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Calculus 1 : Equations

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Example Questions

Example Question #201 : Writing Equations

Identify the inner and the outer functions of the following equation (let f(x) be the outer equation and g(x) be the inner equation:

y=(2x^3+5)^4

Possible Answers:

f(x)=2x^3+5

g(x)=x^4

f(x)=2x^3

g(x)=x+5

Cannot be determined.

f(x)=x^4

g(x)=2x^3+5

Correct answer:

f(x)=x^4

g(x)=2x^3+5

Explanation:

The first section of the equation to be resolved (in this case 2x^3+5) is the innermost function and the second section to solve is the outer function.

So:

f(x)=x^4

g(x)=2x^3+5

becuase g(x) must be performed first before plugging it into f(x).

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Example Question #202 : Writing Equations

Identify the inner and the outer functions of the following equation (let f(x) be the outer equation and g(x) be the inner equation:

$y=\tan\pi x$

Possible Answers:

$f(x)=\tan(x)$

$g(x)=\pi x$

$f(x)=\tan (\pi )$

g(x)=x

Cannot be determined.

$g(x)=\tan(x)$

$f(x)=\pi x$

Correct answer:

$f(x)=\tan(x)$

$g(x)=\pi x$

Explanation:

The first section of the equation to be resolved (in this case pi * x) is the innermost function and the second section to solve is the outer function.

So:

$f(x)=\tan(x)$

$g(x)=\pi x$

becuase g(x) must be performed first before plugging it into f(x).

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Example Question #11 : Other Writing Equations

Derive the equation for the function of the cot(x) .

Possible Answers:

$-\csc(x)^{2}$

$-\csc (x)$

$\sec (x)^{2}$

$\tan (x)$

None of these

Correct answer:

$-\csc(x)^{2}$

Explanation:

$\cot(x)=\frac{\cos (x)}{\sin (x)}$

By the quotient rule, the derivative of

$\frac{f(x)}{g(x)}=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$

The derivative of sin is cos and the derivative of cos is -sin. Thus the derivative of cot(x) is

$\frac{-\sin (x)^2-\cos (x)^2}{\sin (x)^2}$

$=-(1+\cot (x)^2)$

By the trigonometric identities this is equal to $-\csc(x)^{2}$ .

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Example Question #11 : How To Write Equations

Suppose Charlie deposits $\$15$ per month into an account that contains a starting balance of $\$60$ .

Which of the following is a correct initial condition that when coupled with the differential equation from the previous question, will yield a specific solution to the scenario described above?

Possible Answers:

M(0)=0`

M(0)=60

M(0)=15

M(12)=180

M(12)=60

Correct answer:

M(0)=60

Explanation:

Here we are looking for an initial condition that describes the balance of the account at a specific time. We are given the information that the account starts with $\$60$ initially. So at time t=0 years, we know that the amount of money in the account is $\$60$ . Therefore, we can write M(0)=60

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Example Question #204 : Writing Equations

Suppose Charlie deposits $\$15$ per month into an account that contains a starting balance of $\$60$ .

Which of the following is a differential equation that best models the amount of money, , in the account after years?

Possible Answers:

$\frac{dM}{dt}=15$

$\frac{dM}{dt}=180$

$\frac{dM}{dt}=180t$

$\frac{dM}{dt}=180t+60$

$\frac{dM}{dt}=15t+60$

Correct answer:

$\frac{dM}{dt}=180$

Explanation:

Here we are writing a differential equation in the units of dollars per year. This means that we need to reconcile our units to get the amount deposited each year. $\$15/month = \$180/year$ ,

so the yearly rate of change of money in the account,

$\frac{dM}{dt}=180$ .

The starting balance is our initial condition and does not tell us about the change in money in the account, so is not included in our differential equation.

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Example Question #2285 : Calculus

Find the implicit derivative $\frac{\mathrm{d} y}{\mathrm{d} x}$ of 2x^2y+yx-y^3=2x-y at the point (0,1){} .

Possible Answers:

$-\frac{3}{2}$

$-\frac{1}{2}$

$-\frac{3}{4}$

$-\frac{2}{3}$

Correct answer:

$-\frac{1}{2}$

Explanation:

Use implicit differentiation to find $\frac{\mathrm{d} y}{\mathrm{d} x}$ of 2x^2y+yx-y^3=2x-y which means to take the derivative of each term in the function with its respective part.

It is simpfied to

$4xy+2x^2*\frac{\mathrm{d} y}{\mathrm{d} x} + y+x*\frac{\mathrm{d} y}{\mathrm{d} x} - 3y^2*\frac{\mathrm{d} y}{\mathrm{d} x}=2 -\frac{\mathrm{d} y}{\mathrm{d} x}$ .

It is then further simplified to

$2x^2*\frac{\mathrm{d} y}{\mathrm{d} x}+x*\frac{\mathrm{d} y}{\mathrm{d} x}-3y^2*\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{\mathrm{d} y}{\mathrm{d} x}=2-4xy-y$ ,

then to

$\frac{\mathrm{d} y}{\mathrm{d} x}(2x^2+x-3y^2+1)=2-4xy-y$ ,

then to

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2-4xy-y}{2x^2+x-3y^2+1}$ .

