Maths IB standard Calculus Yay Starter activity How

Maths IB standard Calculus Yay!

Starter activity How do we know what the gradient of a line is? n Y = mx + c n If we are given two points how do we find the gradient? n

Gradient/ Slope The gradient or slope of a line describes the steepness. To find the gradient we use the phrase rise run n Remember you rise up in the morning before you go for a run n Click here for a summary of everything n

Gradient/ Slope • To find the gradient/slope between two points you have to find the rise (difference between the y’s) over the run (difference between the x’s) • Look at this applet Maths. Net: A Level Pure C 1 Module Click here for examples of finding the gradient between two points

Formula for the gradient/slope Can you think of a formula to find the gradient between two points (x 1, y 1) and (x 2, y 2)? n M = y 2 -y 1 n x 2 -x 1 n This is the differences of the y’s over the differences of the x’s n

Example 1 Find the gradient between the following points: n 1) (4, 2) (6, 3) n 2) (-1, 3) (5, 4) n 3) (-4, 5) (1, 2) n 4) (2, -3), (5, 6) n 5) (-3, 4) (7, -6) n

More practice Find the gradient between these points n 1) (-2, -4) (10, 2) n 2) (-12, 3) ( -2, 8) n 3) (-2, 7) (4, 5) n 4) (2, 3) (5, 7) n 5) (3 b, -2 b) (7 b, 2 b) n

Example 2 Work out the gradients of the following equations: n 1) y = -2 x+5 n 2) y = -x+7 n 3) –x+2 y -4 = 0 n 4) -3 x+6 y+7 = 0 n 5) 9 x+6 y+2 = 0 n

Looking at the gradient on a curve • • • Let's think the slope of the secant line that goes through two points (a, f(a)) and (a+h, f(a+h)) on y = f(x). The slope is (f(a+h)-f(a))/h. When we make h smaller, what does (f(a+h)-f(a))/h approach ? We call this taking a limit when we investigate h becoming smaller.

Gradients of y = x 2 n Complete this table by looking at this applet X=a 1 2 3 4 5 6 7 x 2 Gradie nt 2 x

Gradients of y= x 3 n Complete this table by using the applet x 1 2 3 4 5 6 7 x 3 gradient

Gradients of y= x 4 n Complete this table by using the applet x 1 2 3 4 5 6 7 x 4 Gradient x

So far y= x 2 has a gradient function of n Y = x 3 the gradient function is n Y = x 4 the gradient function is n

The magic phrase Put the power in front and then drop the power by one!!!!!! n n n We call the gradient function the derivative of the function i. e. dy DON’T WRITE IT LIKE dx THIS dy/dx

The Derivative n So if y = x n n Then dy dx = n x n-1

F (x) or y n n n This is also called the derivative noun Another way to express dy dx the gradient function is to say f (x) or y This is called f dash x or y dash The verb is differentiate These are all names for the GRADIENT FUNCTION

The derivative of a number What is the gradient of the line y = 10? n What is the gradient of the line y = 5? n What does this line look like? n n So what is the derivative of y = x 5 + 10?

Derivative n n So if y = x n where n is a number n Then dy dx = n x n-1 Example 1 Find derivative of A) y = 4 x 2 n dy n dx = n n 8 x b) y= 5 x 3 dy dx= 15 x 2

More examples n Find the derivative of the following: Y = f(x) dy = f’(x) dx x 4 x 5 x 3 x 2 x 7 x X-1 X-2

Answers Find the gradient function (dy when y is: n dx) n A) 2 x 2 – 6 x +3 n B) ½ x 2 +12 x n C) 4 x 2 – 6 n D) 8 x 2+7 x+12 n E) 5+4 x-5 x 2 n

Some rules to follow n n n n n When y = kxn and k is a constant Then dy dx = nkx n-1 Example 1 If y= 4 x 3 then dy dx = 12 x 2 When y = u + v Then dy du dv dx = dx + dx Example 2 If y = 3 x 3 – 5 x 2 then dy/dx = 9 x 2 – 10 x

