Coordinate Geometry in the (x,y) plane

Coordinate Geometry in the (x,y) plane You can define x and y coordinates on a curve using separate functions, x = f(t) and y = g(t) The letter t is used as the primary application of this is in Physics, with t representing time. Draw the curve given by the Parametric Equations: for t -3 -2 -1 1 2 3 x = 2t -6 -4 -2 2 4 6 y = t2 9 4 1 1 4 9 10 𝑥=2𝑡 𝑦= 𝑡 2 -10 10 −3≤ 𝑡 2 ≤3 Work out the x and y co-ordinates to plot by substituting the t values -10 2A

Coordinate Geometry in the (x,y) plane You can define x and y coordinates on a curve using separate functions, x = f(t) and y = g(t) A curve has parametric equations: Find the Cartesian equation of the curve 𝑥=2𝑡 Divide by 2 𝑥 2 =𝑡 𝑥=2𝑡 𝑦= 𝑡 2 𝑦= 𝑡 2 Replace t with x/2 𝑦= 𝑥 2 2 Rewrite t in terms of x, and then substitute it into the y equation… Simplify 𝑦= 𝑥 2 4 2A

Coordinate Geometry in the (x,y) plane 𝑥= 1 𝑡+1 You can define x and y coordinates on a curve using separate functions, x = f(t) and y = g(t) A curve has parametric equations: Find the Cartesian equation of the curve Multiply by (t + 1) 𝑥(𝑡+1)=1 Divide by x 𝑡+1= 1 𝑥 𝑥= 1 𝑡+1 Subtract 1 𝑦= 𝑡 2 𝑡= 1 𝑥 −1 𝑦= 𝑡 2 Rewrite t in terms of x, and then substitute it into the y equation… Replace t 𝑦= 1 𝑥 −1 2 𝑦= 1 𝑥 − 𝑥 𝑥 2 1 = x/x Put the fractions together 𝑦= 1−𝑥 𝑥 2 𝑦= (1−𝑥) 𝑥 2 2 Simplify 2A

Exercise 2A

Coordinate Geometry in the (x,y) plane You need to be able to use Parametric equations to solve problems in coordinate geometry The diagram shows a sketch of the curve with Parametric equations: The curve meets the x-axis at the points A and B. Find their coordinates. A B 𝑥=𝑡−1 𝑦=4− 𝑡 2 𝑦=4− 𝑡 2 y = 0 0=4− 𝑡 2 + t2 At points A and B, y = 0  Sub y = 0 in to find t at these points 𝑡 2 =4 Remember both answers 𝑡=±2 So t = ±2  Sub this into the x equation to find the x-coordinates of A and B 𝑥=𝑡−1 𝑥=𝑡−1 Sub in t = 2 Sub in t = -2 𝑥=(2)−1 𝑥=(−2)−1  A and B are (-3,0) and (1,0) 𝑥=1 𝑥=−3 2B

Coordinate Geometry in the (x,y) plane You need to be able to use Parametric equations to solve problems in coordinate geometry A curve has Parametric equations: where a is a constant. Given that the curve passes through (2,0), find the value of a. 𝑦=𝑎( 𝑡 3 +8) y = 0 0=𝑎( 𝑡 3 +8) The ‘a’ part cannot be 0 or there would not be any equations! 𝑡 3 +8=0 Subtract 8 𝑡 3 =−8 𝑥=𝑎𝑡 𝑦=𝑎( 𝑡 3 +8) Cube root 𝑡=−2 So the t value at the coordinate (2,0) is -2 We know there is a coordinate where x = 2 and y = 0, Sub y = 0 into its equation 𝑥=𝑎𝑡 Sub in x and t values (2)=𝑎(−2) Since we know that at (2,0), x = 2 and t = -2, we can put these into the x equation to find a Divide by -2 −1=𝑎 𝑥=𝑎𝑡 𝑦=𝑎( 𝑡 3 +8) The actual equations with a = -1 𝑥=−𝑡 𝑦=−( 𝑡 3 +8) 2B

