# Unit 4: Contextual Applications of Differentiation

## Interpreting the Derivative

• The derivative tells us the slope of the line tangent to the graph- which tells us the slope of the line at a particular point

• This means that they can tell us the change of a unit over time. This will make sense with straight line motion.

## Straight Line Motion

• We know that position is measured in meters

• We know that velocity is measured in meters per second (m/s)

• Therefore we can derive position to get the rate of change- ie. meters per second!

• We know that acceleration is measured in meters per second squared (m/s^2)

• So we can derive velocity to get the rate of change- ie. meters per second per second!

• You can also take the second derivative of position to get acceleration

Position:

x(t) (sometimes wrote as s(t))

Meters

Velocity:

x’(t) or v(t)

Meters/Second

Acceleration:

x”(t) or v’(t) or a(t)

Meters/Second^2

• Particles will speed up when the sign of velocity and acceleration match

• The must both be negative or positive

• For example, if a particle moves along a straight line with velocity function v(t) = 3t^2 - 4t + 2. Find the acceleration of the particle at time t=2?

• Solution: The acceleration of the particle is the derivative of its velocity function. Thus, we take the derivative of v(t) with respect to t

• a(t) = d/dt v(t) = 6t - 4

• To find the acceleration at t=2, we substitute t=2 into the expression for a(t):

• a(2) = 6(2) - 4 = 8

## Non-Motion Changes

The derivative can also tell us the change of something other than motion

• For example, let’s say the volume of water in a pool is equal to V(t) = 8t^2 -32t +4

• Where V is the volume in gallons and t is the time in hours

• If we want to find the rate that the volume of water is increasing we take the derivative

• dV/dt = 16t - 32 gallons per hour

• At t=2 the volume isn’t changing (equation equals 0)

• Therefore it is increasing for all values >0

• Another example would be where temperature of a cup of coffee is given by the function x(t) = 70 + 50e^(-0.1t), where t is the time in minutes since the coffee was poured. And we need to find the rate of change of the temperature with respect to time at t=5 minutes.

• d/dt of x(t) = -5e^(-0.1t)

• Evaluating this derivative at t=\$ minutes, we get:

• x’(5) = -5e^(-0.1(5)) ≈ -2.27

## Related Rates

• We just saw how the derivative can tell us the change of something but we can also have problems where the change of one thing is related to another- Related Rates!

• Let’s say that a pool of water is expanding at 16π square inches per second and we need to find the rate of the radius expanding when the radius is 4 inches

• We know that we can find the radius using A = πr^2

• Now let’s relate our rates!

• dA/dt = 2πr(dr/dt)

• Notice how we had to follow r with dr/dt, this is because the change in R doesn’t match the change in A (implicit differentiation)

• Now we have the change of the area (dA/dt) and the change of the radius (dr/dt)

• Now we can plug in and solve!

• 16π = 2π(4)dr/dt

• dr/dt = 2

• The radius is changing at a rate of 2 inches per second

• Let’s say a spherical balloon is being inflated at a rate of 10 cubic inches per second. How fast is the radius of the balloon increasing when the radius is 4 inches?

• We know that the volume of a sphere is given by the formula V = (4/3)πr^3.

• Differentiating both sides with respect to time t, we get:

• dV/dt = 4πr^2 (dr/dt)

• We are given that dV/dt = 10 cubic inches per second and r = 4 inches. Substituting these values, we get:

• 10 = 4π(4^2)(dr/dt)

• Simplifying, we get:

• dr/dt = 10/(16π)

• Therefore, the radius of the balloon is increasing at a rate of 10/(16π) inches per second when the radius is 4 inches.

• To solve related rates problems in calculus, follow these steps:

• Read the problem carefully and identify all given information.

• Draw a diagram if possible.

• Determine what needs to be found and assign a variable to it.

• Write an equation that relates the variables involved.

• Differentiate both sides of the equation with respect to time.

• Substitute in the given values and solve for the unknown rate.

• Remember to always include units in your final answer and to check that your answer makes sense in the context of the problem.

### Linearization

• Differentials are very small quantities that correspond to a change in a number. We use Δx to denote a differential.

• They can approximate the value of a function!

• Remember the limit definition of a derivative?

• We just have to replace h with Δx and remove the limit!

• We get that f(x + Δx) ≈ f(x) + f’(x)Δx

• This is no longer equal to the derivative, but an approximation of it

• In simple terms “the value of a function (at x plus a little bit) is equal to the value of the function (at x) plus the product of the derivative of the function (at x) plus a little bit”

• Let’s say we needed a differential to approximate (3.98)^4

• We have to let f(x) = x^4, x= 4, and Δx = -0.02

• Now we have to find f’(x) which is 4x^3

• Now plug into f(x + Δx) ≈ f(x) + f’(x)Δx

• (x + Δx)^4 = x^4 + 4x^3Δx

• When we plug in x= 4 and Δx= -0.02 we get 250.88

• You can check this using your calculator!

### L’Hospital’s Rule

• If a limit gives you 0/0 or ∞/∞, then it is called “indeterminate” and you can use

• L’Hospital’s Rule to interpret it!

• L’Hospital’s Rule says that we can take the derivative of the numerator and denominator and try again

• Let’s say we have the limit of 5x^3 -4x^2 +1/7x^3 +2x - 6 as it approaches infinity

• This equals ∞/∞ so we can take the derivative of the top and bottom

• Then we get 15x^2 -8x/21x^2 +2

• This is still ∞/∞ so we take the derivative again

• Then we get 30x -8/42x which is still ∞/∞

• We take the derivative to get 30/42 or 5/7 which is our answer!