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EXERCISES 3.3 In Exercises 1-18, suppose that x and y are related by the given dy equation and use implicit differentiation to determine 1. x²-y²=1 3. y³ - 3x² = x 2.x²³+y³-6=0 4. x++ (y + 3) = x 5. y-x=y²-x² 7. 2x³ + y = 2y³ + x 9. xy = 5 11. x(y + 2)³=8 6. x² + y² = x² + y²2 8. x² + 4y = x - 4y³ 10. xy³3 =2 12. x²y³ = 6

EXERCISES 3.3 In Exercises 1-18, suppose that x and y are related by the given dy equation and use implicit differentiation to determine 1. x²-y²=1 3. y³ - 3x² = x 2.x²³+y³-6=0 4. x++ (y + 3) = x 5. y-x=y²-x² 7. 2x³ + y = 2y³ + x 9. xy = 5 11. x(y + 2)³=8 6. x² + y² = x² + y²2 8. x² + 4y = x - 4y³ 10. xy³3 =2 12. x²y³ = 6

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Ex 3.3 Q7,9 and Q11 Needed to be solved correctly in 30 minutes and get the thumbs up please show me neat and clean work for it by hand solution needed please do all three in the order to get positive feedback
!!!! <o > O E A₁ X INCORPORATING SOLUTION Assume that p and x are differentiable functions of t, and differentiate the demand equation with respect to t: TECHNOLOGY 1. x²-y²=1 3. 5-3x² = x dx dp dp We want to know at a time when x = 4, p = 6, and -=-.40. (The derivative 18 dt dt dt 3.3 Implicit Differentiation and Related Rates 215 dx negative because the price is decreasing.) We could solve equation (4) for and dt dx dt' then substitute the given values, but since we do not need a general formula for easier to substitute first and then solve: d d d d (p) + (2x) + (xp) = (38) dt dt dt dp dx dp dx dt dt +2 +x +p== 0. dt dt dt Check Your Understanding 3.3 Suppose that x and y are related by the equation 3y2 - 3x² + y = 1. 1. Use implicit differentiation to find a formula for the slope of the graph of the equation. EXERCISES 3.3 In Exercises 1-18, suppose that x and y are related by the given dy equation and use implicit differentiation to determine dx dx dt Thus, sales are rising at the rate of .25 thousand units (or 250 units) per week. 2. x² + y²-6=0 4. x² + (y + 3) = x² dx -40+25 +4(-40) + 6- dt dx 8 = 2 dt dx dt 215 =0 5. Substitute all specified values for the variables and their derivatives. 6. Solve for the derivative that gives the unknown rate. = 25. Suggestions for Solving Related-Rates Problems 1. Draw a picture, if possible. 2. Assign letters to quantities that vary, and identify one variable-say, t-on which the other variables depend. 3. Find an equation that relates the variables to each other. 4. Differentiate the equation with respect to the independent variable t. Use the chain rule whenever appropriate. (4) The graph of an equation in x and y can be easily obtained when y can be expressed as one or more functions of x. For instance, the graph of x² + y² = 4 can be plotted by simultaneously graphing y= V4- x² and y = -√4-x². 5. y* - x^4 = y? - xả 7. 2x³ + y = 2y³ + x 9. xy=5 11. x(y + 2)² = 8 >> Now Try Exercise 39 it is Solutions can be found following the section exercises. 2. Suppose that x and y in the preceding equation are both functions of 1. Differentiate both sides of the equation with dy respect to t, and find a formula for in terms of x, y, and dt dx dt 6. x² + y² = x² + y² 8. x4 + 4y = x - 4y³ 10. xy³ = 2 12. x²y³ = 6
Transcribed Image Text:!!!! <o > O E A₁ X INCORPORATING SOLUTION Assume that p and x are differentiable functions of t, and differentiate the demand equation with respect to t: TECHNOLOGY 1. x²-y²=1 3. 5-3x² = x dx dp dp We want to know at a time when x = 4, p = 6, and -=-.40. (The derivative 18 dt dt dt 3.3 Implicit Differentiation and Related Rates 215 dx negative because the price is decreasing.) We could solve equation (4) for and dt dx dt' then substitute the given values, but since we do not need a general formula for easier to substitute first and then solve: d d d d (p) + (2x) + (xp) = (38) dt dt dt dp dx dp dx dt dt +2 +x +p== 0. dt dt dt Check Your Understanding 3.3 Suppose that x and y are related by the equation 3y2 - 3x² + y = 1. 1. Use implicit differentiation to find a formula for the slope of the graph of the equation. EXERCISES 3.3 In Exercises 1-18, suppose that x and y are related by the given dy equation and use implicit differentiation to determine dx dx dt Thus, sales are rising at the rate of .25 thousand units (or 250 units) per week. 2. x² + y²-6=0 4. x² + (y + 3) = x² dx -40+25 +4(-40) + 6- dt dx 8 = 2 dt dx dt 215 =0 5. Substitute all specified values for the variables and their derivatives. 6. Solve for the derivative that gives the unknown rate. = 25. Suggestions for Solving Related-Rates Problems 1. Draw a picture, if possible. 2. Assign letters to quantities that vary, and identify one variable-say, t-on which the other variables depend. 3. Find an equation that relates the variables to each other. 4. Differentiate the equation with respect to the independent variable t. Use the chain rule whenever appropriate. (4) The graph of an equation in x and y can be easily obtained when y can be expressed as one or more functions of x. For instance, the graph of x² + y² = 4 can be plotted by simultaneously graphing y= V4- x² and y = -√4-x². 5. y* - x^4 = y? - xả 7. 2x³ + y = 2y³ + x 9. xy=5 11. x(y + 2)² = 8 >> Now Try Exercise 39 it is Solutions can be found following the section exercises. 2. Suppose that x and y in the preceding equation are both functions of 1. Differentiate both sides of the equation with dy respect to t, and find a formula for in terms of x, y, and dt dx dt 6. x² + y² = x² + y² 8. x4 + 4y = x - 4y³ 10. xy³ = 2 12. x²y³ = 6
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