Coordinate Geometry in the (x,y) plane You need to be able to use Parametric equations to solve problems in coordinate geometry A curve is given Parametrically by: The line x + y + 4 = 0 meets the curve at point A. Find the coordinates of A. 𝑥= 𝑡 2 𝑦=4𝑡 𝑥+𝑦+4=0 Replace x and y with their equations ( 𝑡 2 )+(4𝑡)+4=0 ‘Tidy up’ 𝑥= 𝑡 2 𝑦=4𝑡 𝑡 2 +4𝑡+4=0 Factorise (𝑡+2) 2 =0 Find t 𝑡=−2 The first thing we need to do is to find the value of t at coordinate A  Sub x and y equations into the line equation So t = -2 where the curve and line meet (point A) 𝑥= 𝑡 2 𝑦=4𝑡 Sub t = -2 Sub t = -2 𝑥= (−2) 2 𝑦=4(−2) We know an equation for x and one for y, and we now have the t value to put into them… 𝑥=4 𝑦=−8 The curve and line meet at (4, -8) 2B

Exercise 2B

Coordinate Geometry in the (x,y) plane You need to be able to convert Parametric equations into Cartesian equations A Cartesian equation is just an equation of a line where the variables used are x and y only A curve has Parametric equations: a) Find the Cartesian equation of the curve b) Sketch the curve 𝑥=𝑠𝑖𝑛𝑡+2 𝑦=𝑐𝑜𝑠𝑡−3 Isolate sint Isolate cost 𝑥−2=𝑠𝑖𝑛𝑡 𝑦+3=𝑐𝑜𝑠𝑡 𝑠𝑖 𝑛 2 𝑡+𝑐𝑜 𝑠 2 𝑡≡1 Replace sint and cost (𝑥−2 ) 2 + (𝑦+3 ) 2 = 1 𝑥=𝑠𝑖𝑛𝑡+2 𝑦=𝑐𝑜𝑠𝑡−3 (𝑥−2 ) 2 +(𝑦+3 ) 2 =1 5 We need to eliminate t from the equations Centre = (2, -3) Radius = 1 -5 5 The equation is that of a circle  Think about where the centre will be, and its radius -5 2C

Coordinate Geometry in the (x,y) plane You need to be able to convert Parametric equations into Cartesian equations A Cartesian equation is just an equation of a line where the variables used are x and y only A curve has Parametric equations: a) Find the Cartesian equation of the curve 𝑥=𝑠𝑖𝑛𝑡 𝑦=𝑠𝑖𝑛2𝑡 Use the double angle formula from C3 Square 𝑥 2 =𝑠𝑖 𝑛 2 𝑡 𝑦=2𝑠𝑖𝑛𝑡𝑐𝑜𝑠𝑡 Replace sint with x 𝑦=2𝑥𝑐𝑜𝑠𝑡 𝑠𝑖 𝑛 2 𝑡+𝑐𝑜 𝑠 2 𝑡≡1 Rearrange 𝑐𝑜 𝑠 2 𝑡=1−𝑠𝑖 𝑛 2 𝑡 Replace sin2t with x2 𝑥=𝑠𝑖𝑛𝑡 𝑦=𝑠𝑖𝑛2𝑡 𝑐𝑜 𝑠 2 𝑡=1− 𝑥 2 Square root 𝑐𝑜𝑠𝑡= 1− 𝑥 2 We need to eliminate t from the equations 𝑦=2𝑥𝑐𝑜𝑠𝑡 We can now replace cos t 𝑦=2𝑥 1− 𝑥 2 Another way of writing this (by squaring the whole of each side) 𝑦 2 =4 𝑥 2 (1− 𝑥 2 ) 2C

Exercise 2C

Coordinate Geometry in the (x,y) plane You need to be able to find the area under a curve when it is given by Parametric equations The Area under a curve is given by: By the Chain rule: Integrate the equation with respect to x 𝑦 𝑑𝑥 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 Integrate the whole expression with respect to t (Remember you don’t have to do anything with the dt at the end!) y multiplied by dx/dt (x differentiated with respect to t) 2D

Coordinate Geometry in the (x,y) plane 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 𝑥=5 𝑡 2 𝑦= 𝑡 3 You need to be able to find the area under a curve when it is given by Parametric equations A curve has Parametric equations: Work out: Differentiate 𝑑𝑥 𝑑𝑡 =10𝑡 1 2 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 𝑥=5 𝑡 2 𝑦= 𝑡 3 Sub in y and dx/dt 1 2 𝑡 3 (10𝑡) 𝑑𝑡 1 2 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 Multiply 1 2 10 𝑡 4 𝑑𝑡 Remember to Integrate now. A common mistake is forgetting to! = 10 𝑡 5 5 1 2 = 2 𝑡 5 1 2 Sub in the two limits = 2(2 ) 5 − 2(1 ) 5 Work out the answer! = 64 − 2 =62 2D

