Concept of the Derivative - AP Calculus AB

Card 0 of 957

Question

The function is differentiable at the point . List which of the following statements must be true about :

1) The limit $\lim_{x->a}\frac{f(x)-f(a)}{x-a}$ exists. 2) $\lim_{x->a^{-}}f(x)=\lim_{x->a^{+}}f(x)$ 3) $\lim_{x->a}\frac{f(x)-f(a)}{x-a}=f(a)$ 4) $\lim_{x->a}f(x)=f(a)$ 5) $\lim_{x->a}f(x)=\lim_{x->a}\frac{f(x)-f(a)}{x-a}$

Answer

1) If a function is differentiable, then by definition of differentiability the limit defined by,

$\lim_{x->a}\frac{f(x)-f(a)}{x-a}$

exists. Therefore (1) is required by definition of differentiability.                                                                                                                                 2) If a function is differentiable at a point then it must also be continuous at that point. (This is not conversely true).

For a function to be continuous at a point we must have:

$\lim_{x->a^-}f(x)=\lim_{x->a^+}f(x)=\lim_{x->a}f(x)=f(a)$

Therefore (2) and (4) are required.

-----------------------------------------------------------------------------------------

3)

$\lim_{x->a}\frac{f(x)-f(a)}{x-a}=f(a)$

This is not required, the left side of the equation is the definition of a derivative at a point for a function . The derivative at a point does not have to equal to the function value at that point, it is equal to the slope at that point. Therefore 3 does not have to be true.

However, we can note that it is possible for a function and its' derivative to be equal for a given point. Sine and cosine, for instance will intersect periodically. Another example would be the exponential function which has itself as its' derivative .

                                                                                                                               4) See 2

$\lim_{x->a}f(x)=f(a)$

5)

$\lim_{x->a}f(x)=\lim_{x->a}\frac{f(x)-f(a)}{x-a}$

Again, the function does not have to approach the same limit as its' derivative. It is possible for a function to behave in this manner, such as in the case of sine and its' derivative cosine, which will both have the same limit at points where they intersect.

Compare your answer with the correct one above

Question

When the limit $\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$ fails to exist,

Answer

By definition of differentiability, $\lim_{h \to 0} \frac{f(2+h)-f(2)}{h} = f'(2)$ when the limit exists. When exists, we say the function is 'differentiable at '.

Compare your answer with the correct one above

Question

Which of the following functions is differentiable at , but not continuous there?

Answer

All of the functions are differentiable at . If you examine the graph of each of the functions, they are all defined at , and do not have a corner, cusp, or a jump there; they are all smooth and connected (Not necessarily everywhere, just at ). Additionally it is not possible to have a function that is differentiable at a point, but not continuous at that same point; differentiablity implies continuity.

Compare your answer with the correct one above

Question

For which of the following functions does a limit exist at , but not a y-value?

Answer

To answer the question, we must find an equation which satisfies two criteria:

(1) it must have limits on either side of that approach the same value and (2) it must have a hole at .

Each of the possible answers provide situations which demonstrate each combination of (1) and (2). That is to say, some of the equations include both a limit and a y-value at , neither, or,in the case of the piecewise function, a y-value and a limit that does not exist.

In the function, $\frac{x^2+x-20}{2x^2+10x-72}$ , the numerator factors to

while the denominator factors to . As a result, the graph of this

function resembles that for $\frac{x+5}{2(x+9)}$ , but with a hole at . Therefore, the limit

at exists, even though the y-value is undefined at .

Compare your answer with the correct one above

Question

Evaluate the following limit:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})$

Answer

When approaches 0 both and will approach . Therefore, L’Hopital’s Rule can be applied here. Take the derivatives of the numerator and denominator and try the limit again:

$\lim_{x\rightarrow 0} (\frac{1-cos2x}{sinx})= \lim_{x\rightarrow 0} (\frac{2sin2x}{cosx})=\frac{0}{1}=0$

Compare your answer with the correct one above

Question

Find the instantaneous rate of change for the function,

at the point .

