3.6 Implicit Differentiation. explicit representation of a function rule. y = f(x) y = f (x) e.g. y = (x 2 + 1) 3 y = 3(x 2 + 1) 2 (2x)

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1 Mathematics 0110a Summar Notes page Implicit Differentiation eplicit representation of a function rule = f() = f () e.g. = ( + 1) 3 = 3( + 1) () implicit representation of a function rule f(, ) = c To find d d : 1. differentiate term b term. appl the chain rule 3. solve for d d Eample 1: Given + = 4. Find d d. solution: Solving for is not convenient. + = (diff. term b term, use chain rule) ( 4) = 4 (collect terms on the left) = 4 4 = (solve for )

2 Mathematics 0110a Summar Notes page 44 Eample : Find the slope of the tangent line to the curve = at the point (1, 1). solution: = 1 = 1 The slope of the tangent line here is m = (1) = 1/. = The relation governed b = is not a function but rather a confluence of two functions: = and =. The point (1, 1) lies on the function =. Implicit differentiation automaticall takes care of the details (placing the tangent line on the proper branch) when the value of is known. Eamples: Find d d in each of the following epressions. 3. = = 5. ( ) 3 = 10 Eample 6: Given =. Find. solution: Solving for is not convenient. Step 1. Find using implicit differentiation = 0 (diff. term b term, use chain rule)

3 Mathematics 0110a Summar Notes page 45 ( + ) = 7 3 (collect terms on the left) = 7 3 ( + ) (solve for ) Note that d d is epressed in terms of both and. Step. Repeat Step 1., to find (in terms of, and ) (use quotient rule) 6 ( + ) (7 3 )(+ ) = ( ) 4( + ) Step 3. Substitute for using the result from Step 1. = ( ) (7 3 ) ( ) + + ( ) + + 4( + ) The procedure should be clear without the need for further simplification. The idea here is simpl to epress in terms of and onl. Man relations with described implicitl are not functions but it is possible to find the slope of a tangent line because values of and identif the branch of the curve that is a function. Eample 7: Find the slope of the tangent line to the curve described b + = 4 at the point (, ). solution: + = 0 = ; so (, ) = 1. We could also have proceeded eplicitl. = 4 identifies the branch we require and is a function. It is eas to show that ( ) = 1 as before although implicit differentiation makes this easier to appl.

4 Mathematics 0110a Summar Notes page 46 Eample 8: Find using the relation in Eample 7. solution: = = = + 3 Related Rates characteristics of the problem 1. Two or more variables are related to one another b an equation.. Each of these variables are also functions of time. 3. The rate of change of a number (usuall all but one) of these variables with respect to time is known. 4. The objective is to determine the rate of change of the remaining dependent variable with respect to time. The steps necessar to solve a related rates problem are best demonstrated using an eample. Eample 9. A spotlight on the ground shines on a wall 1 m awa. If a man m tall walks from the spotlight toward the building at a speed of 1.6 m/s, how fast is his shadow on the building decreasing when he is 4 m from the building? solution: 1. Draw and label a diagram, identifing the variables of the problem. 1. Write down the remaining numerical information in terms of the appropriate variables. d = 1.6 m/s. At 4 m from the wall, = 8 m. When = 8 using similar triangles: 8/ = 1/ = 3.

5 Mathematics 0110a Summar Notes page State the objective. Find d when = 8 m. 4. Write an equation relating the variables. This is b far the most crucial step. Note that the small triangle is similar to the large one. Thus, 1 =. So = Differentiate term b term with respect to t. d + d = 0 6. Substitute for the known quantities and find the required rate. 3(1.6) = 8 d d = 3(1.6)/8 = 0.6 m/s. The shadow is decreasing at a rate of 0.6 m/s when the man is 4 m from the wall. Eample 10. A spherical balloon is being filled with water at the rate of liters per minute. At what rate is the surface area increasing in cm when the radius is 10 cm? solution: 1. Let r = radius, V = volume and S = surface area.. We want ds when r = S = 4πr = 400π at r = ds = 8πr dr. dr Clearl we need. 5. V = 4 3 πr3 dv = 4πr dr = S dr 000 = 400π dr at r = 10.

6 Mathematics 0110a Summar Notes page 48 So dr = π = 5 π at r = 10. Therefore, ds = 80π 5 π = 400 cm /min., when r = 10.