Application of Application of DerivativesDerivatives

Application of Application of DerivativesDerivatives

Derivatives as it Relates to Velocity, Position, and

Acceleration• Velocity = the rate of change of displacement• Displacement is change in position

(x(t)=position)• Instantaneous Velocity= Derivative of

Position(x’(t)= position)• x(t)= position

• x ‘(t)= v(t)= velocity• x “(t)= v (t) = a (t)= acceleration

• Velocity is speed with direction

Derivatives as it Relates to Velocity, Position, and Acceleration (cont’d)

• Example 1• The position x of a particle at time t is given by the

function x(t) = kt^2 where k =  −5 m s −2. Find (a) the velocity as a function of time; (b) the velocity at time t = 3 s.

• (a) The velocity vx is the derivative with respect to time of the position function x(t) which is of the form At^n , with A = k and n = 2.

• (b) Remembering that k is given as −5 m s−2, the velocity at time t = 3 s is now easily obtained from the equation above, as follows

• vx(3 s) = 2kt = 2 × (−5 m s−2) × (3 s) = −30 m s−1• Note that we have multiplied m s^−2 by s to give

m s^−1

Mean Value Theorem• Suppose f(x) is a function that satisfies both of

the following.• f(x) is continuous on the closed interval [a,b]. • f(x) is differentiable on the open interval (a,b). *Then there is a number c such that a < c < b and• f’(c)= f(b)-f(a)/ b-

a                                                     

                                                 

Mean Value Theorem (cont’d)

• f(x)= x^3+2x^2-x on [-1,2]• 1st: find the derivative• 3x^2+4x-1• 2nd: now we need to do is plug this into

the formula given by the Mean Value Theorem.

• f’(c)=f(2)-f(-1)/ 2-(-1)• 3c^2+4c-1= 14-2/ 3=12/3=4

Mean Value Theorem (cont’d)

• 3rd: Quadratic Formula

Mean Value Theorem (cont’d)

• 4th: Results from the Quadratic Formula

Mean Value Theorem (cont’d)

• 5th: We now get two values from the quadratic formula. However, only one value fits within the integral, therefore, we dismiss the second one.

Connecting f(x), f’(x), and f’’(x)

• If f is continuous on a closed interval, [a,b], then f has both a maximum value and a minimum value on the interval.

Connecting f(x), f’(x), Connecting f(x), f’(x), and f’’(x) (cont’d)and f’’(x) (cont’d)

• Critical Points:• A point in the interior domain of a

function f at which f’=0 or f’ does not exist is a critical point.

• Examples include maximum, minimum, and inflection points

Connecting f(x), f’(x), Connecting f(x), f’(x), and f’’(x) (cont’d)and f’’(x) (cont’d)

• f’(x) > 0= f(x) is increasing• f’(x)<0= f(x) is decreasing• Ex. Y= x(sqrd) – 4x + 7• Find all critical points• 1st: Find the derivative-y’=2x – 4• Solve y’=0• 2x-4=0; x=2

Connecting f(x), f’(x), Connecting f(x), f’(x), and f’’(x) (cont’d)and f’’(x) (cont’d)

• 2nd: Do a “Sign Line”• --------|--------• Set the “2” in the middle of the line.• 3rd: Plug in a number before “2” into the derivative: 2(0)-4= -

4• 4th: Plug in a number after “2”: 2(3)-4= 2• Thus, on the sign line, you would put a negative sign before

the “2” and a positive sign after the “2”.• Minimum at x=2

• Decreasing: (-inf.,2)• Increasing: (2,inf.)

• Whenever it goes from negative to positive: Minimum• Whenever it goes from positive to negative: Maximum

Connecting f(x), f’(x), Connecting f(x), f’(x), and f’’(x) (cont’d)and f’’(x) (cont’d)

• Concavity:• The second derivative of a function tells

the concavity of the function.• f’’(x)<0=concaves down• f’’(x)>0=concaves up• Where concavity changes, there is an

inflection point (where the signs change).

Connecting f(x), f’(x), Connecting f(x), f’(x), and f’’(x) (cont’d)and f’’(x) (cont’d)

Example 1• 3x^2-6x=0• X^2-2x=0• x(x-2)=0• X=0,2• Possible local extrema at x=0,2

Connecting f(x), f’(x), Connecting f(x), f’(x), and f’’(x) (cont’d)and f’’(x) (cont’d)

• Y’=3x^2-6x• Y’’=6x-6• In the first derivative, it had a possible local

extrema at x=0,2• Y’’(0)=6 x 0-6=-6• Because the 2nd derivative at x=0 is negative,

the graph is concave down.• Y’’(2)=6 x 2 – 6=6• Because the 2nd derivative at x=2 is positive,

it concaves up

Related Rates• A spot light is on the ground 20 ft away

from a wall and a 6 ft tall person is walking towards the wall at a rate of 2.5 ft/sec.  How fast the height of the shadow changing when the person is 8 feet from the wall?  Is the shadow increasing or decreasing in height at this time?

 

Related Rates (cont’d)• In this case we want to determine y’ when

the person is 8 ft from wall or x=12 ft. Also, if the person is moving towards the wall at 2.5 ft/sec then the person must be moving away from the spotlight at 2.5 ft/sec and so we also know that x’=2.5.

• In this case the equation we’re going to work with is, y/6=20/x

• Y=120/x• Now all that we need to do is differentiate and

plug values into solve to get y’.

Related Rates (cont’d)

• The height of the shadow is then decreasing at a rate of 2.0833 ft/sec.  

Related Rates (cont’d)• Ex. 2• Two people are 50 feet apart.  One of them

starts walking north at a rate so that the angle shown in the diagram below is changing at a constant rate of 0.01 rad/min.  At what rate is distance between the two people changing when theta=0.5  radians?

Related Rates (cont’d)• We can then relate all the known

quantities by one of two trig formulas.

Related Rates (cont’d)• We want to find x’ and we could

find x if we wanted to at the point in question using cosine since we also know the angle at that point in time.  However, if we use the second formula we won’t need to know x as you’ll see.  So, let’s differentiate that formula.

Related Rates (cont’d)• Just plug in and solve. As noted,

there are no x’s in this formula.

Related Rates (cont’d)• Ex. 3• Air is being pumped into a spherical

balloon at a rate of 5 cm3/min.  Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.

• We know that air is being pumped into the balloon at a rate of 5 cm3/min.  This is the rate at which the volume is increasing.

Related Rates (cont’d)• We want to determine the rate at

which the radius is changing.

Related Rates (cont’d)• In this case we can relate the

volume and the radius with the formula for the volume of a sphere.

Related Rates (cont’d)• we will need to do implicit

differentiation on the above formula.  Doing this gives,

Related Rates (cont’d)• Now all that we need to do is plug

in what we know and solve for what we want to find.