Math 34A Practice Final Solutions Fall 2007

Transcription

1 Mat 34A Practice Final Solutions Fall 007 Problem Find te derivatives of te following functions:. f(x) = 3x + e 3x. f(x) = x + x 3. f(x) = (x + a) 4. Is te function 3t 4t t 3 increasing or decreasing wen t =? 5. Find a nonzero function f(x) suc tat f (x) 4f(x) = 0. Solution. f (x) = 6x + 6e 3x. f (x) = x 3. f (x) = x + 4a 4. f (x) = 6t 4 3t. So, f () = = < 0, wic means f is decreasing. 5. f(x) = e 4x Problem (a) Wat is te equation for te line tangent to f(x) = x at x =? Solution First, we find te slope of te line by finding f (): f (x) = f () = x Next, we find a point on te line by substituting x = into f(x): (, f()) = (, ) Ten we can use point-slope format to find te equation of te line: y = (x ) y = x + (b) Use te tangent line approximation at x = of f(x) to approximate.. Solution To do tis, we simply substitute x =. into te equation for line tangent to f(x) at x = : y = (.) + y =.05

2 Problem 3 If te alf life of some element is 0 years, ow long does it take until % of te element remains? Solution Call te initial starting amount of te element A 0. Ten according to te alf life formula, te amount left of te element at year t is given by t/0 A(t) = A 0 Ten to find wen % of te element is left, we set A(t) equal to 0.0 A 0, and solve for t: t/0 0.0 A 0 = A 0 log 0.0 = t log log 0.0 t = log(0.5) Problem 4 A jet airliner flies at 300 mp for te first alf our and last alf our of a fligt. Te rest of te time it flies at 600 mp. How long does it take to fly from LA to NY, a distance of 00 miles? Solution Let T be te time, in ours, spent flying at 600 mp. In te middle of te fligt, after te first alf our and before te last alf our, te plane travels a distance of 600T miles. In te first alf our, te plane travels 300( ) = 50 miles. In te last alf our, te plane travels 300( ) = 50 miles. So te total distance traveled is 00 = T + 50 = T. Solving tis equation for T, we get = 600T 800 = 600T ( ) = T 3 = T So te middle of te fligt takes 3 ours. In total, te fligt took 3+ + = 4 ours. Problem 5 Solve for x (a) 5 x = 3 9 x Solution First, rewrite te rigt and side: 3 9 x = 3 (3 ) x 3 (3 ) x = 3 (3 x ) 3 (3 x ) = 3 x+

3 So now we ave te equation 5 x = 3 x+. Take te log of bot sides and ten solve for x: log(5 x ) = log(3 x+ ) xlog(5) = (x + )log(3) xlog(5) = xlog(3) + log(3) xlog(5) xlog(3) = log(3) x(log(5) log(3)) = log(3) x = log(3) log(5) log(3) (b) Solution ln(x ) = ln(x) ln(x) = ln(x) ln(x) ln(x) = 0 ln(x) = 0 x = Problem 6 For f(x) = x + x, wats te average rate of cange of te function from to +? Wats te instantaneous rate of cange at x =? Solution Te rate of cange formula in general is f(x ) f(x ) x x. In our case x = +, x =, so f(x ) f(x ) x x = f( + ) f() + = ( + + ( + ) ) ( + ) = = = 3 + Te instantaneous rate can be found in one of two ways, eiter we can take te limit as 0 or take te derivative of f(x) at x =. From te previous part, taking te limit gives te instantaneous rate to be 3. We can ceck tis by noticing f (x) = + x and f () = + = 3. Problem 7 An object is dropped out of an airplane. Te eigt of te object t seconds after being dropped is (t) = 000 5t meters.. Wat is te eigt of te plane wen te object was dropped?. Wat is e velocity of te object at t = 3 3

4 3. Wat is te acceleration of te object? 4. Wen did te object it te ground? 5. How fast was it going wen it it te ground? Solution. Te object is dropped at time = 0. At wic its eigt is (0) = = Te velocity is te derivative of position, in our case eigt. So v(t) = 0t. At time = 3, our velocity is v(3) = 30. Te velocity is negative because we are falling. 3. Te acceleration is te derivative of te velocity (or te second derivative of position). So a(t) = 0. Hence our acceleration is always -0, regardless of te time. 4. An object its te ground wen its eigt is 0. So we must solve 0 = 000 5t wic gives t = From te previous part we know wen te object its te ground. Tis is critical. It its te ground after 0 seconds. So its velocity after 0 seconds is v(0) = 0 0 = 00. We re not done yet toug, te question asks for speed, wic is te magnitude of velocity. Tere s no suc ting as a negative speed. Te speed is te absolute value, or 00. Problem 8 A cylindrical metal can is to ave no lid. It is to ave a volume of 8πin 3. Wat eigt minimizes te amount of metal used (ie surface area). Solution Let = eigt of te can, and r = radius of te can. We try to minimize te surface area (SA), and te equation of surface area is SA = πr + πr. We d like to take te derivative of tis function and set it equal to 0, owever, te function is in two variables. We must eliminate one of tem, so we use te extra information given in te problem. V = 8π = πr = 8 r Now we can plug tis relationsip into our area function to ave it in terms of one variable. SA = πr + πr = πr( 8 r ) + πr = 6π 8 r + πr 4

5 Now we take te derivative and set it equal to 0. 0 = 6πr + πr = 6π = πr r 6π r r 3 = 8 r =. = πr But we re not done yet. We meerly found te radius at wic te eigt is minimized. But we know ow te eigt related to te radius: = 8. So wen r = r = 8/4 =. Problem 9 A rectangular field will ave one side made of a brick wall and te oter tree sides made of wooden fence. Brick wall costs $0 per meter. Wooden fence costs $40 for 4 meters. Te area of te field is to be 400 m. Wat lengt sould te brick wall be to give te lowest total cost of te wall plus fence? Solution Let L = lengt of te rectangle, H = eigt. Ten te area of te field = LH = 400. We need to figure out te cost of te fence in terms of tese variables. Suppose te wall made of brick is along te lengt (tis assumption won t matter in te end). Ten Cost = 40 (L + H + H) + 0(L) 4 = 30L + 0H Again tis is an equation wit two variables. We use te fact tat LH = 400 H = 400/L to eliminate a variable. Plugging in gives, Cost = 30L + 0( 400 L ) = 30L L Taking te derivative and setting it equal to zero gives, 0 = L L = 30 L = = L = 40 Since te brick wall was along te lengt, our answer is 40. 5