 2 (x 2 3x + 2) 2 = 2x2 + 6x 5 (x 2 3x + 2) 2 f (x) is always negative, so the function is concave down everywhere. (d) Sketch the graph of f(x). We can compute the vertical asymptotes of f(x) as: f(x) = = f(x) x 2 + x 1 Moreover as x ±, the input x 2 3x + 2 for ln goes to and hence ln(x 2 3x + 2) goes to infinity.
4 4 PRACTICE FINAL Problem 6. Two people start walking from the same point. One walks east at 3 miles/hr and the other walks north at 2 miles/hr. How fast is the distance between them increasing after 15 minutes? Solution. Let x be the distance travelled by eastbound person and let y be the distance travelled by northbound person. We are given the information dx dy = 3 and = 2 (in miles per hour). The dt dt distance between them is given by s 2 = x 2 + y 2 Taking derivative of this equation relative to t gives 2s ds dt = 2xdx dy + 2y dt dt After 15 minutes, we have x = 3/4 and y = 2/4. At that time we have s = 13/4. Substituting these values in the equation above, we get ( 13/4) ds dt Thus ds dt = 13 miles per hour. = (3/4)(3) + (2/4)(2) Problem 7. Consider the region bounded between the curves y = x 2 and y = x. Find the volume of the solid obtained by rotating this region around y axis. Solution. The curves y = x 2 and y = x meet at (, ) and (1, 1). We find the volume by integrating relative to y. A typical y crosssection is the region bound between two concentric circles, with the radius of the outer circle y and the radius of the inner circle y 2. Thus the area of a typical y crosssection is Thus the volume of the solid is given by A(y) = π(y y 4 ) [ ] y Volume = π(y y ( ) dy = π 2 y5 1 = π ) = π Problem 8. Show that the sum of the x and y intercepts of any tangent line to the curve x + y = c is equal to c. Solution. Let us compute the equation of the tangent line to the given curve at a point (x, y ). Taking the derivative of x + y = c with respect to x gives 1 2 x dy y dx = dy y dx = x Thus the slope of the tangent line is y / x. The equation is y y y = (x x ) x The x intercept of this line is when y = and the y intercept is when x = : x intercept = x + x y y intercept = y + x y
5 The sum of these intercepts can be computed as: Problem 9. Compute the following: PRACTICE FINAL 5 x + 2 x y + y = ( x + y ) 2 = ( c) 2 = c (a) d dx ((cos(x))x ) Solution. Let y = cos(x) x. Therefore ln(y) = x ln(cos(x)). Taking derivative with respect to x gives: Hence y = cos(x) x (ln(cos(x)) x tan(x)) y y = ln(cos(x)) + x sin(x) cos(x) (b) 2 t 2 t + 3 dt Solution. Using the substitution x = 2 t + 3 we get dx = 2 t (ln(2))dt and hence 2 t 2 t + 3 dt = 1 dx ln(2) x = 1 ln(2) ln( x ) = 1 ln(2) ln(2t + 3) Problem 1. The acceleration of a car is given by a(t) = 3t 5m/s 2. Find the distance traveled by the car from t = to t = 3 s, if the initial velocity is 8 3 m/s. Since a(t) = 3t 5, we get that the velocity is given by v(t) = a(t)dt = 3t 2 /2 5t + constant The value of constant is determined by v() = constant = 8/3. Thus v(t) = 3t2 2 5t = 1 ( 9t 2 3t + 16 ) = 1 ((3t 8)(3t 2)). 6 6 This function changes sign at t = 8/3 and t = 2/3. Namely we have v(t) > for < t < 2/3 and 8/3 < t < 3. v(t) < for 2/3 < t < 8/3. Therefore the distance travelled is given by Distance = 2/3 v(t)dt 8/3 2/3 3 v(t)dt + v(t)dt 8/3 An antiderivative of v(t) = 3/2t 2 5t + 8/3 is given by s(t) = t 3 /2 5t 2 /2 + 8t/3. The distance travelled is given by Distance = s(2/3) s() s(8/3) + s(2/3) + s(3) s(8/3) = s(3) 2s(8/3) + 2s(2/3) s() The value of the function s(t) at these points is s(3) = 27/2 45/2 + 8 = 1 s(8/3) = 256/27 16/9 + 64/9 = 32/27 s(2/3) = 4/27 1/9 + 16/9 = 22/27
6 6 PRACTICE FINAL Thus the distance travelled is = = 3 m Problem 11. Use linear approximation to find the approximate value of sec(.1). Solution. We linearize the function f(x) = sec(x) at x =. f (x) = sec(x) tan(x) f() = 1 f () = Thus the linearization is given by L(x) = f() + f ()(x ) = 1. Thus sec(.1) 1. Problem 12. Find the absolute minimum of f(x) = x2 + 1 over x > and use it to prove that x x for every x >. x Solution. The derivative of f(x) is computed as follows. ( d x 2 ) + 1 = d ( x + 1 ) = 1 1 dx x dx x x 2 Over the domain (, ) there is only one critical point of f(x), namely x = 1. For < x < 1 the derivative is negative and for x > 1 the derivative is positive. Hence the function takes absolute minimum value at x = 1, namely f(1) = 2. Therefore for any x > we have f(x) f(1) = 2
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