Chapter 4 Review Prof. Townsend Fall 016 Sections covered: 4.1 Tangents and Normals 4.3 Curvilinear Motion 4.4 Related Rates 4.5 Using Derivatives in Curve Sketching 4.7 Applied Maximum and Minimum Problems Note that in the examples below, you don't see much algebra or derivative detail. On the exam, you are free to use your calculator for those calculations. Follow the details of the solutions below to show me your work. For example, state the derivative you are taking, given y = 1 7 x + 51 7, = 1. That is all you have to do to show your work. For algebra, say 7 something like "using Solver:" or "using Factor:", or "using Expand:". Note don't plug in values of variables until the derivatives are taken. All problems involve differentiation. 4.1 Tangents and Normals Notation: a given point will be denoted as ( x 1, y 1 ) As developed in section 3., the slope of the tangent line at a point is the first derivative of the function evaluated at that point. m = x=x1 Since we will be dealing with both tangents and normal, I will use subscripts to distinguish between them when necessary. m tan = and m = 1. x=x1 m tan We solved four types of problems in this section. A) Find the equation of the line tangent to the given curve, given the point x 1, y 1 ( ). ( ). B) Find the equation of the line normal to the given curve, given the point x 1, y 1 C) Find the equation of the line tangent to the given curve, given the slope m. D) Find the equation of the line normal to the given curve, given the slope m. Following are examples of each. Page 1 of 19
A) Find the equation of the line tangent to the curve y = x x +1 at the point x =. 1) Take the derivative of the function. = 4x 1 ) Evaluate the slope. m = x= = 4 ( ) 1 = 7 3) Find the y value at the point. y = ( ) +1 = 7 4) Find the value of the y intercept. b = y 1 mx 1 = 7 7 = 7 5) Write the equation of the tangent line. y = 7x 7 Check by graphing both the original function, y = x x +1, and the equation of the tangent line, y = 3x +1. 1) Define the functions ( F1). ) Window make sure the point of interest is within the window of the graph. I also like to make sure the two axes show. Note that I first graphed the function using ZoomStd to see what it looks like. Finally, I chose the following window after playing with some of the values. 3) The graph. Page of 19
If instead of graphing the tangent line as a second equation, I just use F5/A when looking at the graph, I get the following, thus verifying my tangent line equation. B) Find the equation of the line normal to the curve y = x x +1 at the point x =. 1) We know the tangent slope from the above. m tan = x= = 7 Here we want the normal slope. m = 1 m tan = 1 7 ) Find the value of the y intercept. b = y 1 mx 1 = 7 1 7 = 51 7 3) Write the equation of the normal line. y = 1 7 x + 51 7 4) Check by graphing both the original function, y = x x +1, and the equation of the normal line, y = 1 7 x + 51. Since the original function and the point are the same 7 as above, I will just include the normal line equation, then graph both. The graph The line does not look perpendicular (normal). So, F/ZoomSqr. Page 3 of 19
Now I will add the tangent line (F5/A) to make sure the line is perpendicular. For this example, the above graph looks perpendicular but for some graphs it will not be so obvious. C) Find the equation of the line tangent to the curve y = x x +1 given the slope, m = 7. You are given the slope so you need to find the point on the curve where the slope is 7. 1) Take the derivative of the function. = 4x 1 ) Find the value of x. = 4( x) 1 = m = 7 x = 8 4 = 3) You know both m and x so follow example A above to finish the problem. D) Find the equation of the line normal to the curve y = x x +1 given the slope, m = 1 7. 1) Find the tangent line slope so you can equate it to the derivative to find x. m tan = 1 = 1/ ( 1/ 7) = 7 m ) Find the value of x given the tangent slope. = 4 x 3) You know both m and x so follow example B above to finish the problem. ( ) 1 = m = 7 x = 8 4 = Page 4 of 19
4.3 Curvilinear Motion In PHY 10 you learned that x = x 0 + v 0 t + 1 at and v = v 0 + at A more general form is given an equation for x, v x = and given an equation for y, v y =. The acceleration is no longer constant, it is the derivative of the velocity. a x = dv x and a y = dv y In class we solved two forms of problems. A) Given x and y as functions of time, i.e. x t acceleration. ( ) and y t B) Given y as functions of x and v x, find the velocity. Examples: ( ), find the velocity and A) Find the velocity and acceleration given x = 1 t +1 and y = t 3 +1 at t =. 1) Straight TI-89 solution - velocity. a) Find the derivative The notation is v! = d( [ x, y],t). Note that the TI removes the comma from the vector brackets when it shows the result. b) Plug in the time Page 5 of 19