Coordinate Geometry in the (x,y) plane 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 You need to be able to find the area under a curve when it is given by Parametric equations The diagram shows a sketch of the curve with Parametric equations: The curve meets the x-axis at x = 0 and x = 9. The shaded region is bounded by the curve and the x-axis. Find the value of t when: x = 0 x = 9 b) Find the Area of R R 9 𝑥= 𝑡 2 𝑦=2𝑡(3−𝑡) 𝑡≥0 i) 𝑥= 𝑡 2 ii) 𝑥= 𝑡 2 x = 0 x = 9 0= 𝑡 2 9= 𝑡 2 Square root Square root 0=𝑡 ±3=𝑡 t ≥ 0 3=𝑡 Normally when you integrate to find an area, you use the limits of x, and substitute them into the equation When integrating using Parametric Equations, you need to use the limits of t (since we integrate with respect to t, not x) The limits of t are worked out using the limits of x, as we have just done 2D

Coordinate Geometry in the (x,y) plane 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 𝑦=2𝑡(3−𝑡) You need to be able to find the area under a curve when it is given by Parametric equations The diagram shows a sketch of the curve with Parametric equations: The curve meets the x-axis at x = 0 and x = 9. The shaded region is bounded by the curve and the x-axis. Find the value of t when: x = 0 x = 9 b) Find the Area of R  Limits of t are 3 and 0 𝑥= 𝑡 2 Differentiate R 𝑑𝑥 𝑑𝑡 =2𝑡 9 𝑥= 𝑡 2 𝑦=2𝑡(3−𝑡) 𝑡≥0 0 3 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 Sub in y and dx/dt 0 3 2𝑡 3−𝑡 2𝑡 𝑑𝑡 Multiply 0 3 12 𝑡 2 −4 𝑡 3 𝑑𝑡 INTEGRATE!! (don’t forget!) = 12 𝑡 3 3 − 4 𝑡 4 4 0 3 = 4 𝑡 3 − 𝑡 4 0 3 Sub in 3 and 0 = 4 (3) 3 − (3) 4 − 4 (0) 3 − (0) 4 Work out the answer = 108−81 =27 2D

Coordinate Geometry in the (x,y) plane 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 You need to be able to find the area under a curve when it is given by Parametric equations The diagram shows a sketch of the curve with Parametric equations: Calculate the finite area inside the loop… 8 𝑥= 2𝑡 2 𝑦=𝑡(4− 𝑡 2 ) We have the x limits, we need the t limits 𝑥= 2𝑡 2 𝑥= 2𝑡 2 x = 0 x = 8 0= 2𝑡 2 8= 2𝑡 2 Halve and square root Halve and square root 0=𝑡 ±2=𝑡 Our t values are 0 and ±2 We have 3 t values. We now have to integrate twice, once using 2 and once using -2 One gives the area above the x-axis, and the other the area below (in this case the areas are equal but only since the graph is symmetrical) 2D

Coordinate Geometry in the (x,y) plane 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 You need to be able to find the area under a curve when it is given by Parametric equations The diagram shows a sketch of the curve with Parametric equations: Calculate the finite area inside the loop… 𝑥= 2𝑡 2 𝑦=𝑡(4− 𝑡 2 ) Differentiate 𝑑𝑥 𝑑𝑡 =4𝑡 0 2 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 Sub in y and dx/dt 𝑥= 2𝑡 2 𝑦=𝑡(4− 𝑡 2 ) 0 2 𝑡(4− 𝑡 2 ) (4𝑡) 𝑑𝑡 Multiply out 0 2 16 𝑡 2 −4 𝑡 4 𝑑𝑡 The t limits are 0 and ±2 INTEGRATE!!!! = 16 𝑡 3 3 − 4 𝑡 5 5 0 2 Sub in 2 and 0 = 16 (2) 3 3 − 4 (2) 5 5 − 16 (0) 3 3 − 4 (0) 5 5 8 =17 1 15 At this point we can just double the answer, but just to show you the other pairs give the same answer (as the graph was symmetrical) 2D

