Question and Answers #93 Derivatives

Questions and Answers Sheet 93 Derivatives Question #1 Given x = cos t , y = sin 2t , how do you find the dy terms parameter t and find the values parameter t points dx dy = 0? dx Answer: Answer: dy  = 0 at t = ( 2m + 1) dx 4 Explanation: dy dy dt = In parametric equations x = x ( t ) and y = y ( t ) dx dx dt As y = sin 2t dy = cos 2t  2 = 2 cos 2t dt and as x = cos t , Thus dx = − sin t dt dy −2 cos 2t = dx sin t As sin t  0 , when t = n dy  = 0 , when cos 2t = 0 but t  n i.e. 2t = ( 2m + 1) dx 2 or 2t = ( 2m + 1)  2 , where m is an integer But note that x and y both are sinusoidal functions and hence their domain is limited to [-1,1] and hence as x = cos t dy 1 = 0 at x =  dx 2 Question #2 A street light is at the top of a 12 ft tall pole. A woman 6 ft tall walks away from the pole with a speed of 5 ft/sec along a straight path. How fast is the tip of her shadow moving when she is 45 ft from the base of the pole? Answer: In t sec the lady would be at a distance of (speed)(time) = (5 ft/sec)(t sec) = 5t ft. According to similar triangles, the tip of the shadow would be at a distance of x = 2 (5t) ft = 10t because the height of the pole is twice that of the lady. The distance equation is x = 10t ft So, the speed is x' = 10 ft/sec. Like the lady's pace, the speed remains consistent. Question #3 f ( x ) = sec x Find the second derivative of Answer: d ( sec x ) = sex tan x dx d d d ( sec x tan x ) = sec x ( tan x ) + tan x ( sec x ) dx dx dx = sec x sec2 x + tan x sec x tan x = sec3 x + tan 2 x sec x Question #4 What is differentiation of e x + b... ? a 2 Answer: According to how it is written, the query requests the derivative of y = ea x 2 + b a Assuming that a and b are constants, e is a constant, so we can simply use the power rule to get dy = 2e a x dx Question #5 Find the second derivative of the following function: y = e A. e B. e 3x+2 3x+2 9y C. 3e 3x+2 D. 9 y Answer: We have, y = e3 x + 2 We must employ the chain rule in order to distinguish this function. First derivative of this function is: dy d = e3 x + 2 . (3 x + 2) dx dx dy = 3e3 x + 2 = 3 y dx Second derivative of this function is: d 2 y d 3x+2 d = 3e = 3 y. (3 x + 2) 2 dx dx dx d2y = 9y dx 2 Question #6 Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank? Answer: 3 Let V be the volume of water in the tank, in cm ; let h be the depth/height of the water, in cm; and let r be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that The volume of the inverted cone of water is then V = Now differentiate both sides with respect to time h 6 = = 3 so that h = 3r r 2 1 2  r h =  r3 3 t (in minutes) to get dV dr = 3 r 2  (the Chain Rule is used in dt dt this step). 2 dV dVi  200  = − 10000 = 3     20 (when the dt dt  3  200 height/depth of water is 2 meters, the radius of the water is cm). 