c) Convert the answer to polar coordinates. You can see from the bottom of the screen that the angle mode is DEG i.e. degrees. ) Straight TI-89 solution - acceleration. a) Find the derivative of the velocity found above or enter the following The notation is a! = d b) Plug in the time ( ) = d x, y v x,v y,t ([ ],t,). c) Convert the answer to polar coordinates. Page 6 of 19
1) By hand solution - velocity. a) Find the derivative v x = = d 1 t +1 = d ( t +1) 1 = t +1 ( ) ( ) = 1 so the third factor in the power rule, Note that d t +1 v y = = d ( t 3 +1) = d ( t 3 +1) 1/ = 1 t 3 +1 ( ) 1/ ( 3t ) = d du un = nu n 1, was 1. 3t t 3 +1 b) Plug in the time v x = ( +1) = 1 3 = 1 9 v y = 3( ) 3 +1 = 3 9 = c) Convert the answer to polar coordinates. v = v x + v y = 1 9 + =.003 tan(θ v ) = v y v x θ v = tan 1 v y v x = tan 1 1/ 9 = tan 1 18 ( ) = 86.8! But the velocity is in the second quadrant since the x component is negative and the, component is positive. Whenever the x component is negative, add 180! to the calculator answer. θ v = 86.8! +180! = 93.18! ) By hand solution - acceleration. a) Find the derivative of the velocity found above. v x = ( t +1) a x = dv x { } = = ( t +1) 3 ( t +1) 3 Note I prefer the product rule to the quotient rule as u and v are interchangeable in the product rule and not in the quotient rule. ( ) 1/ v y = 3 t t 3 +1 Page 7 of 19
a y = dv y a y = 3 t 3 +1 = 3 t t 3 +1 ( ) 1/ + t 1 ( ) 3/ t ( t 3 +1 ) + t 1 ( ) ( ) 3/ a y = 3t t 3 + 4 4 t 3 +1 ( ) 3/ ( 3t ) t 3 +1 ( 3t ) = 3 t 3 +1 t ( ) 3/ 4 + t b) Plug in the time, t=. a x = ( +1) = 3 7 ( ) ( ) ( ) a y = 3 3 + 4 4 3 +1 = 3 1 3/ ( 9) = 3 1 3/ 7 ( ) ( ) = 3 c) Convert the answer to polar coordinates using a = a x + a y = 0.671 θ a = tan 1 a y a x = / 3 tan 1 / 7 = tan 1 9 ( ) = 83.7! a x is positive so do not add 180! to the calculator answer. Page 8 of 19
B) Given v x and y as functions of x, find the y component of the velocity. Find the velocity given y = x 3 + x and v x = 3 at t =. Since, in general, we want the velocity, v! =, but y is given as a function of x, not t, so we need to use the chain rule. v y = = = v x Solution. v y = v = d x ( x3 + x) v x = ( 6x +1) ( 3) = 18x + 3 Since v x is constant, we know that x = v x t = 3t. v y = 18 3t ( ) + 3 = 18( 6) + 3 = 651. Note, if v x is not constant, we need integration (MTH 41) to find x as a function of time. Once you have both components of velocity, you can find their polar representation by methods used above. 4.4 Related Rates Every problem in related rates is essentially the same. You are given one rate (either or ) and the value of x. To find the other rate use the chain rule: = Plug in the value of x into the derivative. For volume and surface area formulas, look in the front cover of your book. A couple of good examples you might want to download are found at http://www.austincc.edu/pintutor/pin_mh/_source/handouts/geometry_formulas/geome try_formulas_d_3d_perimeter_area_volume.pdf and http://www.science.co.il/formula.asp In the following examples, I will switch the notation to x and y alongside using the actual variables so you can see the pattern better. The examples below are from pages 730-733.. Page 9 of 19
Actual Problem V = 0.030 I r = 0.030 I r r = 0.040in dv = dv dr dr di = 0.00 A s x and y version y = 0.030 x r = 0.030 x r = 0.00 r = 0.040in = dv di = 0.030 r dv = 0.030 r di di = 0.030 r di = 0.030 ( 0.040) 0.00 ( ) = 0.375 V s = 0.030 r = 0.030 r = 0.030 r = 0.030 ( 0.040) 0.00 ( ) = 0.375 Actual Problem V = πr h dr mm cm = 0.100 = 0.010 min min r = 9.5 cm = 4.75cm h = 15.0cm x and y version y = π x h x = 4.75cm = 0.010 cm min h = 15.0cm Page 10 of 19
dv = dv dr dr h is constant = dv dr = πh dr dr = πhr dv = dv dr dr = ( πhr) dr = ( π 15 4.75) ( 0.010) = 4.477 cm3 min = ( π xh ) = = ( πhx) = ( π 15 4.75) ( 0.010) = 4.477 cm3 min Actual Problem dw D = w + h = 0.5 in s h = 4.0in dd = dd dw dw x and y version y = x + h = 0.5 in s h = 4.0in = dd dw = d dw w + h = d ( dw w + h ) 1/ = = d x + h = d ( x + h ) 1/ = 1 ( w + h ) 1/ ( w) dw = dd = w dw w + h w dw w + h = 6.5 ( 0.5) = 6.5 + 4 1 ( x + h ) 1/ ( x) = = x x + h x x + h = 6.5 ( 0.5) = 6.5 + 4 0.13 in s 0.13 in s Page 11 of 19