Answer

Find the instantaneous rate of change for the function,

at the point .

1) First compute the derivative of the function, since this will give us the instantaneous rate of change of the function as a function of .

$\frac{dy}{dx}=\frac{d}{dx}(2x^3-x-4)$

$\frac{dy}{dx}=\frac{d}{dx}(2x^3)-\frac{d}{dx}(x)-\frac{d}{dx}(4)$

$\frac{dy}{dx}= 6x^2-1$

2) Now evaluate the derivative at the value ,

$\frac{dy}{dx}_{|x=-1}= 6(-1)^2-1 = 5$

Therefore, is the instantaneous rate of change of the function

at the point .

Compare your answer with the correct one above

Question

A particle is traveling in a straight line along the x-axis with position function $x(t) = t+\sin(t)$ . What is the instantaneous rate of change in the particle's position at time $\pi$ seconds?

Answer

To find the instantaneous rate of change of the particle at time $t=\pi$ , we have to find the derivative of and plug $\pi$ into it.

$x'(t) :=v(t) = 1+\cos(t)$ .

And

$v(\pi) = 1+\cos(\pi) = 1+-1 = 0$ .

Hence the instantaneous rate of change in position (or just 'velocity') of the particle at $t=\pi$ is . (At that very instant, the particle is not moving.)

Compare your answer with the correct one above

Question

Find the function values and as well as the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1): : : : : : : x =\frac{\pi}{2}$

$2): : : : : : : x =\frac{\pi}{4}$

$3): : : : : : : x =\pi$

Answer

Find the instantaneous rate of change for the function $g(x)=x\sin(x)$ corresponding to the following values of

$1): : : : : : : x =\frac{\pi}{2}$

$2): : : : : : : x =\frac{\pi}{4}$

$3): : : : : : : x =\pi$

Evaluate the function at each value of

$g\left(\frac{\pi}{2} \right )=\frac{\pi}{2}sin\left(\frac{\pi}{2} \right )=\frac{\pi}{2}$

$g\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\sin\left(\frac{\pi}{4} \right )=\frac{\pi}{4}\frac{\sqrt{2}}{2}=\frac{\pi\sqrt{2}}{8}$

$g(\pi)=\pi \sin(\pi)=0$

The instantaneous rate of change at any point will be given by the derivative at that point. First compute the derivative of the function:

$g(x)=x\sin(x)$

Apply the product rule:

$g'(x)=x\frac{d}{dx}\sin(x)+\sin(x)\frac{d}{dx}(x)$

$= x\cos(x)+\sin(x)$

Therefore,

$g'(x)=x\cos(x)+\sin(x)$

Now evaluate the derivative for each given value of :

$1): : : : : : : x =\frac{\pi}{2}: : : :\rightarrow: : : :g'\left(\frac{\pi}{2} \right ): : =: : \frac{\pi}{2}\cos\left(\frac{\pi}{2} \right )+\sin\left(\frac{\pi}{2} \right ): :=:1$

$2): : : : : : : x =\frac{\pi}{4}: : : :\rightarrow: : : :g'\left(\frac{\pi}{4} \right ): : =: : \frac{\pi}{4}\cos\left(\frac{\pi}{4} \right )+\sin\left(\frac{\pi}{4} \right ): :=:\frac{\sqrt{2}}{2}\left(1+\frac{\pi}{4} \right )=\frac{\sqrt{2}}{8}(4+\pi)$

$3): : : : : : : x =\pi: : : :\rightarrow: : : :g'(\pi): : =: : \pi\cos(\pi) +\sin(\pi) : :=:\pi$

Therefore, the instantaneous rate of change of the function at the corresponding values of are:

Compare your answer with the correct one above

Question

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

Answer

Given that v(t) is the velocity of a particle, find the particle's acceleration when t=3.

We are given velocity and asked to find acceleration. Our first step should be to find the derivative.

We can use our standard power rule for our 1st and 3rd terms, but we need to remember something else for our second term. Namely, that the derivative of is simply

With that in mind, let;s find v'(t)

Now, for the final push, we need to find the acceleration when t=3. We do this by plugging in 3 for t and simplifying.

