Implicit Differentiation and Related Rates

Transcription

1 Implicit Differentiation an Relate Rates Up until now ou have been fining the erivatives of functions that have alrea been solve for their epenent variable. However, there are some functions that cannot be easil solve for the epenent variable so we nee to have a wa of still fining the erivative. This process is calle implicit ifferentiation. Fining the erivative of a function b implicit ifferentiation uses the same erivative formulas that were covere earlier. The important part to remember is that when ou take the erivative of the epenent variable ou must inclue the erivative notation /x or in the erivative. The notation that is use epens on which is easier for ou. Let s take the function + 3x as an example. This function can easil be solve for the epenent variable, but lets look at how implicit ifferentiation works. The first term is the prouct of an so we woul appl the prouct rule. First we woul take the erivative of each term an then substitute into the prouct rule. ( x) ( ) or x x x x ( x) ( x)( ) + ( )() x + Taking the erivatives of 3x an woul be one in the same manner as before. So the implicit erivative woul be: + 3x ( x) + (3 x) () x x x Now we can solve for Geral Manahan SLAC, San Antonio College, 008

2 Therefore, the steps involve in using implicit ifferentiation to fin /x are:. Differentiate each term on both sies of the equals sign with respect to the inepenent variable x. When taking the erivative of the epenent variable on t forget to inclue /x or in the erivative.. Solve for /x or b grouping onl the terms with /x or on one sie of the equals sign 3. Factor out /x or 4. Divie to isolate /x or Let s look at a few examples. Example : Fin the erivative of x 3 + 4x using implicit ifferentiation. Solution: Step : Differentiate each term on both sies of the equals sign + 3 x 4x x x x 3 ( x ) + ( 4x) ( ) Appl prouct rule to both terms on the left x x x x 3 3 ( x ) ( ) + ( ) ( x ) + ( 4x) ( ) + ( ) ( 4x) Appl the chain rule x x 3 ( x ) ( 3)( ) + ( ) + ( 4x) ( ) + ( ) ( ) Step : Group all terms with /x on one sie + x + x + x x 3x + 4x x x x x Step 3: Factor /x x ( ) 3x + 4x Geral Manahan SLAC, San Antonio College, 008

3 Example (Continue): Step 4: Divie to isolate /x 3 4 x 3x + 4x Example : Fin the erivative of x 3 6x using implicit ifferentiation. Solution: Step : Differentiate each term on both sies of the equals sign 4+ 5x x 3 x 6x We can simplif the right sie b using the properties of exponents x 6 x 3 x ( x) Appl quotient rule to the left sie an the generalize power rule to the right sie 4+ 5x ( 6x) x 3 x x x x x x ( 3) ( 4+ 5x) ( 4+ 5x) ( 3) ( 3) x x ( ) ( x) ( 3) ( )( 6 ) ( 6 ) ( )( 6x) ( 6) In this example we will replace the /x notation with ( 3)( 4 + 5) ( 4+ 5x)( 3 ) ( 3) ( )( 6x) ( 6) Geral Manahan SLAC, San Antonio College, 008 3

4 Example (Continue): Step : Group all terms with on one sie ( 3)( 4 + 5) ( 4+ 5x)( 3 ) ( )( 6x) ( 6) ( 3) ( ) ( 5x ) 6 ( 3) ( 6x) x 6 ( 3) ( 3) x 6 ( ) ( ) x x Step 3: Factor ( ) x ( ) ( 4+ 5x) + Step 4: Divie to isolate ( ) x ( ) x ( ) x ( ) ( 4 + 5x) We can also fin the equation of a tangent line at a given point using implicit ifferentiation. The steps involve in o this are:. Fin the erivative using implicit ifferentiation. If both the x an coorinates are not known fin the missing coorinate 3. Substitute the x an coorinates into the erivative to fin the slope of the tangent line 4. Fin the equation of the tangent line using the point-slope formula Geral Manahan SLAC, San Antonio College, 008 4

5 x Example 3: Fin the equation of the tangent line when x 4 for the curve + 3. Solution: Step : Fin the erivative using implicit ifferentiation x + 3 x x x x x x 0 ( ) ( 3) x 0 x 0 x 0 + x 0 + x ( + x ) + x ( ) + x Geral Manahan SLAC, San Antonio College, 008 5

6 Example 3 (Continue): Step : Fin an missing coorinates. In this problem we know the value for x but not. So we must fin the corresponing value when x. + x or 0 ( )( ) Step 3: Fin slope of tangent line x 4, x 4, 4 4 () + ( ) + 8 ()( ) x + x x + x 4 4 ( ) + ( ) + 8 ( )( ) Geral Manahan SLAC, San Antonio College, 008 6