Actual Problem T v = 331 73 = 331 73 T dt! =.0 C! h =.0 K since! K =! C + 73 h dv = dv dt dt dv dt = 331 d T 73 dt = 331 dt 1/ 73 dt = 331 1 73 T 1/ 331 = T 73 dv = 331 73T dt = 331 303 73.0 ( ) = x and y version y = 331 =.0 = = 331 73 x 73 = 331 73 x 1/ = 331 x 73 = 331 73 303 = 1.151 1.151 m s h = 1.151 m s h 1h 3600s = 4.144E3 m s 4.5 Using Derivatives in Curve Sketching d y >0 >0 =0 =0 or <0 <0 The maxima and minima are found where the slope is zero and the concavity is positive for minima and negative for maxima. Page 1 of 19
4.7 Applied Maximum and Minimum Problems There are two types of max/min problems we have met. 1) The first is y is a function of x, find the max/min value of y. ) The second involves two equations. u is a function of x and y, and v is a function of x and y. Find the max/min of one of them given the value of the other one. The examples below are from pages 730-733. Type 1 Max/Min The first max/min type is where y is a function of x, find the max/min value of y. i) The equation: y = k( x 4 5Lx 3 + 3L x ) where both k and L are constant. ii) Find the two derivatives of y. = k 8x3 15Lx + 6L x ( ) d y = k 4x 30Lx + 6L ( ) iii) Solve for the location of the minimum. = k ( 8x3 15Lx + 6L x) = 0 x = 0 clearly not the answer physically. x = 1.3L x cannot be larger than L. x = 0.578L must be the answer iv) Check the second derivative. d y = k 4 ( 0.578L ) 30L 0.578L so x = 0.578L ( ( ) + 6L ) = 3.3kL < 0 maximum Page 13 of 19
144r i) The equation: P = ( r + 0.6) ii) Find the two derivatives of P. dp 144 = ( r 0.6 ) dr r + 0.6 ( ) 3 ( ) ( ) 4 d P 88 r 1. = dr r + 0.6 iii) Solve for the location of the minimum. dp 144 = ( r 0.6 ) dr r + 0.6 ( ) 3 = 0 r = 0.6 Plug in r into the second derivative to make sure the solution is for a maximum or just observe that if r<1., d P 88 = ( r 1. ) dr r + 0.6 maximum. then is negative so the solution is at a ( ) 4 88 = ( 0.6 ) ( 0.6 + 0.6) < 0 maximum 4 d P dr Page 14 of 19
Type Max/Min The second max/min type involves two equations. u is a function of x and y, and v is a function of x and y. Find the max/min of one of them given the value of the other one. i) The equations: V = 3x + y 6 = 3x + y The second equation has the number so solve for x or y and plug it into the first equation. y = 3 3 x V = 3x + 3 3 x ii) Find the two derivatives of V. dv = 15x 18 d V = 15 > 0 minimum = 15 x 18x +18 iii) Solve for the location of the minimum. dv = 15x 18 = 0 x = 18 15 = 1. y = 3 3 1. ( ) = 1. iv) Graph to check your answer. Since the point of interest is x=1.. What is y? Note this is the y value representing V for the graph, not the y of the problem. Page 15 of 19
Go back to the HOME screen to find y. The calculator knows the function y1 since you just typed it in so perform the following calculation. So we need YMAX at least 7.. I will choose 15. I will pick the following window. Now graph then take the minimum (F5/3). Page 16 of 19
i) The equations. We need to create two formulas. To do this we need to label some more sides. First note that u = v since the wih of the box will be the same as the length of the lid. The full wih is W = x + v + x + u = x + v = 15 The full height is H = x + w + x = x + w = 10 The volume of the box is V = x w v What do we actually need? Since V = x w v, we need w in terms of x from x + w = 10 and v in terms of x from x + v = 15. The equations are then V = x w v v = 15 Plugging into V we get V = x 10 x ( ) 15 x x w = 10 x = x 3 5x + 75x Page 17 of 19
where I used the TI expand function to get the product. ii) Find the two derivatives of V. dv = 6x 50x + 75 d V = 1x 50 iii) Solve for the location of the minimum. dv = 6x 50x + 75 = 0 Which one is it? Plug the answers into one of the equations above. w = 10 6.37 ( ) =.74 A dimension cannot be negative so x = 1.961. iv) Check the second derivative. Is this a maximum or a minimum? d V = 1 1.961 It's a maximum. ( ) 50 = 6.5 < 0 The question was to find x so we are done. Page 18 of 19
i) The equations: Height: d where d is the diameter of the semicircles Perimeter: P = x + πr = x + πd = 400 where r = d Area: A = xd + πr Solve the perimeter equation for d: πd = 400 x d = 400 x π Plug into the area equation of just the rectangle 400 x A = xd = x π = ( 00x x ) π ii) Find the two derivatives of A. da = ( 00 x ) π = 4 ( π 100 x) d A dr = 4 π < 0 maximum iii) Solve for the location of the maximum. da = 4 ( π 100 x) = 0 x = 100m Page 19 of 19