Some rules to follow When y = u + v n Then dy = du + dv n dx dx dx n

Example 2 n If y = 3 x 3 – 5 x 4 then dy dx = 9 x 2 – 20 x 3 n n n Ditto for addition

Example 3 n (i) What about the derivative of y = 1 x 3 n (ii) Find the derivative of Y= 20 – 1/3 x 4+10 x dy = - 4/3 x 3 +10 dx Rules: 1) derivative of a number (constant) is zero 2) derivative of 10 x is 10 why? n n

Example 4 n Find the derivative of Y= 3 x - 1+1/5 x 2 +x 4 n The derivative is 3 + 2/5 x + 4 x 3

Example 5 Find the gradient of the curve with equation y = (2 x-1)(3 x+2) at the point x= 3 n Here we multiply the brackets out first by using the moon! n Y= 6 x 2+x-2 n dy dx = 12 x+1 n Now the gradient at x=3 is 12(3)+1= 37 n

Starter Quick questions n Write down the magic phrase to find the derivative of a function. n Write down at least three different types of notation used for the derivative of a function. n What is another name for derivative? n

Quick questions- find the derivative

In the next few lessons we will… A) Find out how to find the second derivative. n B) Use first derivative as a rate of change for real life situations n C) Review: Finding gradients at particular points on a curve n D) Find the equation of a tangent and normal to a curve n

Applications of the first derivative n Some movies but got to get correct format

An online quiz Rates of change n Distance, velocity and acceleration n n

Second derivative The second derivative means just differentiate dy dx again! n The second derivative looks like this: n d 2 y dx 2 or y ‘’ or f ‘’(x) n Download livemath plugin now to view inter active pages n

An example of second derivative n Find the second derivative:

Quiz n A MC quiz on differentiating polynomials

Gradient as a rate of change n n n Remember that the first derivative is the gradient function or the rate of change of one variable y with respect to x. What if we were given an equation of displacement s with respect to time? For example s = t 2+3 t If I differentiate this I get a rate of change of distance with respect to time- this is called Velocity! So v = ds dt

The derivative as a rate of change The derivative tells us the gradient function of any curve. If we have a distance time graph what does the gradient tell us? n The gradient tells us the rate of change of the distance with respect to time which is also called speed! n Click here for the surfer n Click here for AS guru Game n

Example 1 n n n n The displacement of a body at a time t seconds is given in metres by s = t 2+3 t Find a) the velocity of the body at time t b) the initial velocity of the body Solution A)The velocity = ds dt = 2 t+3 B) The initial velocity is when t=0 so v = 2(0)+3 = 3 m/s So the initial velocity is 3 m/s

The rate of change- gradient n Example The volume of an expanding spherical balloon is related to its radius, r cm by the formula V = 4/3 r 3. Find the rate of change of volume with respect to the radius at the instant when the radius is 5 cm. n Ex 7 G

Example 1 n Find the gradient of the curve f(x) = x(x+1) at the point P(0, 0).

Example 2 n Find the gradient of the curve y = 3 x 2 +x – 3 at the point (1, 1)

Example 3 n Find the gradient of the curve y = ½ x 2 + 3/2 x at the point (1, 2)

Drag and drop n Put these tangents onto the curve

Equations of tangents and normals n Some definitions first:

Finding the equation of a Tangent What does the curve y = (x-3)(x+2) look like? n At the point x = 2 we can draw a tangent to the curve. n How do we find the equation of this line? n What information do we know already? n

Tangents If we know the gradient, m at any point on a curve and we have the coordinates of that point (x , y) we can easily find the equation of the tangent line at that point. n What is the equation of a straight line? n

The steps Example 1 n Find the equation of the tangent of the curve y = (x-3)(x+2) at the point x = 1 y = - 6 n

The steps Complete these sentences: n 1) Find the gradient of the curve at the point x = 1 by……. n 2) The equation of a straight line is…… n 3) Now substituting x = 1 and y = -6 and m = ……… to obtain the equation of the tangent. n

