Guidelines for implicit differentiation

Transcription

1 Guidelines for implicit differentiation Given an equation with x s and y s scattered, to differentiate we use implicit differentiation. Some informal guidelines to differentiate an equation containing x s and y s with respect to x are as follows. Implicit Differentiation (with respect to x) 1 Differentiate both sides as usual, 2 Whenever you do the derivative of terms with y, multiply by dy dx. The second step is essentially the chain rule thinking of y as a function of x. With respect to t the guidelines are: Implicit Differentiation (with respect to t) 1 Differentiate both sides as usual, 2 Whenever you do the derivative of terms with x, multiply by dx, 3 Whenever you do the derivative of terms with y, multiply by dy. Calculus I (University of Calgary) Fall / 17

2 Example Thinking of x and y as functions of t, differentiate the following equation with respect to t: x 2 + y 2 = 100. Solution: Using the Chain Rule we have: 2x dx Example If y = x 3 + 5x and dx = 7, find dy when x = y dy = 0. Solution: Differentiating each side with respect to t gives: When x = 1 and dx dy = 7 we have: dy = 3x 2 dx + 5dx. = 3(1 2 )(7) + 5(7) = = 56 Calculus I (University of Calgary) Fall / 17

3 Related Rates Problem: A balloon is being filled with air so that the radius is increasing at the rate of. How fast is the volume changing when the radius is 2.5 inches? Blow up a balloon for two seconds. Calculus I (University of Calgary) Fall / 17

4 Related Rates Problem: A balloon is being filled with air so that the radius is increasing at the rate of. How fast is the volume changing when the radius is 2.5 inches? Blow up a balloon for two seconds. Measure the change in radius (using string). Calculus I (University of Calgary) Fall / 17

5 Related Rates Problem: A balloon is being filled with air so that the radius is increasing at the rate of. How fast is the volume changing when the radius is 2.5 inches? Blow up a balloon for two seconds. Measure the change in radius (using string). In this example, the radius increased by 2 inches over 2 seconds. Calculus I (University of Calgary) Fall / 17

6 Related Rates Problem: A balloon is being filled with air so that the radius is increasing at the rate of 2 inches/2 seconds. How fast is the volume changing when the radius is 2.5 inches? Blow up a balloon for two seconds. Measure the change in radius (using string). In this example, the radius increased by 2 inches over 2 seconds. This gives the rate of change of radius over time. Calculus I (University of Calgary) Fall / 17

7 Example A balloon is being filled with air so that the radius is increasing at the rate of 1 inch per second. How fast is the volume changing when the radius is 2.5 inches? Solution: The volume of a sphere is V = 4 3 πr 3. The word fast means we need compute a rate of change. In particular, we are asked to compute dv. Differentiating implicitly with respect to t (time) gives: dv = 4 3 π 3r 2 dr Given: r = 2.5 inches and dr = 1 inch / second: dv = 4 3 π 3(2.5 inches)2 (1 inch/second) = 78.5 inches 3 /second Therefore, the volume is changing at a rate of 78.5 cubic inches per second when the radius is 2.5 inches. Calculus I (University of Calgary) Fall / 17

8 Steps to solve related rates problems The following guidelines help to solve related rates problems. Related Rates 1 Draw a diagram if possible. 2 Introduce notation and assign symbols to quantities depending on time. 3 Write an equation relating the various quantities (perhaps use geometry). 4 Differentiate both sides of the equation with respect to t implicitly. 5 Substitute given information into the resulting rate equation. Note on Units: The units are important: make sure that the quantities are expressed in compatible units. For example, if r is in cm, and t is in minutes, then dr should be in cm/min. Calculus I (University of Calgary) Fall / 17

9 Example A 10 ft long ladder rests against a wall. Suppose the bottom of the ladder slides away from the wall at a rate of 1 ft/s. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? Solution: We first draw a picture: Calculus I (University of Calgary) Fall / 17

10 Introduce notation using variables for quantities that change over time: Let x be the distance from the bottom of the ladder to the wall (measured in feet). Let y be the distance from from the top of the ladder to the floor (measured in feet). Note that both x and y are changing over time as the ladder is sliding down the wall. Since the length of the ladder does not change over time, we do not assign a variable to it. Let us label what is known in our triangle. We know that the ladder is 10 ft long, thus we have: Calculus I (University of Calgary) Fall / 17

11 We now write an equation relating x and y. By using the Pythagorean Theorem we get: Original Equation: x 2 + y 2 = 10 2 We differentiate this equation implicitly with respect to t: Rate Equation: 2x dx + 2y dy = 0 Calculus I (University of Calgary) Fall / 17

12 Original Equation: x 2 + y 2 = 10 2 Rate Equation: 2x dx + 2y dy = 0 A 10 ft long ladder rests against a wall. Suppose the bottom of the ladder slides away from the wall at a rate of 1 ft/s. How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? The bottom of the ladder slides away from the wall at a rate of 1 ft/s The above sentence is saying that the rate of change in x is 1, in other words: dx = 1. Note that we use positive 1 since x gets bigger over time (increases). How fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall? dy We want when x = 6. Calculus I (University of Calgary) Fall / 17

13 Original Equation: x 2 + y 2 = 10 2 Rate Equation: We want to solve for dy, so we still need y. x = 6 2x dx dy + 2y = 0 dx = y 2 = 10 2 y 2 = = 64 We now plug it all in dy y = ±8 y = 8 (y is a length). = 0 16dy dy = 0 = 12 dy = = 3 4 ft/sec Therefore, the top of the ladder is sliding down at a rate of 3/4 ft/s. Calculus I (University of Calgary) Fall / 17

14 Example An oil tank has the shape of an inverted circular cone with base radius 2 m and height 6 m. If oil is being pumped into the tank at a rate of 3 m 3 /min, find the rate at which the oil level is rising when the oil is 4 m deep. Solution: We start with a diagram: Calculus I (University of Calgary) Fall / 17

15 Now we introduce notation using variables for quantities that change over time: Let V be the volume of the oil at time t (t measured in minutes). Let r be the radius of the surface at time t (t measured in minutes). Let h be the height of the oil at time t (t measured in minutes). We now label our diagram. One equation we have is the volume of the smaller cone: V = 1 3 πr 2 h. It is easier to do this question if we find another equation to eliminate one of our variables. Calculus I (University of Calgary) Fall / 17

16 Note that our figure has two triangles: By similar triangles we have: r h = 2 6 r = h 3. If you don t like similar triangles you can also compute tan θ in both triangles to get: Hence we have the equation: tan θ = r h = 2 6 V = 1 3 π ( h 3 ) 2 h V = π 27 h3 Calculus I (University of Calgary) Fall / 17

17 Differentiating both sides of V = π 27 h3 with respect to t gives: dv = π 9 h2 dh. The question states oil is being pumped into the tank at a rate of 3 m 3 /min. Thus, dv = 3 (positive rate since volume increases over time). We are asked find the rate at which the oil level is rising when the oil is 4 m deep : That is, compute dh when h = 4. Substituting the given information gives: 3 = π 9 (42 ) dh, Solving for dh gives: dh = 27 16π Since the rate dh is positive, the oil level is rising. Therefore, the oil level is rising at a rate of 27/(16π) m/min 0.54 m/min. Calculus I (University of Calgary) Fall / 17