14. Functions and Derivatives

1. 14. Functions and Derivatives Objectives: Tangents to curves Rates of Change - Applications of the Chain Rule Refs: B&Z 10.3.

2. Tangents to Curves Find the tangent to the curve y = e2x-1 at the point (x,y) = (1/2, 1). What does this mean? We must determine the equation of a straight line through the point(1/2, 1) which has the same slope as the curve at (1/2, 1).

3. y = e2x-1 Step 1: Differentiate = e2x-1. 2 = 2 e2x-1. dx dx dx dx dy dy dy dy Step 2: Evaluate at the point x=1/2 (1/2) = 2 e2(1/2)-1 = 2e0 = 2. = 2 e2x-1 so This will tell us the gradient of the tangent to the curve at any point x.

4. We have already calculated m=2. So the equation for our tangent is: ytan = 2x + c. Step 3:Remember that the equation for a straight line is y = mx + c. Step 4: We now calculate c. To do this we need to know a point on the line. That’s OK since we were asked to find the tangent at (1/2,1) - so we can use this point. ytan = 2x + c at (1/2,1) so 1 = 2(1/2) + c  c = 0. So ytan = 2x.

5. For a function y = f(x) we may interpret the quotient to be the ratio of the change in y (f(x+h)-f(x)) to the change in x (x+h-x=h). dy dt A familiar example is velocity. If y is the distance travelled in time t, then is the rate of change of distance with respect to time. Rate of Change In this way the derivative is the instantaneous rate of change (of y with respect to x).

6. A plane travels a distance of y kms in time t hours. For any 0 ≤ t ≤ 2, the distance is calculated according to the formula y(t) = 800(t2-1/3 t3). Example What is the velocity of the plane at time t=1? We need to calculate the rate of change of distance with respect to time. y'(t) = 800(2t-t2)for 0 ≤ t ≤ 2. This tells us the velocity at any time0 ≤ t ≤ 2. So y'(1) = 800(2(1)-12) = 800 km/hr when t=1.

7. x dx dt dV dt We want to know (rate of change of volume)when x= 3.0 cm. = 0.1 cm for any t. We know that Applications of the chain rule. A cubic crystal grows so that its side length increases at a constant rate of 0.1cm per day. When the crystal has side length 3.0 cm, at what rate is its volume increasing? V(volume) = x3. The length of the side is changing with time and so is the volume.

8. Now, V=x3  = 3x2 and = 0.1 cm. Now applying the chain rule we have =3x2(0.1) = 0.3 x2 . At x=3.0 cm we have dV dx dx dt dV dt dV dt = 0.3(3.0)2=2.7 cm3 According to the chain rule So when x=3cm, the volume is increasing by 2.7 cm3 per day.

9. Example A 6m ladder is placed against a wall (which is perpendicular to the ground). The top of the ladder is sliding down the wall at a rate of 2/3 m/sec. At what velocity is the bottom of the ladder moving away from the wall when the bottom of the ladder is 3m from the wall?

10. wall 6m y x We want to know dx dt dy dt when x=3m. We know = -2/3 m/sec and by Pythagorus, 62=x2+y2. So when x=3, y2= 27. Also x= (36-y2)1/2. Solution: Both x (distance from ladder to wall) and y (height of ladder from ground) are changing with time (t).

11. Now . = So 1/2(36-y2)-1/2 . (-2y) = -y(36-y2)-1/2. = dx dy dx dt dx dt dx dy dx dt dy dt and -y(36-y2)-1/2 . (-2/3) =2y/3(36-y2)-1/2 (by the chain rule) = Now when x=3, y2= 27 so 2√3 = 2(27)1/2 m/sec (36-27)-1/2 =2√3(9)-1/2 = 3 3

12. dP dt when V = 2.4. We want to know Example Gas in a large container is being compressed by an increasing load on a piston. When the volume of gas is 2.4 m3 the volume is being decreased by 0.05 m3/minute and the pressure is 320 kpa. Assuming Boyles law (PV = constant), what is the rate of increase in pressure (kpa/min) at that time? Solution

13. and that P=320 when V=2.4. dV We know that = -0.05 m3/min at V=2.4 dt Now dP dV dV dt . by the chain rule = So when V=2.4 dP . dV dP dt = -0.05 We know that PV=constant, so when V=2.4, P=320 gives the constant as 768. Hence P=768V-1 and dP = -768V-2. dV dP dt Hence dP dt = -0.05 . -768 (2.4)2 =6.7 kpa/min.

14. You should now be able to complete Example Sheet 5 from the Orange Book.