AP Calculus AB - Differentiable Rate

Find the rate of change of $y=4x^{3}-12x^{2}+6x+7$ at .

Explanation

To find the rate of change of a polynomial at a point, we must find the first derivative of the polynomial and evaluate the derivative at that point.

For this problem,

$y=4x^{3}-12x^{2}+6x+7$

the first derivative of this expression is

$\frac{dy}{dx}=12x^{2}-24x+6$

at the rate of change is

$\frac{dy}{dx}=12*(2)^{2}-24(2)+6=48-48+6=6$

The find the change of volume of a spherical balloon that is growing from to

$50\pi; cubic ;inches$

$25\pi; cubic ;inches$

$5\pi; cubic ;inches$

$100\pi; cubic ;inches$

$10\pi;cubic;inches$

Explanation

This is a related rate problem. To find the rate of change of volume with respect to radius, we need to take the derivative of the volume of a sphere equation $(\frac{4}{3}\pi r^{3}).$

Then, we will plug in the relevant information. The initial radius will be substituted in for , and , since that is the change from the initial to final radius of the balloon.

$dV=4\pi r^{2}dr.$

$dV=4\pi (5in)^{2}(0.5in)=50\pi in^{3}.$

For the relation $\small x^2y+2x-y^2=x$ , compute $\small y'$ using implicit differentiation.

$\small \small \small \small y'=-\frac{2xy+1}{x^2-2y}$

$\small \small \small \small y'=\frac{2xy+1}{x^2-2y}$

$\small \small \small \small y'=-\frac{2x+1}{x^2-2y}$

$\small \small \small \small y'=-\frac{2xy+1}{y(x^2-2y)}$

Explanation

Computing $\small y'$ of the relation $\small x^2y+2x-y^2=x$ can be done through implicit differentiation:

$\small \small \frac{d}{dx}(x^2y+2x-y^2)=\frac{d}{dx}x$

$\small \small \implies 2xy+x^2y'+2-2yy'=1$

Now we isolate the $\small y'$ :

$\small \small \small \iff x^2y'-2yy'=1-2-2xy=-2xy-1$

$\small \iff y'(x^2-2y)=-(2xy+1)$

$\small \small \implies y'=-\frac{2xy+1}{x^2-2y}$

Let $y = x^{xlnx-e}.$ Use logarithmic differentiation to find .

Explanation

The form of log differentiation after first "logging" both sides, then taking the derivative is as follows:

$\frac{1}{y}y' = \frac{d}{dx}(lnf(x))$ which implies

$y' = f(x)\frac{d}{dx}(lnf(x))$

So:

$y'(e)=e^{elne-e}(ln^{2}e+2lne-\frac{e}{e})= 1(1+2-1)= 2$

In chemistry, rate of reaction is related directly to rate constant .

$\frac{dC}{dt}=k$ , where is the initial concentration

Give the concentration of a mixture with rate constant $k=\frac{3}{sec}$ and initial concentration , seconds after the reaction began.

$40e^{(-30)}$

$30e^{(-40)}$

Explanation

This is a simple problem of integration. To find the formula for concentration from the formula of concentration rates, you simply take the integral of both sides of the rate equation with respect to time.

$\int \frac{dC}{dt} dt=\int (k)dt$

Therefore, the concentration function is given by

, where is the initial concentration.

Plugging in our values,

We can interperet a derrivative as $\frac{dy}{dx}= \lim_{\Delta \rightarrow 0} \frac{\Delta y}{\Delta x}$ (i.e. the slope of the secant line cutting the function as the change in x and y approaches zero) but these so-called "differentials" ( $\Delta x$ and $\Delta y$ ) can be a good tool to use for aproximations. If we suppose that $\frac{dy}{dx}=f'(x)$ , or equivalently . If we suppose a change in x (have a concrete value for $\Delta x$ ) we can find the change in with the afore mentioned relation.