3 If Vi is the volume of water that has been pumped in, then Hence dVi 800000 cm3 = + 10000  847758 dt 3 min Question #7 How to find parametric equations for the tangent line to the curve with the given parametric equations x = 7t − 4 and y = 7t + 4 and 2 Answer: Answer: x = 3 + 14t 2 z = 6t + 5 and (3,11,11)? y = 11 + 14t z = 11 + 6t The point (3,11,11) is for t = 1 , as you can see substituting it in the three equations of the curve. Let us now look for the generic vector tangent to the curve: x = 14t y = 14t z = 6 Therefore, for t = 1 it is: v (14,14, 6 ) Thus, remembering that given a point P ( xP , yP , zP ) and a direction v ( a, b, c ) the line that passes from that point with that direction is: x = xP + at y = yP + bt z = zP + ct Then the tangent is: x = 3 + 14t y = 11 + 14t z = 11 + 6t N.B. The direction (14,14,6) is the same that (7,7,3), so the line could be written also: x = 3 + 7t y = 11 + 7t z = 11 + 3t Question #8 How to find the linearization of f ( x ) = e5x at the point x=0? Answer: The linearization of f ( x ) is its approximation using the tangent line: y ( x ) = f ( x0 ) + f  ( x0 )( x − x0 ) In our instance: f ( x ) = e5x f  ( x ) = 5e5 x So for x0 = 0 f ( 0) = 1 f  ( 0) = 5 and the approximation is: e5 x  1 + 5 x Question #9 How to find the parametric equations of line that passes through the points (1, 3, 2) and ( -4, 3, 0)? Answer:  x   1   −5        The equation is y = 3 + t 0        z   2   −2        Solution: Let the points be A = (1,3, 2 ) B = ( −4,3, 0 ) The vector  −4   1   −5        AB =  3  −  3  =  0   0   2   −2        The parametric equation of the line is  x   1   −5         y  =  3 + t  0   z   2   −2        Question #10 How to find the linearization of f ( x ) = x 4 + 5 x 2 at the point a=1? Answer: Given solution: f ( x ) = x4 + 5x2 At x = 1 , we have y = f (1) = 6 f  ( x ) = 4 x3 + 10 x so at x = 1 , the slope of the tangent line is m = f  (1) = 14 Equation of tangent line in point-slope form: y − 6 = 14 ( x − 1) Linearization at a = 1 (in a form I am used to): L ( x ) = 6 + 14 ( x − 1) Free illustration of the linearization process f (1.1)  L (1.1) = 6 + 14 (1.1 − 1) = 6 + 14 ( 0.1) = 6 + 1.4 = 7.4 (The point on the tangent line with x -coordinate 1.1 has y-coordinate 7.4) The exact value is f (1.1) = 1.7541 If you're very careful and patient, you can use the graph to determine these two values: graph{( y − x 4 − 5 x 2 )( y − 14 x + 8) = 0[0.1604,1.846, 6.859, 7.7024]} Question #11 A ball thrown vertically upwards with a speed of 19.6 m/s from the top of a tower returns to the earth in 6s. Find the height of tower: A)60 B) 58 C) 58.8 D) None of these Answer: The correct option is C. 58.8 m. Let the height of the tower be h. Here, u=19.6 m/s t=6 s s=−h (The height of the tower is the magnitude of the displacement. Its negative value is used because it is in the opposite direction as the initial velocity.). 𝑎 = −𝑔 = −9.8𝑚/𝑠 2 (It is regarded as negative because the direction of gravity's acceleration is downwards, whereas the initial velocity is upwards). Using the second equation of motion s = ut + ∴ −ℎ = (19.6 × 6) − 1 2 at we get, 2 9.8 × 6 × 6 2 =-58.8 m Therefore, the height of the tower is 58.8 m. Question #12 How to find the linear approximation of a function? Answer: The tangent line to the function graph is used to approximate a function linearly, or L ( x ) . The equation of the tangent line at x = a is given by y = f  ( a )( x − a ) + f ( a ) , the approximate linear model is L ( x ) = f  ( a )( x − a ) + f ( a ) Question #13 Find the second derivative of f ( x ) = x x +1 2 Answer: f ( x) = x x +1 2  f ( x) = f ( x) = g  ( x ) .h ( x ) − h ( x ) .g ( x ) ( ( f ( x )) ) 1. x 2 + 1 − 2 x ( x ) f  ( x ) = f  ( x ) = f  ( x ) = (x 2 ) +1 2 ( −2 x ) . ( x 2 + 1) 2 = − x2 + 1 (x 2 ) +1 2 ( f ( x) = )( − x2 + 1 (x 2 ) +1 2 ) − 2 ( 2x ) x2 + 1 − x2 + 1 2 (x 2 ) +1 4 ( −2 x ) . ( x4 + 2 x2 + 1) − 4 x (1 − x4 ) (x ) +1 2 4 −2 x5 − 4 x3 − 2 x − 4 x + 4 x5 f  ( x ) == (x 2 ) +1 4 = −2 x5 − 4 x3 − 2 x − 4 x + 4 x5 (x 2 ) +1 4 = 2 x5 − 4 x3 − 6 x (x 2 ) +1 4 2 x5 − 4 x3 − 6 x (x 2 ) +1 4 Question #14 The radius of a sphere is increasing at a rate of 4 mm/s. How fast is the volume increasing when the diameter is 40 mm? Answer: You can express the first rate as follows by using r to stand for the radius and t for time: dr mm =4 dt s Or r = r ( t ) = 4t The volume of a solid sphere is given by the formula: 4 V = V (r ) =  r3 3 When both sides' derivatives with regard to time are taken... dV 4  dr  =  3r 2   dt 3  dt  ( ) ...remember the Chain Rule for implicit differentiation. The general format for this is: dV ( r ) dt with = dV ( r ) dr ( t )  dr ( t ) dt V = V ( r ) and r = r ( t ) .  dV ( r )  , but we want to do it  dr ( t )    So, when you take the derivative of the volume, it is with respect to its variable r   dV ( r )   . Since r = r ( t ) and r ( t ) is implicitly a function of t , to make the equality work, you dt    dr ( t )  have to multiply by the derivative of the function r ( t ) with respect to t   as well. That way, you're taking  dt  a derivative along a chain of functions, so to speak V → r → t ). with respect to t  Now what you can do is simply plug in what r is (note you were given diameter) and what dV dr is, because dt dt describes the rate of change of the volume over time, of a sphere. ( ) dV 4 2  mm  =  3 ( 20mm )  4  dt 3  s  = 6400 mm3 s We cannot simply state that the two rates are equal because time simply rises, the radius increases as a function of time, and the volume increases as a function of a constant times the radius squared. Question #15 A model used for the yield Y of an agricultural crop as a function of the nitrogen level N in the soil (measured in appropriate units) is 𝑦 = 𝑘𝑁 1+𝑁2 where k is a positive constant. What nitrogen level gives the best yield? Answer: Take the first derivative with respect to N: 𝑦′ = (1 + 𝑁 2 )𝑘 − 𝑘𝑁(2𝑁) 1 + 𝑁2 𝑦′ = 𝑘 + 𝑘𝑁 2 − 2𝑘𝑁 2 1 + 𝑁2 𝑦′ = 𝑘 − 𝑘𝑁 2 1 + 𝑁2 2 2 2 Equate to 0 and find N by solving: 2 𝑘 − 𝑘𝑁 2 =0 1 + 𝑁2 𝑘(1 − 𝑁 2 ) = 0 1 − 𝑁2 = 0 𝑁2 = 1 𝑁 = ±1 → 𝑁 = 1 is the only viable response because a nitrogen level that is negative is not possible. The "best yield" would be y at its highest value. Ascertain that y has a maximum when N=1 by evaluating y' over the following intervals: [0,1), (1, ∞) to determine whether y' is positive (y is increasing) or y' is negative (y is decreasing) in each interval. If N=1 is a maximum, then y' will be positive before we reach N=1 and negative afterwards: [0,1): 𝑦 ′ (0) = 𝑘−𝑘(0) 12 = 𝑘 > 0 Therefore, y is increasing on [0, N) (1, ∞): 