So, our answer is 123.5

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\&\text{What is its velocity at time }t=6\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function }6t - sin(\frac{(5\cdot \pi t)}{2}) + 5\&\text{We find the velocity function: }6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\&\text{Plugging in values we then get:}\&v(6)=6 -\frac{ (5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\&v(6)=13.85\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\&\text{What is its velocity at time }t=3\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }- 3t - cos(5\cdot \pi t) - sin(\frac{(7\cdot \pi t)}{2}) - 5\&\text{We find the velocity function: }5\cdot \pi sin(5\cdot \pi t) -\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi t)}{2}))}{2}- 3\&\text{Plugging in our value of time we then get:}\&v(3)=5\cdot \pi sin(5\cdot \pi (3)) -\frac{ (7\cdot \pi cos(\frac{(7\cdot \pi (3))}{2}))}{2}- 3\&v(3)=-3\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\&\text{What is its velocity at time }t=1\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }cos(\frac{(9\cdot \pi t)}{2}) - 2t + cos(\frac{(17\cdot \pi t)}{2}) - 1\&\text{We find the velocity function: }-\frac{ (9\cdot \pi sin(\frac{(9\cdot \pi t)}{2}))}{2}-\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi t)}{2}))}{2}- 2\&\text{Plugging in our value of time we then get:}\&v(1)=-\frac{ (9\cdot \pi sin(\frac{(9\cdot \pi (1))}{2}))}{2}-\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi (1))}{2}))}{2}- 2\&v(1)=-42.84\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{At any given time, a particle has the position :}\&p(t)=6t - cos(\frac{(13\cdot \pi t)}{2}) + 1\&\text{Calculate its velocity at time }t=7\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }6t - cos(\frac{(13\cdot \pi t)}{2}) + 1\&\text{We find the velocity function: }\frac{(13\cdot \pi sin(\frac{(13\cdot \pi t)}{2}))}{2}+ 6\&\text{Plugging in our value of time we then get:}\&v(7)=\frac{(13\cdot \pi sin(\frac{(13\cdot \pi (7))}{2}))}{2}+ 6\&v(7)=-14.42\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=3t - cos(\frac{(\pi t)}{2}) + sin(4\cdot \pi t) + 5t^{2} + 6t^{3} + 6\&\text{What is its velocity at time }t=6\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }3t - cos(\frac{(\pi t)}{2}) + sin(4\cdot \pi t) + 5t^{2} + 6t^{3} + 6\&\text{We find the velocity function: }10t + 4\cdot \pi cos(4\cdot \pi t) +\frac{ (\pi sin(\frac{(\pi t)}{2}))}{2}+ 18t^{2} + 3\&\text{Plugging in our value of time we then get:}\&v(6)=10(6) + 4\cdot \pi cos(4\cdot \pi (6)) +\frac{ (\pi sin(\frac{(\pi (6))}{2}))}{2}+ 18(6)^{2} + 3\&v(6)=723.57\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{A particle at any point in time has the position :}\&p(t)=- cos(\frac{(11\cdot \pi t)}{2}) - 6\&\text{Determine its velocity at time }t=2\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }- cos(\frac{(11\cdot \pi t)}{2}) - 6\&\text{We find the velocity function: }\frac{(11\cdot \pi sin(\frac{(11\cdot \pi t)}{2}))}{2}\&\text{Plugging in our value of time we then get:}\&v(2)=\frac{(11\cdot \pi sin(\frac{(11\cdot \pi (2))}{2}))}{2}\&v(2)=0\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=sin(\frac{(5\cdot \pi t)}{2}) + 2\&\text{What is its velocity at time }t=6\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }sin(\frac{(5\cdot \pi t)}{2}) + 2\&\text{We find the velocity function: }\frac{(5\cdot \pi cos(\frac{(5\cdot \pi t)}{2}))}{2}\&\text{Plugging in our value of time we then get:}\&v(6)=\frac{(5\cdot \pi cos(\frac{(5\cdot \pi (6))}{2}))}{2}\&v(6)=-7.85\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=3t + cos(\frac{(7\cdot \pi t)}{2}) + sin(\frac{(9\cdot \pi t)}{2}) - t^{2} - 6t^{3} + 3\&\text{What is its velocity at time }t=10\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }3t + cos(\frac{(7\cdot \pi t)}{2}) + sin(\frac{(9\cdot \pi t)}{2}) - t^{2} - 6t^{3} + 3\&\text{We find the velocity function: }\frac{(9\cdot \pi cos(\frac{(9\cdot \pi t)}{2}))}{2}- 2t -\frac{ (7\cdot \pi sin(\frac{(7\cdot \pi t)}{2}))}{2}- 18t^{2} + 3\&\text{Plugging in our value of time we then get:}\&v(10)=\frac{(9\cdot \pi cos(\frac{(9\cdot \pi (10))}{2}))}{2}- 2(10) -\frac{ (7\cdot \pi sin(\frac{(7\cdot \pi (10))}{2}))}{2}- 18(10)^{2} + 3\&v(10)=-1831.14\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{A particle at any point in time has the position :}\&p(t)=cos(\frac{(\pi t)}{2}) - 4\&\text{Determine its velocity at time }t=3\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }cos(\frac{(\pi t)}{2}) - 4\&\text{We find the velocity function: }-\frac{(\pi sin(\frac{(\pi t)}{2}))}{2}\&\text{Plugging in our value of time we then get:}\&v(3)=-\frac{(\pi sin(\frac{(\pi (3))}{2}))}{2}\&v(3)=1.57\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{A particle at any point in time has the position :}\&p(t)=cos(2\cdot \pi t) - 6t + sin(7\cdot \pi t) + 3t^{2} + 5\&\text{Determine its velocity at time }t=5\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }cos(2\cdot \pi t) - 6t + sin(7\cdot \pi t) + 3t^{2} + 5\&\text{We find the velocity function: }6t + 7\cdot \pi cos(7\cdot \pi t) - 2\cdot \pi sin(2\cdot \pi t) - 6\&\text{Plugging in our value of time we then get:}\&v(5)=6(5) + 7\cdot \pi cos(7\cdot \pi (5)) - 2\cdot \pi sin(2\cdot \pi (5)) - 6\&v(5)=2.01\end{align}$