An example n n n n n Find the equation of the tangent on the curve Y= (x+3)(x-1) at the point (1, 5). How do I find the equation of a line given a gradient and point? Y = mx + c How do I find the gradient? Y = x 2+3 x-x-3 Y= x 2+2 x-3 dy dx = 2 x+2 Gradient at x = 1 dy dx = 2(1) + 2= 4 Y-5 = 4(x-1) Y = 4 x+1 is the equation of the tangent at the point x = 1

Your example Find the equation of the tangent to the curve with equation y = (2 x-1)(x+1) at the point x = 3 y =20 n Y= 2 x 2+x-1 n So dy dx = 4 x+1 n So the gradient is dy dx = 4(3) +1 = 13 n So using y = mx+c n Y- 20 = 13(x-3) n Simplifying this gives y = 13 x - 19 n

Example 3 n n n n n Find the equation of the tangent at the point x = -2 on the curve y = 3 x 3 +4 x 2 -7 Step 1 dy dx = 9 x 2 +8 x Step 2 x = -2 into dy dx = 9(-2)2+8(-2) = 20 Step 3 m=20 x = -2 find the y coordinate y = 3(-2)3+4(-2)2 -7 y = -24+16 -7= -15 One point (-2, -15) and gradient m = 20 y = mx+c n n Y = 20 x+25

Follow these steps to find the equation of a tangent Step 1 Find dy/dx n Step 2 Sub the x value into dy/dx to find m, the gradient n Step 3 Use y = mx+b or y-y 1 = m(x-x 1)to find the equation n A game for tangents As guru n

Example n Here are some guided examples. n So we use y = mx+c to find the equation of the tangent

Tangents You will be given a card. n Order these steps to find the tangent n

What is a normal? n A normal is perpendicular to the tangent

Equations of normals n To find the equation of a normal we follow the steps but we use -1/m instead of m. Find the equation of the normal on the curve Y= (x+3)(x-1) at the point (1, 5). n Check the steps here to find the normal n Ex 7 H, 7 L n

Quiz n A quiz on differentiating polynomials

A beautiful application problem n Jordan wants to design a drink container for orange juice in the shape of a cylinder with NO lid. The volume of the orange juice is 1000 cc. Show that the surface area of this container with NO lid is given by n Show that when the rate of the change of the area with respect to r is zero then r 3= 500/

Solution

Card A n. When x = 1 n dy/dx = 5

Card B n 2 = 5(1)+ C

Card C n. A curve has equation 2 Y = 3 x -x

Card D n. Second find the gradient at x = 1

Card E n. Simplify

Card F n. Use y = mx + C

Card G n. When x = 1 y=2

Card H n. First find a point on the curve

Card I n. Y = 5 x-3

Card J n. The tangent at x = 1 is to be found

Card K n. The derivative is: dy/dx = 6 x-1

A lot of activities n Click here for mathsnet. net

Derivatives n What does a derivative look like?

C) Gradient as a rate of change n n n Remember that the first derivative is the gradient function or the rate of change of one variable y with respect to x. What if we were given an equation of displacement s with respect to time? For example s = t 2+3 t If I differentiate this I get a rate of change of distance with respect to time- this is called Velocity! So v = ds dt

Example 1 n n n n The displacement of a body at a time t seconds is given in metres by s = t 2+3 t Find a) the velocity of the body at time t b) the initial velocity of the body Solution A)The velocity = ds dt = 2 t+3 B) The initial velocity is when t=0 so v = 2(0)+3 = 3 m/s So the initial velocity is 3 m/s

What is dv dt? If we differentiate displacement w. r. t. to time we get velocity n What do we get if we differentiate velocity? n What is the rate of change of velocity? n This is called acceleration! n So dv dt = acceleration, a. n

Example 2 The displacement s metres of body at time, t is given by s = 2 t 3 –t 2+2. Find n A) the velocity after 1 second n B) the acceleration after 1 second n C) the time at which the acceleration is zero. n