𝑦 ′ (2) = ( 𝑘−4𝑘 2 1+4 )=− 3𝑘 25<0 Then, y is decreasing on (1, ∞) and the maximum possible crop yield happens with N=1. Question #16 How to find the equation of the tangent to the curve x = t + 1 y = t + t at the point where t 4 Answer: To find the equation of the tangent line, we need the slope m= dy and the point of tangency ( xo , yo ) . dx Then the equation is the usual y − yo = m ( x − xo ) . We have the parametric curve x = t + 1, y = t + t , 4 Therefore, we compute 3 dx dy = 4t 3 and = 3t 2 + 1. dt dt dy dy dy dx dy dt =  = . The chain rule says that dt dx dt dx dx dt Thus, we use the derivatives of the parametric equations: 3 = −1 ? dy 3t 2 + 1 = . Now put in t = −1 and find: dx 4t 3 dy 3  ( −1) + 1 3 + 1 m= = = = −1. 3 dx −4 4  ( −1) 2 Also at t = −1 the original equations give ( xo , yo ) = ( ( −1) 4 ) + 1, ( −1) + ( −1) = (1 + 1, ( −1) + ( −1) ) = ( 2, −2 ) 3 Now we put in the info for the tangent line: y − yo = m ( x − xo ) y − ( −2 ) = ( −1)( x − 2 ) or y + 2 = − x + 2 or just plain old y = − x . Question #17 How to use linear approximation to the square root function to estimate square roots 8.95 ? Answer: The function we want to approximate is f ( x ) = We want to estimate f ( 8.95 ) = x. 8.95 so we need a value for a that is close to 8.95 and for which we can find f ( a ) . We can readily find f (1) or f ( 4 ) , but f ( 5) would be more challenging. Although we can easily find f ( 25 ) , we'd like a to be as close as we can get to 8.95 and still calculate f ( a ) . By now you may be anxious for me to say, "we'll use So f ( a ) = f (9) = 3 For the linearization we'll need f ( x) = 1 2 x , so f  ( 9 ) = The linearization of f ( x ) = L ( x) = 3 + f  ( 9 ) as well. 1 6 x at a = 9 is 1 ( x − 9) 6 Our estimate is f ( 8.95 )  L ( 8.95 ) = 3 + 1 (8.95 − 9 ) 6 a = 9 ". So I will. We'll use a = 9 . = 3− 0.05 = 3 − 0.0083 6 = 2.9916 Round to the desired accuracy. Keep in mind that this is an estimate for the desired number too much accuracy. 8.95 so don't claim 2.99 or perhaps even 2.991 seems reasonable. Question #18 What is the second derivative of 𝑥 2 + 𝑦 2 = 25 evaluated at (−3, −4)? Answer: We can treat y as a constant if we want to find the derivative with respect to x. (derivative of a constant is zero). 𝑑 2 (𝑥 + 𝑦 2 = 25) 𝑑𝑥 The sum rule allows us to divide this up (𝑎 + 𝑏)′ = 𝑎′ + 𝑏 ′ 𝑑 2 𝑑 2 (𝑥 ) + (𝑦 = 25) 𝑑𝑥 𝑑𝑥 Using the power rule, 𝑑 𝑑𝑥 (𝑥 2 ) becomes 2x, and if we treat 𝑦 2 as a constant, the derivative of that and 25 becomes 0. We're just left with 2x. 𝑑 = 2𝑥 𝑑𝑥 Finding the Second Derivative: 𝑑 (2𝑥) = 2 𝑑𝑥 We arrive at 2 by finding the second derivative. Question #19 d2y 2 2 3 How to find for the curve x = 4 + t , y = t + t ? 2 dx Answer: Step 1 First Derivative dy dy dt y ( t ) 2t + 3t 2 3 = = = = 1+ t dx dx x ( t ) 2t 2 dt Step 2 Second Derivative d dy d 1 + 3 t  3 d y d dy dt dx dt  2  2 3 = = = = = 2 dx dx dx dx x ( t ) 2t 4t dt 2 Question #20 If a function is continuous along the interval [1,3], would it be differentiable at x=1 and x=3? Answer: The function f ( x ) = x 2 − 4 x + 3 is continuous everywhere, including [1,3], but is not differentiable at 1 or at 3.