Related Rates section 3.9

Transcription

1 Related Rate ection 3.9 Iportant Note: In olving the related rate proble, the rate of change of a quantity i given and the rate of change of another quantity i aked for. You need to find a relationhip between the two quantitie to be able to olve the proble. Exaple : The area of a quare i decreaing at the rate 1 when it ide i 570 c. Find the rate of change of it perieter at that oent; decribe it in ter of. Solution. Let u denote the ide by x, the area by A, and the perieter by P. So: P = 4x A = x Along the path, we change all quantitie of length to eter and all quantitie of tie to ute. Then A = x da = x dx (1) For a particular oent we are given da = 1 = ( 1)(60) x = 570 c = 5.7 (1) 60 = (5.7) dx dx = Next Step: P = 4x dp = 4 dx dp = (4) ( ) 60 =

2 Exaple (fro the textbook): A an walk along a traight path at a peed of 1.5. A earchlight i located on the ground 6 fro the path and i kept focued on the an. At what rate i the earchlight rotating when the an i 8 fro the point on the path cloet to the earchlight?. Solution. According to the figure, we are given that dx = 1.5 dθ and we want to find. x 6 = tan θ x = 6 tan θ ( ) dx = (6 ec θ) dθ Subtitute the value dx = 1.5 : 1.5 = (6 ec θ) dθ dθ = ec θ = 1.5 co θ 6 = co θ 4 = ( ) adjacent hypotenue 4 = ( ) = = Page

3 Exaple (fro the textbook): Air i being puped into a phere balloon o that it volue increae at a rate of 100 c3. How fat i the radiu of the balloon i increaing when the diaeter i 50 c?. Solution. We are given that : = 100 c3 We want to find : when r = 5 c So, here i what we do for it : V = 4 3 πr3 = = 4πr ubtitute 100 = 4π(5) = 100 4π(5) = 1 5 π c and thi value i poitive, eaning that the radiu i increaing, a expected. Page 3

4 Exaple : Air i being puped out of a phere balloon o that it volue decreae at a rate of 50 c3. How fat i the radiu of the balloon i decreaing when the diaeter i 10 c?. Solution: Solution. We are given that : = 50 c3 be careful about the negative ign We want to find : when r = 5 c So, here i what we do for it : V = 4 3 πr3 = = 4πr ubtitute 50 = 4π(5) = 50 4π(5) = 1 π c and thi value i negative, eaning that the radiu i decreaing, a expected. Page 4

5 Exaple : The area of a quare i decreaing at the rate 1 when it ide i 570 c. Find the rate of change of it perieter at that oent; decribe it in ter of. Solution: Let u denote the ide by x, the area by A, and the perieter by P. So: P = 4x A = x Along the path, we change all quantitie of length to eter and all quantitie of tie to ute. Then A = x da For a particular oent we are given = x dx (1) da = 1 = ( 1)(60) x = 570 c = 5.7 (1) 60 = (5.7) dx dx = Next Step: P = 4x dp = 4 dx dp = (4) ( ) 60 = Page 5

6 Iportant note 1. Suppoe that f(t) i a function of tie t. When we ay f in decreaing at the rate of 0.3 per econd at tie t = t 0 we ean that the derivative of f with repect to t at tie t 0 i 0.3, i.e. f (t 0 ) = 0.3. and, when we ay that f i increaing at the rate of 1.7 at tie t = t 0, we ean that f (t 0 ) = 1.7. Iportant note. Suppoe that f(t) i a function of tie t and that the unit of eaureent for the value f(t) i eter. Suppoe that the unit of eaureent for t i econd. Then the value f(t) f(t 0 ) are alo eaured in eter, and the value t t 0 are eaured in econd, therefore the unit of eaureent for the quotient f(t) f(t 0 ) t t 0 i eter econd f (t 0 ) = li t t0 f(t) f(t 0 ), and then the unit of eaureent for the liit t t 0 i eter econd. And o on,... Iportant note 3. Ue negative nuber to denote decreaing rate of change. See the following exaple for thi note. Page 6

7 Exaple (FALL ier exa). Sand i poured into a conical pile at a rate of 0 3 per ute. The diaeter of the cone i alway equal to it height. How fat i the height of the conical pile increaing when the pile i 10 high?. Solution: Given: = 0 and r = h dh? when h = 10 V = 1 3 πr h V = 1π ( ) h 3 h = π 1 h3 = π 1 (3h ) dh = π dh h 4 0 = π dh (10) dh = 4 4 5π Page 7

8 Exaple (FALL ier exa). The altitude of a triangle i increaing at a rate of 1 c, while the area i increaing at a rate of c (ee figure below). At what rate i the bae of the triangle changing when the altitude i 10 c and the area i 100 c Solution: Given: dh = 1 and da = db? when h = 10 and A = 100 Step 1. We find the value of all variable at that oent: At that oent we have h = 10 and A = 100 o then: A = 1 b h 100 = 1 (b)(10) b = 0 Step. We differentiate the equation with repect to the tie; we u the prie to refer to the derivative with repect to the tie: A = 1 bh A = 1 {b h + b h } = 1 (b )(10) + 1 (0) (1) b = 1.6 (at that oent) db = 1.6 c Page 8

9 Exaple : A particle i oving along the curve y = + x + 1. At the intant that the oving particle pae through the point (3, 4), it x-coordinate i increaing at a rate of 8 c. How fat i the y-coordinate changing at that oent?. Solution: Step 1. We find the value of all variable at that oent: At that oent we have x = 3 and y = 4. Step. We differentiate the equation with repect to the tie; we u the prie to refer to the derivative with repect to the tie: y = + (x + 1) 1 y = { 1 (x + 1) 1 } (x + 1) = { 1 (x + 1) 1 } (x + 0) = x x + 1 = = So, at that oent we have: dy = Page 9

10 Exaple (Winter Quiz 6). The length of a rectangle i increaing at a rate of 8 c and it wih i increaing at a rate of 3 c. How fat i the area of the rectangle increaing, when the length i 0 c and the wih i 10 c. Solution: Step 1. We find the value of all variable at that oent: At that oent we have x = 10 and y = 0. Step. Let the area be denoted by A. Then A = xy We differentiate the equation with repect to the tie; we u the prie to refer to the derivative with repect to the tie: A = x y + xy A = (3)(0) + (10)(8) = 140 So, at that oent we have: da = 140 c Page 10