Interpretation of the derivative as a rate of change - AP Calculus AB

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Question

Find the rate of change of y if $y=5x^{-3}+3e^{x}$

Answer

The rate of change of y is also the derivative of y.

Differentiate the function given.

You should get $y'=-15x^{-4}+3e^{x}$

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Question

Given j(k), find the rate of change when k=5.

Answer

Given j(k), find the rate of change when k=5

Let's begin by realizing that a rate of change refers to a derivative.

So, we need to find the derivative of j(k)

We find this by multiplying each term by the exponent, and decreasing the exponent by 1

Next, plug in 5 to find our answer:

So, our rate of change is -221.

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Question

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

Answer

If p(t) gives the position of an asteroid as a function of time, find the function which models the velocity of the asteroid as a function of time.

Begin by recalling that velocity is the first derivative of position. So all we need to do is find the first derivative of our position function.

Recall that the derivative of sine is cosine, and that the derivative of polynomials can be found by multiplying each term by its exponent, and decreasing the exponent by 1.

Starting with:

We get:

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Question

If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.

Answer

If p(t) gives the position of a planet as a function of time, find the function which models the planet's velocity.

Velocity is the first derivative of position.

Therefore, all we need to do to solve this problem is to find the first derivative.

We can do this via the power rule and the rule for differentiating sine.

1) $\frac{d}{dx}x^n=(n)x^{n-1}$

$\frac{d}{dx}(sin(x))=cos(x)$

So, we these rules in mind, we get:

So our final answer is:

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Question

If p(t) gives the position of a planet as a function of time, find the function which models the planet's acceleration.

Answer

If p(t) gives the position of a planet as a function of time, find the function which models the planet's acceleration.

Velocity is the first derivative of position. Acceleration is the first derivative of velocity.

Therefore, all we need to do to solve this problem is to find the second derivative of p(t).

We can do this via the power rule and the rule for differentiating sine and cosine.

1) $\frac{d}{dx}x^n=(n)x^{n-1}$

$\frac{d}{dx}(sin(x))=cos(x)$

So, we these rules in mind, we get:

So our velocity function is:

Next, differentiate v(t) to get a(t).

So, our acceleration function is

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Question

If p(t) gives the position of a planet as a function of time, find the planet's acceleration when t=0.

Answer

If p(t) gives the position of a planet as a function of time, find the planet's acceleration when t=0.

Velocity is the first derivative of position. Acceleration is the first derivative of velocity.

Therefore, all we need to do to solve this problem is to find the second derivative of p(t) and then plug in 0 for t and solve.

We can do this via the power rule and the rule for differentiating sine and cosine.

1) $\frac{d}{dx}x^n=(n)x^{n-1}$

$\frac{d}{dx}(sin(x))=cos(x)$

So, we these rules in mind, we get:

So our velocity function is:

Next, differentiate v(t) to get a(t).

So, our acceleration function is

Next, plug in 0 and simplify.

Our answer is 12. Since we are not given any units, we can leave it as 12

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Question

If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.

Answer

If p(t) gives the position of a planet as a function of time, find the planet's velocity when t=0.

Velocity is the first derivative of position. Therefore, all we need to do to solve this problem is to find the first derivative of p(t) and then plug in 0 for t and solve.

We can do this via the power rule and the rule for differentiating sine and cosine.

1) $\frac{d}{dx}x^n=(n)x^{n-1}$

$\frac{d}{dx}(sin(x))=cos(x)$

So, we these rules in mind, we get:

So our velocity function is:

Now, plug in 0 and simplify.

So, our answer is -11. We are not given any units, so we do not need to worry about them.

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Question

A factory producing pens wants to maximize its output; to do so, it needs to find the rate of change of pen production. Find this rate if pens are produced according to the following function:

$p(t)=5000t^2+400\cos(t)$

Answer

The rate of change of a function is given by the derivative of that function. So, to find the rate of change of production, we must take the first derivative of the function for production, which is equal to

$p'(t)=10000t-400\sin(t)$

found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(t)=-\sin(t)$

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Question

Concrete at a factory flows according to the following theoretical model:

$c(x)=2x^2-5\cos(x)+5\sin(x)$

What is the rate of change of the concrete flow?

Answer

The rate of change of the concrete flow is given by the first derivative of the concrete flow function:

$c'(x)=4x+5\sin(x)+5\cos(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Question

Find the speed of the car at t=5 if its position is given by

$s(t)=4t^2+\frac{t}{5}$

Answer

To determine the speed of the car, we must take the first derivative of the position function, which gives us the rate of change of the position of the car - in other words, speed.