Coordinate Geometry in the (x,y) plane 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 You need to be able to find the area under a curve when it is given by Parametric equations The diagram shows a sketch of the curve with Parametric equations: Calculate the finite area inside the loop… 𝑥= 2𝑡 2 𝑦=𝑡(4− 𝑡 2 ) Differentiate 𝑑𝑥 𝑑𝑡 =4𝑡 −2 0 𝑦 𝑑𝑥 𝑑𝑡 𝑑𝑡 Sub in y and dx/dt 𝑥= 2𝑡 2 𝑦=𝑡(4− 𝑡 2 ) −2 0 𝑡(4− 𝑡 2 ) (4𝑡) 𝑑𝑡 Multiply out −2 0 16 𝑡 2 −4 𝑡 4 𝑑𝑡 The t limits are 0 and ±2 INTEGRATE!!!! = 16 𝑡 3 3 − 4 𝑡 5 5 −2 0 Sub in 0 and -2 = 16 (0) 3 3 − 4 (0) 5 5 − 16 (−2) 3 3 − 4 (−2) 5 5 Here you end up subtracting a negative… 8 =17 1 15 Double =34 2 15 2D

Exercise 2D and 2E

Differentiation

You can find the gradient of a curve given in parametric coordinates Differentiation You can find the gradient of a curve given in parametric coordinates When a curve is given in parametric equations: Differentiate y and x with respect to the parameter t We can then use the following calculation, a variation of the chain rule 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 ÷ 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 ÷ 𝑑𝑥 𝑑𝑡 Rewrite as a multiplication Written differently 𝑑𝑦 𝑑𝑡 × 𝑑𝑡 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑥 Cancel dts = 𝑑𝑦 𝑑𝑥 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Differentiation You can find the gradient of a curve given in parametric coordinates 𝑥= 𝑡 3 +𝑡 Find the gradient at the point P where t = 2, on the curve given by parametric equations: 𝑦= 𝑡 2 +1 𝑥= 𝑡 3 +𝑡 𝑦= 𝑡 2 +1 Differentiate Differentiate 𝑑𝑥 𝑑𝑡 =3 𝑡 2 +1 𝑑𝑦 𝑑𝑡 =2𝑡 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Write in terms of t using dy/dt and dx/dt  This is the gradient function for the curve, but in terms of t 𝑑𝑦 𝑑𝑥 = 2𝑡 3 𝑡 2 +1 Sub in t = 2 𝑑𝑦 𝑑𝑥 = 2(2) 3 (2) 2 +1 Work out 𝑑𝑦 𝑑𝑥 = 4 13 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Differentiation You can find the gradient of a curve given in parametric coordinates Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations: 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 We need to find the gradient at point P by differentiating and substituting θ in We then need to find the gradient of the normal We also need the coordinates of x and y at point P Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Differentiate Differentiate 𝑑𝑥 𝑑𝜃 =3𝑐𝑜𝑠𝜃 𝑑𝑦 𝑑𝜃 =−5𝑠𝑖𝑛𝜃 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝜃 𝑑𝑥 𝑑𝜃 −5 3 3 Gradient of the tangent at P: Sub in dx/dθ and dy/dθ 3 3 5 𝑑𝑦 𝑑𝑥 = −5𝑠𝑖𝑛𝜃 3𝑐𝑜𝑠𝜃 Gradient of the normal at P: Sub θ = π/6 in to obtain the gradient at P 𝑑𝑦 𝑑𝑥 = −5𝑠𝑖𝑛 𝜋 6 3𝑐𝑜𝑠 𝜋 6 Work out the fraction 𝑑𝑦 𝑑𝑥 = −5 3 3 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Differentiation You can find the gradient of a curve given in parametric coordinates Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations: 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Gradient of the normal at P: = 3 3 5 We need to find the gradient at point P by differentiating and substituting θ in We then need to find the gradient of the normal We also need the coordinates of x and y at point P Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Sub in P = π/6 Sub in P = π/6 𝑥=3𝑠𝑖𝑛 𝜋 6 𝑦=5𝑐𝑜𝑠 𝜋 6 Work out Work out 𝑦= 5 3 2 𝑥= 3 2 3 2 , 5 3 2 The coordinates of P are: 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Differentiation You can find the gradient of a curve given in parametric coordinates Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations: 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Gradient of the normal at P: = 3 3 5 We need to find the gradient at point P by differentiating and substituting θ in We then need to find the gradient of the normal We also need the coordinates of x and y at point P Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line 3 2 , 5 3 2 The coordinates of P are: 𝑦 − 𝑦 1 =𝑚(𝑥 − 𝑥 1 ) Sub in y1, m and x1 − 5 3 2 3 3 5 𝑥 − 3 2 𝑦 = Multiply by 2 6 3 5 𝑥 − 3 2 2𝑦 − 5 3 = Multiply by 5 (You could multiply by 10 in one step if confident) 𝑥 − 3 2 10𝑦 − 25 3 = 6 3 Multiply the bracket out 10𝑦 − 25 3 = 6𝑥 3 − 9 3 Add 25√3 10𝑦 = 6𝑥 3 + 16 3 Divide by 2 5𝑦 =3𝑥 3 +8 3 4A