Compare your answer with the correct one above

Question

$\begin{align}&\text{The position of a particle at a given time is given by the function:}\&p(t)=- 6t - cos(\frac{(7\cdot \pi t)}{2}) - cos(\frac{(15\cdot \pi t)}{2}) - 4t^{2} - 3\&\text{What is its velocity at time }t=4\text{?}\end{align}$

Answer

$\begin{align}&\text{Velocity is the time rate of change of position. In lay terms,}\&\text{it is an instantaneous measure of speed and direction. However,}\&\text{going back to that first definition, note that it is a rate,}\&\text{and whenever the term rate is used, derivatives should come}\&\text{to mind. The rate of change of position with respect to time}\&\text{is the derivative of position with respect to time:}\&v(t)=\frac{dp(t)}{dt}\&\text{Taking the derivative of our position function, }- 6t - cos(\frac{(7\cdot \pi t)}{2}) - cos(\frac{(15\cdot \pi t)}{2}) - 4t^{2} - 3\&\text{We find the velocity function: }\frac{(7\cdot \pi sin(\frac{(7\cdot \pi t)}{2}))}{2}- 8t +\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi t)}{2}))}{2}- 6\&\text{Plugging in our value of time we then get:}\&v(4)=\frac{(7\cdot \pi sin(\frac{(7\cdot \pi (4))}{2}))}{2}- 8(4) +\frac{ (15\cdot \pi sin(\frac{(15\cdot \pi (4))}{2}))}{2}- 6\&v(4)=-38\end{align}$

Compare your answer with the correct one above

Tap the card to reveal the answer