$s'(t)=8t+\frac{1}{5}$

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x}ax=a$

Evaluating the derivative function at t=5, we get our speed as

$s'(5)=8(5)+\frac{1}{5}=40.2$

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Question

A group of scientists use the following code for describing velocity and acceleration of particles:

$\alpha$ , when both the velocity and acceleration are positive;
$\beta$ , when the velocity is positive, but the acceleration is negative;
$\gamma$ , when the velocity is negative, but the acceleration is positive;
$\delta$ , when both the velocity and acceleration are negative.

What code - at t=2 - will the scientists use when describing a particle moving with a position function given by the following equation:

Answer

To determine which code the scientists will use, we must find the velocity and acceleration of the particle, given by the first and second derivatives of the position function, respectively, evaluated at the given point.

The velocity and acceleration functions, therefore, are

The derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Evaluated at t=2, we find that

When the velocity is positive, but the acceleration is negative, the code the scientists use is $\beta$ .

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Question

Find the velocity of the particle at x=0, its position given by the following function:

Answer

To determine the velocity of the particle, we must take the first derivative of the position function - in other words, the rate of change of the position is the velocity:

The derivative was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

To find the velocity at the given point, we simply plug in the value into the velocity function:

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Question

The velocity of a particle in feet per second can be modeled by the function

$v(t) = t^{3} - \ln (t^{2}+ 4)$

What is the acceleration of this particle at 6 seconds?

Answer

The acceleration of a particle is its change in velocity; therefore, to find its acceleration at a particular point in time, it is necessary to take the derivative of the velocity function and evaluate it for that value. In short, we want .

Differentiate :

$v(t) = t^{3} - \ln (t^{2}+ 4)$

$v'(t) =\frac{\mathrm{d} }{\mathrm{d} t} \left ( t^{3} - \ln (t^{2}+ 4) \right )$

Apply the sum rule:

$v'(t) =\frac{\mathrm{d} }{\mathrm{d} t} t^{3} - \frac{\mathrm{d} }{\mathrm{d} t} \ln (t^{2}+ 4)$

$v'(t) = 3t^{2} - \frac{\mathrm{d} }{\mathrm{d} t} \ln (t^{2}+ 4)$

Apply the chain rule by setting $u = t^{2}+4$ :

$v'(t) = 3t^{2} - \frac{\mathrm{d} }{\mathrm{d} t} \ln u$

$v'(t) = 3t^{2} - \frac{\mathrm{d} u }{\mathrm{d} t} \cdot \frac{\mathrm{d} }{\mathrm{d}u} \ln u$

$v'(t) = 3t^{2} - 2t \cdot \frac{1}{u}$

$v'(t) = 3t^{2} - \frac{2t }{t^{2}+4}$

This is the acceleration function; substitute 6 for :

$v'(6) = 3 (6)^{2} - \frac{2 (6) }{6^{2}+4}$

$= 108- \frac{12 }{40}$

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Question

The velocity of a particle in feet after seconds can be modeled by the equation

$v(t ) = t ^{2} \cos \frac{\pi }{5}t$ .

At seconds, give the acceleration of this particle in feet per second squared.

Answer

The acceleration function of a particle with respect to time is the derivative of the velocity function of the particle.

Find this derivative by first using the product rule:

$v(t ) = t ^{2} \cos \frac{\pi }{5}t$

$v'(t ) = \frac{\mathrm{d} }{\mathrm{d} t} \left (t ^{2} \cos \frac{\pi }{5}t \right )$

Apply the product rule:

$v'(t ) =t ^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} t} \left (\cos \frac{\pi }{5}t \right ) + \cos \frac{\pi }{5}t \cdot \frac{\mathrm{d} }{\mathrm{d}t} \left (t ^{2} \right )$

$v'(t ) =t ^{2} \cdot \frac{\pi }{5} \left (- \sin \frac{\pi }{5}t \right ) + \cos \frac{\pi }{5}t \cdot 2t$

$v'(t ) =- \frac{\pi }{5} t ^{2} \sin \frac{\pi }{5}t +2t \cos \frac{\pi }{5}t$

Evaluate :

$v'(5) =- \frac{\pi }{5} \cdot 5 ^{2} \cdot \sin\left ( \frac{\pi }{5} \cdot 5 \right ) +2 \cdot 5 \cdot \cos\left ( \frac{\pi }{5} \cdot 5 \right )$

$v'(5) =- 5 \pi \sin \pi +10 \cos \pi$

$v'(5) =- 5 \pi (0)+10 (-1)$

feet per second squared.

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Question

Determine the velocity of a particle at x=10 when the position function is given by

Answer

To determine the velocity function - the rate of change of position - we must take the derivative of the position function:

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

To find the velocity at a certain value of x - the instantaneous rate of change of position - we simply plug in the given value of x into the velocity function:

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Question

A crystal forms at a rate given by the following equation:

$c(t)=10e^{\frac{t}{2}}$

What is the rate of change of the crystal growth rate at t=5?