Plugging in (0,1){} into the equation gives the value of $-\frac{1}{2}$ .

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Example Question #2286 : Calculus

If f(x)=3g(x) and g(x)=f(x) , then what is f''(x^2) ?

Possible Answers:

3g(x^2)+12xf(x^2)

3g(x^2)+6xf(x^2)

6g(x^2)+12x^2f(x^2)

6g(x^2)+12xf(x^2)

6g(x)+12f(x)

Correct answer:

6g(x^2)+12x^2f(x^2)

Explanation:

First we need to find f'(x^2) , which is f'(x^2)=3g(x^2)2x=6xg(x^2) .

Then we find f''(x^2) , which equals to

$f''(x^2)=\frac{\mathrm{d} (6xg(x^2))}{\mathrm{d} x}=6g(x^2)+6xf(x^2)2x=6g(x^2)+12x^2f(x^2)$ .

We used the power rule and the product rule to find the derivatives.

Power Rule: $x^n \rightarrow nx^{n-1}$

Product Rule: $f(x)g(x)\rightarrow f(x)g'(x)+g(x)f'(x)$

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Example Question #2287 : Calculus

$y=x^{x^3+2x^2}$ , find $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^3+4x)]*(x^{x^3+2x^2})$

$\frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^2+2x)]*(x^{x^3+2x^2})$

$\frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^3+2x^2)]*(x^{x^3+2x^2})$

$\frac{\mathrm{d} y}{\mathrm{d} x}=2[(3x^2+4x)\ln (x)+(x^2+2x)]*(x^{x^3+2x^2})$

$\frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^2+2x)]$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^2+2x)]*(x^{x^3+2x^2})$

Explanation:

Our starting function is $y=x^{x^3+2x^2}$ in order to create an easier function we will want to take the natural log of both sides. Taking the natural log will bring the exponet infront creating a product for which we can more easily differentiate.

Next take the natural log of both sides which results in the following,

$\ln(y)=\ln (x)^{x^3+2x^2}=(x^3+2x^2)\ln (x)$ .

Now differentiate both sides with respect to x.

$\frac{1}{y}*\frac{\mathrm{d} y}{\mathrm{d} x}=(3x^2+4x)\ln (x)+\frac{1}{x}*(x^3+2x^2)=(3x^2+4x)\ln (x)+(x^2+2x)$ , and when one isolates $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

After isolating $\frac{\mathrm{d} y}{\mathrm{d} x}$ make sure to replace the y term in terms of x.

$\\ \frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^2+2x)]*y\\ \\=[(3x^2+4x)\ln (x)+(x^2+2x)]*(x^{x^3+2x^2})$

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Example Question #2288 : Calculus

Find the limit:

${} \lim_{x\rightarrow \infty} \frac{3x^2+5x+2}{2x^2+x+1}$

Possible Answers:

$\large \infty$

undefined

$\large \frac{3}{2}$

$\large \frac{5}{2}$

Correct answer:

$\large \frac{3}{2}$

Explanation:

This is an indeterminate limit of the form ${} \frac{\infty}{\infty}$ so we can use L'Hopital's rule:

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}$

(this only works when our original limit is an indeterminate form!)

We take the derivative of the top and bottom and this becomes

${} \lim_{x \to \infty}\frac{6x+5}{4x+1}$ .

But yet again, this is indeterminate, of the same type. So we apply L'Hopital's rule one more time and get

${} \lim_{x \to \infty}\frac{6}{4}$

which reduces to 3/2.

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Example Question #2289 : Calculus

Given the acceleration of a baseball is given by $a(t)=8t^{2}+2t$ , what is the average value of the acceleration function over the interval t=1 to t=5 ?

Possible Answers:

$35 \frac{meters}{sec^{2}}$

$88.6667 \frac{meters}{sec^{2}}$

$-14.2 \frac{meters}{sec^{2}}$

$19.86 \frac{meters}{sec^{2}}$

$29.33333 \frac{meters}{sec^{2}}$

Correct answer:

$88.6667 \frac{meters}{sec^{2}}$

Explanation:

To find the average value of this function, we must use the formula:

$Average Value=\frac{\int_{a}^{b}a(t)}{b-a}=\frac{\int_{1}^{5}8t^{2}+2t}{5-1}$

To integrate this function, we must use these formulas:

$\int cf(x)dx=c\int f(x)dx, \int x^{n}dx=\frac{1}{n+1}x^{n+1}$

$\\ \int_{1}^{5} a(t)dt=\frac{8t^{3}}{3}+\frac{2t}{2}\\=\left [ \frac{8t^{3}}{3}+t\right ]^{5}_{1} =\left(\frac{8(5)^{3}}{3}+5\right)-\left(\frac{8(1)^{3}}{3}+1\right)$

=358.33-3.67=354.667

$=\frac{354.667}{4}=88.67 \frac{meters}{sec^{2}}$

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #201 : Writing Equations

Example Question #202 : Writing Equations

Example Question #11 : Other Writing Equations

Example Question #11 : How To Write Equations

Example Question #204 : Writing Equations

Example Question #2285 : Calculus

Example Question #2286 : Calculus

Example Question #2287 : Calculus

Example Question #2288 : Calculus

Example Question #2289 : Calculus

All Calculus 1 Resources

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