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

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).

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

Step 3: Remember that the equation for a straight line is y = mx + c. We have already calculated m=2. So the equation for our tangent is: ytan = 2x + 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.

Rate of Change 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). In this way the derivative is the instantaneous rate of change (of y with respect to x). 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. dy dt

Example 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). 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 time 0 ≤ t ≤ 2. So y'(1) = 800(2(1)-12) = 800 km/hr when t=1.

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? x V(volume) = x3. The length of the side is changing with time and so is the volume. (rate of change of volume) when We want to know dV dt x= 3.0 cm. We know that dx dt = 0.1 cm for any t.

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

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?

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

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

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 dP dt We want to know when V = 2.4.

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

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