Exercise 4A

Differentiation 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 4B You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Implicit effectively means all the terms are mixed up, not necessarily in the form ‘y = …’  This technique is useful as some equations are difficult to arrange into this form… 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 Write dy/dx after differentiating the y term This is written differently  The reason is that as the equation is not written as ‘y =…’, y is not a function of x Differentiate the y term as you would for an x term 6𝑦 𝑑𝑦 𝑑𝑥 For example: 3 𝑦 2 8 𝑑𝑦 𝑑𝑥 8𝑦 This is what happens when you differentiate an equation which starts ‘y =…’ 𝑑𝑦 𝑑𝑥 𝑦 4B

Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find dy/dx in terms of x and y for the following equation: 𝑥 3 + 𝑥 + 𝑦 3 + 3𝑦 = 6 𝑥 3 + 𝑥 + 𝑦 3 + 3𝑦 = 6 Differentiate each part one at a time It would be very difficult to rearrange this into the form ‘y = …’ 3𝑦 2 𝑑𝑦 𝑑𝑥 3 𝑑𝑦 𝑑𝑥 3𝑥 2 + 1 + + = Isolate the terms with dy/dx in 3𝑦 2 𝑑𝑦 𝑑𝑥 3 𝑑𝑦 𝑑𝑥 + = − 3𝑥 2 − 1 Factorise by taking out dy/dx 𝑑𝑦 𝑑𝑥 3𝑦 2 + 3 = − 3𝑥 2 − 1 Divide by (3y2 + 3) 𝑑𝑦 𝑑𝑥 −3 𝑥 2 −1 3𝑦 2 + 3 = You now have a formula for the gradient, but in terms of x AND y, not just x 4B

Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find the value of dy/dx at the point (1,1) where: 4𝑥𝑦 2 −5𝑥=10 4𝑥𝑦 2 −5𝑥=10 Differentiate each one at a time – remember the product rule for the first term!!!! 𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 𝑢=4𝑥 𝑣= 𝑦 2 8𝑥𝑦 𝑑𝑦 𝑑𝑥 + 4𝑦 2 − 5 = 0 𝑑𝑦 𝑑𝑥 = 8𝑥𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑢 𝑑𝑥 =4 𝑑𝑣 𝑑𝑥 =2𝑦 𝑑𝑦 𝑑𝑥 + 4 𝑦 2 Add 5, subtract 4y2 8𝑥𝑦 𝑑𝑦 𝑑𝑥 =5−4 𝑦 2 You can substitute x = 1 and y = 1 now (or rearrange first)  Do not replace the terms in dy/dx 8 𝑑𝑦 𝑑𝑥 =5 − 4 Divide by 8 𝑑𝑦 𝑑𝑥 = 1 8 Keep your workings tidy – you may need to use the product rule on multiple terms as well as factorise – show everything you’re doing! 4B

Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find the value of dy/dx at the point (1,1) where: 𝑒 2𝑥 𝑙𝑛𝑦 = 𝑥+𝑦−2 Give your answer in terms of e. 𝑒 2𝑥 𝑙𝑛𝑦 = 𝑥+𝑦−2 𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 𝑢= 𝑒 2𝑥 𝑣=𝑙𝑛𝑦 Differentiate (again watch out for the product rule! 𝑒 2𝑥 𝑦 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛𝑦 𝑒 2𝑥 =1+ 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑒 2𝑥 𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑢 𝑑𝑥 =2 𝑒 2𝑥 𝑑𝑣 𝑑𝑥 = 1 𝑦 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛𝑦 𝑒 2𝑥 Sub in x = 1 and y = 1 𝑒 2 1 𝑑𝑦 𝑑𝑥 =1+ 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛1 𝑒 2 ln1 = 0 since e0 = 1, cancelling the term out 𝑒 2 𝑑𝑦 𝑑𝑥 =1+ 𝑑𝑦 𝑑𝑥 Subtract dy/dx 𝑒 2 𝑑𝑦 𝑑𝑥 − 𝑑𝑦 𝑑𝑥 =1 Factorise 𝑑𝑦 𝑑𝑥 ( 𝑒 2 −1)=1 Divide by (e2 – 1) 𝑑𝑦 𝑑𝑥 = 1 𝑒 2 −1 4B

Exercise 4B