Answer

To find the rate of change of crystal growth at a given time, we must take the derivative of the growth rate function and evaluate it at a specific time.

The derivative of the function is

$c'(t)=5e^{\frac{t}{2}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u \frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Now, to find the growth rate at t=5, simply plug in this value for t into the derivative function:

$c'(t)=5e^{\frac{5}{2}}$

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Question

A car is moving with a velocity that can be modeled by the equation

What is the cars acceleration at

Answer

The car's acceleration is the instantaneous rate of change in its velocity with respect to time ( $a(t)=\frac{\mathrm{d}v }{\mathrm{d} t}$ ), so we can find the value of the cars acceleration at any time by taking the derivative of the velocity equation

Evaluating at , we get

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Question

A body's position "s" is given by the equation:

, $0\leq t \leq 2$

a) Find the body's speed at the endpoints of the given interval

b) Find the body's acceleration at the endpoints of the given interval

Answer

We are given the function describing the position of the body given a time "t":

We are also given the interval that the function can be applied over:

$0\leq t \leq 2$

First, we are tasked to find the speed of the body at each of the endpoints (0 and 2 seconds, respectively).

To figure this, we must understand that speed is the absolute value of velocity.

To find the velocity function, we must take the derivative of the position function with respect to time:

Now that we have the function of velocity given a time "t", we can find the speed of the body given a time "t" by simply taking the absolute value of the velocity function.

$speed(t)=\left |v(t)\right |=\left |2t - 3 \right |$

Now, to find the speed of the endpoints:

$speed(0)=\left |v(0)\right |=\left |2(0) - 3 \right |$

$speed(0)=\left |v(0)\right |=\left | - 3 \right |$

$speed(0)=\left |v(0)\right |=3 [m/s]$

We can repeat the same process to find the speed at 2 seconds.

$speed(t)=\left |v(t)\right |=\left |2t - 3 \right |$

$speed(2)=\left |v(2)\right |=\left |2(2) - 3 \right |$

$speed(2)=\left |v(2)\right |=\left |4 - 3 \right |$

$speed(2)=\left |v(2)\right |=\left |1 \right |$

$speed(2)=\left |v(2)\right |=1[m/s]$

For part a, the speeds at 0 and 2 seconds are 3 and 1 \[m/s\], respectively.

Part b asks us to find the acceleration at the endpoint times (0 and 2 seconds). To do this, we must first understand that to find acceleration at a time "t", we must take the derivative, with respect to time, of the velocity function.

From part a, we found the velocity function:

Thus, to find acceleration, we derive this function with respect to time.

The result of our derivation tells us that no matter what time is plugged into the function, the acceleration shall always return 2\[m/s^2\].

So, for part b, the acceleration at 0 and 2 seconds are 2 and 2 \[m/s^2\], respectively.

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Question

At any time t, the position of a body is given by the equation:

Find the body's acceleration at each time the velocity is zero.

Answer

We are given the position equation:

To find the acceleration when the velocity is equal to zero, we must first find the function to describe the velocity given a time "t".

We know that the velocity function is found by deriving the position function with respect to time.

Thus, the velocity function is given by the equation:

To find the acceleration when the velocity is equal to zero, we must set the velocity function equal to zero and solve for the times.

(multiply c factor by the a coefficient)

(divide out the original a coefficient)

Thus, the velocity is zero at both 3 and 1 seconds. The question asks for the acceleration at these times.

To do the next part, we must understand that the acceleration function is the derivative of the velocity function, with respect to time. We already know the velocity function:

So, if we derive this function with respect to time and plug in our times, we will know the acceleration when the velocity is equal to zero.

Our derivation tells us that given any time "t", the acceleration of the body can be found by the function:

So, to find the acceleration at times 3 and 1, we simply plug in the values to our acceleration function above.

So, the acceleration at time 1 is -6\[m/s^2\] and the acceleration at time 3 is 6\[m/s^2\].

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Question

Given the function describing an object's position with respect to time:

a) Find the function describing the object's velocity with respect to time

and

b) Find the function describing the object's acceleration with respect to time

Answer

This problem, although not technically complex, requires some understanding of the derivative and how it relates to elementary physics.

For part a)

Given a function that returns an objects position at time t, we can take the derivative of the position function to find the velocity function

We are given the position function:

We can apply the general power rule for derivatives to find the derivative of this function:

And from our understanding that the derivative of the position function is the velocity function, we can make the substitution

Thus, our part a answer is:

For part b)

Furthermore, we can take the derivative of the velocity function to find our acceleration function.

We found the velocity function to be:

Thus, the derivative of the velocity function (the acceleration function) is simply:

The full answer should be:

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