7 Example 3 (Continue): Step 4: Fin the equation of the tangent lines x 4,, / ( ) m x x 4 x 3 ( x ) x + 3 x+ 3 x 4,, /8 ( ) m x x 4 8 x 8 x x+ 8 ( x ) Another application for implicit ifferentiation is the topic of relate rates. Relate rates are use to etermine the rate at which a variable is changing with respect to time. We use the concept of implicit ifferentiation because time is not usuall a variable in the equation. For example, if we were aske to etermine the rate at which the area of a square is changing then implicit ifferentiation must be use because the equation for the area of a square onl contains the variables for the length, with, an area. Time is not a variable in the equation so the onl wa to etermine the rate at which the area is changing (A/t) is to take the erivative implicitl. Geral Manahan SLAC, San Antonio College, 008 7

8 The steps involve in solving a relate rates problem can be summarize as:. Ientif all given information an what we must fin.. Draw a sketch if it is possible 3. Determine the equation that relates the variables 4. Fin the erivative using implicit ifferentiation 5. Solve the erivative for the unknown rate 6. Substitute in the given information an solve Example 4: A 50-ft laer is place against a builing. The top of the laer is sliing own the builing at the rate of ft/min. Fin the rate at which the base of the laer is moving awa from the builing at the instant that the base is 30 ft from the builing. Solution: Step : Ientif all given information an what we must fin Length of laer (c) 50 ft Initial height of laer (a)? Rate laer is sliing own (a/t) - ft/min* Distance of base from builing (b) 30 ft Rate base is moving awa (b/t)? * The rate the laer is sliing own is shown as a negative value to inicate the irection in which the top of the laer is moving. Step : Draw a sketch Geral Manahan SLAC, San Antonio College, 008 8

9 Example 4 (Continue): Step 3: Determine the equation that relates the variables In this problem are variables are the sies the triangle forme b the laer (c), the builing (a), an the groun (b). An equation that woul relate these three variables is the Pthagorean theorem, a + b c. Step 4: Fin the erivative using implicit ifferentiation a + b c t t a b c a + b c t t t ( a + b ) ( c ) Step 5: Solve the erivative for the unknown rate a b c a + b c t t t b c a b c a t t t c a c a b t t t b Step 6: Substitute in the given information an solve Since c represents the length of the laer an the length is not going to change the relate rate c/t woul be equal to zero. Also we o not know the value of a so we must fin it first before fining b/t. a a a + b c ( 30) ( 50) a 600 a 600 a 40 Geral Manahan SLAC, San Antonio College, 008 9

10 Example 4 (Continue): Now we can solve for b/t. c a c a b t t t b ( )( ) ( )( ) 30 ( ) b/t is approximatel equal to.67 ft/min. Example 5: A rock is thrown into a still pon. The circular ripples move outwar from the point of impact of the rock so that the raius of the circle forme b a ripple increases at the rate of ft per minute. Fin the rate at which the area is changing at the instant the raius is 4 ft. Solution: Step : Ientif all given information an what we must fin Rate the raius is changing (r/t) ft/min Raius (r) 4 ft Rate the area of the circle is changing (A/t)? Step : Draw a sketch Step 3: Determine the equation that relates the variables In this example we are ealing with the area of a circle, therefore the equation that we will use woul be A Π r Geral Manahan SLAC, San Antonio College, 008 0

11 Example 5 (Continue): Step 4: Fin the erivative using implicit ifferentiation t A π r ( A) ( π r ) t A r π r t t Step 5: Solve the erivative for the unknown rate The erivative is alrea solve for the unknown rate so we can go to the next step. Step 6: Substitute in the given information an solve A r π r t t π 4 ( )( ) 6π A/t is approximatel equal to 50.3 ft/min. Example 6: The revenue function for a compan is 60x -.5x an its cost function is 8x +, where x is the ail prouction an sales. If the ail prouction is currentl 0 units an the rate of change for prouction is 6 units per a, fin the rate at which the compan s profit is changing. Solution: Step : Ientif all given information an what we must fin Revenue function R(x) 60x -.5x Cost function C(x) 8x + x 0 x/t 6 Profit function? P/t? Geral Manahan SLAC, San Antonio College, 008

12 Example 6 (Continue): Step : Draw a sketch when it is possible With this tpe of problem a sketch is not use. Step 3: Determine the equation that relates the variables Since we nee to fin the rate at which the profit is changing we nee to fin the Profit function. The profit function woul be equal to the ifference between the revenue an cost functions. P(x) R(x) C(x) 60x -.5x (8x + ) 60x -.5x 8x -.5x + 5 Step 4: Fin the erivative using implicit ifferentiation P x + x.5 5 P x + x t t P x x x t t t P x x x + 5 t t t (.5 5 ) Step 5: Solve the erivative for the unknown rate The equation is alrea solve for P/t. Step 6: Substitute in the given information an solve P x x x + 5 t t t + ( 0)( 6) 5( 6) Geral Manahan SLAC, San Antonio College, 008