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Calculus 1 : Calculus

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Example Questions

Example Question #72 : Rate

Let $f(x) = ln(e^{x}-sin(\pi -\pi x))$ . Use linear approximation to estimate $f(\frac{3}{2})$ .

Possible Answers:

$2-\frac{\pi }{e^{3/2}}$

$1-\frac{\pi }{e^{3/2}}$

$\frac{\pi }{e^{3/2}}$

$\frac{3}{2}$

Correct answer:

$2-\frac{\pi }{e^{3/2}}$

Explanation:

Note that:

f(1)= ln(e-sin(0))= lne =1

Therefore, for values relatively close to 1, we can use the formula for dy (the differential form of the derivative) to estimate f at close values.

$dy= \frac{e^{x}+\pi cos(\pi -\pi x)}{e^{x}-sin(\pi -\pi x)}dx$ From log derivative with chain rule.

Since $\frac{3}{2}$ lies $\frac{1}{2}$ to the right of , $dx = \frac{1}{2}$ and x=1 for the estimation, so:

$dy = \frac{e^{3/2}+\pi cos(-\pi )}{e^{3/2}} = 1-\frac{\pi }{e^{3/2}}$

So then:

$f(3/2) \approx1+1-\frac{\pi }{e^{3/2}} = 2-\frac{\pi }{e^{3/2}}$

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Example Question #3 : How To Find Approximation Of Rate

Using $(1+x)^{k}\approx1+kx$ , approximate the value of $(0.985)^{50}.$

Possible Answers:

0.25

1.25

1.75

1.0

0.75

Correct answer:

0.25

Explanation:

First, we need to rearrange the given to match the approximation formula. 0.985 = 1-0.015. Therefore, $(0.985)^{50}=(1-0.015)^{50}=1-50(0.015)=0.25.\textup{}$

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Example Question #4 : How To Find Approximation Of Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{7}{30}$ ?

Possible Answers:

$\frac{49}{2700}$

$\frac{49}{900}$

$\frac{49}{300}$

$\frac{7}{900}$

$\frac{7}{300}$

Correct answer:

$\frac{49}{300}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

V=s^3

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{7}{30})^2=\frac{49}{300}$

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Example Question #5 : How To Find Approximation Of Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{13}{15}$ ?

Possible Answers:

$\frac{169}{225}$

$\frac{169}{75}$

$\frac{13}{75}$

$\frac{169}{675}$

$\frac{13}{225}$

Correct answer:

$\frac{169}{75}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

V=s^3

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{13}{15})^2=\frac{169}{75}$

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Example Question #6 : How To Find Approximation Of Rate

A cube is diminishing in size. What is the ratio of the rate of loss of the cube's volume to the rate of loss of its sides when its sides have length $\frac{1}{30}$ ?

Possible Answers:

$\frac{1}{2700}$

$\frac{1}{810}$

$\frac{1}{90}$

$\frac{1}{900}$

$\frac{1}{300}$

Correct answer:

$\frac{1}{300}$

Explanation:

Begin by writing the equations for a cube's dimensions. Namely its volume in terms of the length of its sides:

V=s^3

The rates of change of the volume can be found by taking the derivative of each side of the equation with respect to time:

$\frac{dV}{dt}=3s^2\frac{ds}{dt}$

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

$3s^2\frac{ds}{dt}=(\phi)\frac{ds}{dt}$

$\phi=3s^2$

$\phi =3(\frac{1}{30})^2=\frac{1}{300}$

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Example Question #1 : Differentiable Rate

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

Possible Answers:

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

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

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

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

Correct answer:

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

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 $\small y'$ :

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

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

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

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Example Question #2 : Differentiable Rate

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 C(0)=40M , seconds after the reaction began.

Possible Answers:

$30e^{(-40)}$

30M

70M

$40e^{(-30)}$

Correct answer:

70M

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

C=kt+C_0 , where C_0 is the initial concentration.

Plugging in our values,

C=3*10+40= 70M

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Example Question #2901 : Calculus

and are related by the function $y=3x^{4}-x+5$ . Find $\frac{dy}{dt}$ at t=2 if x=1 and $\frac{dx}{dt}=6$ at t=2 .

Possible Answers:

$\frac{dy}{dt}=12$

$\frac{dy}{dt}=66$

$\frac{dy}{dt}=6$

$\frac{dy}{dt}=60$

$\frac{dy}{dt}=73$

Correct answer:

$\frac{dy}{dt}=66$

Explanation:

We will use the chain and power rules to differentiate both sides of this equation.

Power Rule:

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

Chain Rule:

$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$ .

Applying the above rules to our function we find the following derivative.

$\frac{dy}{dt}=\frac{d}{dt}[3x^{4}-x+5]$

$\frac{dy}{dt}=12x^{3}\frac{dx}{dt}-1\frac{dx}{dt}$

at t=2 , $\frac{dx}{dt}=6$ and x=1 .

Therefore at t=2

$\frac{dy}{dt}=12(1)^{3}(6)-1(6)=72-6=66$

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Example Question #4 : How To Find Differentiable Of Rate

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

Possible Answers:

Correct answer:

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:

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

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Example Question #1 : How To Find Differentiable Of Rate

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 dy=f'(x)dx . 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.

Let f(x)= 3x^2 + x -1 . Find f'(3) and, given $\Delta x = 0.01,$ and find $\Delta y$ .

Possible Answers:

1.9

0.19

-0.19

0.1

0.034

Correct answer:

0.19

Explanation:

Taking the derivative of the function:

$f'(x)= \frac{d}{dx}[3x^2 + x -1]$

$\frac{dy}{dx}= 6x + 1$

Evaluating at x=3 :

$\frac{dy}{dx}|_{x=3} = 6(3) + 1$

$\frac{dy}{dx}|_{x=3} = 6(3) + 1=19$

Manipulating the equation:

dy =19dx

Allowing dx to be .01:

dy=19*0.01

dy=0.19

Which is our answer.

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #72 : Rate

Example Question #3 : How To Find Approximation Of Rate

Example Question #4 : How To Find Approximation Of Rate

Example Question #5 : How To Find Approximation Of Rate

Example Question #6 : How To Find Approximation Of Rate

Example Question #1 : Differentiable Rate

Example Question #2 : Differentiable Rate

Example Question #2901 : Calculus

Example Question #4 : How To Find Differentiable Of Rate

Example Question #1 : How To Find Differentiable Of Rate

All Calculus 1 Resources

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