Homework Quiz A quiz on differentiating polynomials n Here you must print this out with your test results n

D) Second derivative The second derivative means just differentiate dy dx again! n The second derivative looks like this: n d 2 y dx 2 or y ‘’ or f ‘’(x) n Download livemath plugin now to view inter active pages n

More practice with the second derivative n n n n Find the second derivative of the following functions. A) f(x) = x 4 -3 x 3+1 B) y = 8 x 2 -7 x C) f(x) = ½ x 2 D) y = (x 3+2 x)2 E) f(x) = 7 x 3 -5 x 2 F) y = x+1 x

More practice with the second derivative n n n Find the second derivative of the following functions. A) f(x) = x 2+4 x-3 B) y = x 3 -3 x C) f(x) = x 2 +5 D) y = 2 x 3 -2 x E) f(x) = 3 x 3

Finding the second derivative

Second derivative n Look at this applet to understand the second derivative

Stationary points A stationary point is when the gradient is zero or when the curve is flat. n There are three types of stationary points n

Stationary points n A turning point is when the gradient changes in sign. Look at this diagram

Maximum and minimums For a maximum you can see that we walk uphill , flat and then downhill. This means the gradient is positive, zero and then negative. A maximum moves to a negative gradient n For a minimum we walk downhill, flat and then uphill. This means the gradient is negative, zero and then positive. For a minimum it moves to a positive gradient. n

Stationary points n An good explanation A little game with a stationary point

Second derivative n Look at this applet to understand the second derivative and stationary points n Here are some more practice questions

How do we find out where the stationary points are? n n n n n We firstly find the first derivative dy/dx and see when this equals zero! Example 1 Find the where the stationary point is for the curve y = x 2 dy dx = 2 x When does dy dx = 0? 2 x = 0 so x = 0 and y = 0 Example 2 Find where the stationary point is for the curve y = x 2 – 5 x+6 Find the first derivative and let this equal zero.

Follow these steps to test for stationary points- maximum or minimum n Step 1 Find the values of x for which dy dx = 0 n Step 2 Find n d 2 y dx 2 Step 3 Substitute the x values you got for dy/dx = 0 into the second derivative to test what kind of stationary point it is (the nature). n Now substitute x value into function to find corresponding y value n

Second derivative n The second derivative tells us whether a stationary point is a maximum, minimum or inflexion point A sad negative maximum A happy positive minimum t t

Complete this table when dy dx = 0 d 2 y/dx 2 > 0 second derivative is positive d 2 y/dx 2 < 0 second derivative is negative Minimum Maximum Happy face positive Sad face negative

An example n n Find the coordinates of the stationary points on the curve y= x 3 -6 x 2 -15 x+1, using the second derivative determine their nature. Two stat points are min (5, -99) and max (-1, 9)

Another example Find the coordinates of the stationary points on the curve y= x 4 -4 x 3 and determine their nature. n Min (3, -27) n

Drag and drop n Put these tangents onto the curve

Maxima and minima problems n Look at this link for lots of examples n Look at this applet to see maximization

Maxima and minima n We can use calculus to solve many practical problems such as finding the maximum area or volume.

Example 2 A closed right cylinder of base radius r cm and height h cm has a volume of 54. Show that S the total surface area of the cylinder is given by n S = 108 +2 r 2 r n Find the maximum surface area n

Hints! n n n Steps: 1) Find the volume of the cylinder and let this equal 54. Make h the subject of this formula. 2) Find an expression for the total surface area. 3) Substitute what you got in 1) into the total surface area expression from 2) to give you S in terms of r. 4) Find the derivative of this expression to find the stationary values and determine their nature by using the second derivative. 5) Find the corresponding height.

More examples n The profit, $y, generated from the sale of x items of a certain luxury product is given by the formula y = 600 x+15 x 2 -x 3. Calculate the value of x which gives a maximum profit, and determine that maximum profit. Solution n X = 20 profit = $10, 000 n

Quiz n A MC quiz on differentiating polynomials