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Calculus 1 : How to write equations

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Example Question #2289 : Calculus

Find the limit:

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

Possible Answers:

undefined

$\large \infty$

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

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

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 #2290 : 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:

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

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

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

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

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

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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Example Question #21 : Other Writing Equations

Find the equation of the line tangent to $f(x)=3x^{3}-x^{2}+1$ at x=-2 .

Possible Answers:

y=x+53

y=20x-17

y=-40x+53

The tangent line does not exist at this point.

y=40x+53

Correct answer:

y=40x+53

Explanation:

To find the equation for the tangent line, we must first find the derivative of f(x) .

In order to evaluate this derivative, we must use these formulae:

$\frac{d}{dx}(c)=0, \frac{d}{dx}(x)=1, \frac{d}{dx}cx=c$

$\frac{d}{dx}(x^{n}))=nx^{n-1}, \frac{d}{dx}(cx^{n})=ncx^{n-1}$

$f'(x)=3*3x^{3-1}-2x^{2-1}+0=9x^{2}-2x$

Next we find the slope of the tangent line at this point:

$f'(-2)=9(-2)^{2}-2(-2)=36+4=40$

Then we need to plug the given value back into the original function in order to find the corresponding value.

$f(-2)=3*(-2)^{3}-(-2)^{2}+1=-24-4+1=-27$

Now that we have a point (-2,-27) and a slope of , we can find the equation for the line by plugging these values into y=mx+b form to solve for :

-27=40*-2+b ==> b = 53

So, the equation for the line is y=40x+53 .

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Example Question #22 : Other Writing Equations

Give the position equation p(t) for an object falling towards Earth with a constant acceleration of $-9.8 \frac{m}{s^2}$ , with initial velocity of $13 \frac{m}{s}$ , and initial position at 35m above the ground.

Possible Answers:

p(t)=-9.8t^2+13t+35

p(t)=-4.9t^2+35t+13

p(t)=-4.9t^2+13t+35

p(t)=4.9t^2+13t+35

Correct answer:

p(t)=-4.9t^2+13t+35

Explanation:

For this equation, let's start with acceleration and move backwards towards position. We know that the object is accelerating at a constant rate of -9.8

p''(t)=-9.8

To solve for velocity , we integrate both sides with respect to

$\int p''(t)dt=\int -9.8 dt$

p'(t)=-9.8t+B , where is a constant

Since we know that the initial velocity is , p'(0)=13, B=13 .

p'(t)=-9.8t+13

To solve for position we have to take the integral of both sides with respect to .

$\int p'(t)dt=\int (-9.8t+13)dt$

p(t)=-4.9t^2+13t+C , where is a constant.

Since we know that the initial position is , P(0)=35 , C=35

The formula for this motion can be modelled by:

p(t)=-4.9t^2+13t+35

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Example Question #23 : Other Writing Equations

Inverse Function

What is the inverse function of the following:

4y-x=4

Possible Answers:

4x+y+4=0

4x=y-4

4x-y=1

None of the above

$x-{\frac14}y=1$

Correct answer:

$x-{\frac14}y=1$

Explanation:

4y-x=4 can be written as $y=\frac{1}{4}x+1$ with slope of $\frac{1}{4}$ and y-intercept of 1. So the inverse function would be $x=\frac{1}{4}y+1$ or, $x-{\frac14}y=1$

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Example Question #24 : Other Writing Equations

Logarithm Functions

Solve for ,

$\ln x=-9$

Possible Answers:

9-e

$-e\sqrt{9}$

$e^{-9}$

e-9

Correct answer:

$e^{-9}$

Explanation:

$\ln x=-9$ implies $e^{lnx}=e^{-9}$ implies $x=e^{-9}$

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Example Question #25 : Other Writing Equations

Write the integral expression for how to find the area under the curve f(x) from the point (-3,4) to (5,6) .

f(x)=4x^3-2x+7

Possible Answers:

$\int_{4}^{6}(4x^3-2x+7)dx$

$\int_{4}^{5}(4x^3-2x+7)dx$

$\int_{-3}^{6}(4x^3-2x+7)dx$

$\int_{-3}^{5}(4x^3-2x+7)dx$

Correct answer:

$\int_{-3}^{5}(4x^3-2x+7)dx$

Explanation:

To write the expression, simply use the bounds laid out by the x-coordinates of the points given. Thus, your answer is:

$\int_{-3}^{5}(4x^3-2x+7)dx$

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Example Question #26 : Other Writing Equations

$f(x) = \left\{\begin{matrix} cx^2-2x & x<3\\ x& x \geq 3 \end{matrix}\right.$

For what values of the constant is the function f(x) continuous on $(-\infty, \infty)$ ?

Possible Answers:

Correct answer:

Explanation:

Both cx^2-2x and are continuous in the intervals they are defined. In order for the function to be continuous on the entire real line, we need to have cx^2 -2x =x at x =3 . Plugging the value in, we get $c \cdot 3^2 - 2 \cdot 3 = 3$ $\Rightarrow 9c-6 =3 \Rightarrow c=1.$

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Example Question #211 : Writing Equations

White the expression for the derivative of the following function.

p(t)=2t^2-1

Possible Answers:

p'(t)=4t

p'(t)=2t^2-1

$p'(t)=\frac{2}{3}t^3-t$

p(t)=4t

Correct answer:

p'(t)=4t

Explanation:

To solve, simply differentiate and don't forget to indicate the differentiation by the "apostrophe" between the function and dependent variable. Thus,

p'(t)=4t

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Example Question #28 : Other Writing Equations

Find the equation for the velocity whose position function is below:

p(t)=5t^2-6t-3

Possible Answers:

v(t)=10

v(t)=10t+6

v(t)=5t^2-6t-3

v(t)=10t-6

Correct answer:

v(t)=10t-6

Explanation:

To solve, simply differentiate p(t) .

Remember to use the power rule.

Recall the power rule:

f(x)=x^n

$f'(x)=nx^{n-1}$

Apply this to our situation to get

v(t)=p'(t)=10t-6

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Calculus 1 : How to write equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #2289 : Calculus

Example Question #2290 : Calculus

Example Question #21 : Other Writing Equations

Example Question #22 : Other Writing Equations

Example Question #23 : Other Writing Equations

Example Question #24 : Other Writing Equations

Example Question #25 : Other Writing Equations

Example Question #26 : Other Writing Equations

Example Question #211 : Writing Equations

Example Question #28 : Other Writing Equations

All Calculus 1 Resources

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