Lines - AP Calculus AB

Card 0 of 672

Question

Find the equation of the line tangent to at .

Answer

To get the slope, find the derivative of and plug in the desired point for , giving us an answer of $\frac{2}{5}$ for the slope.

$f'(x)=\frac{2}{2x+1}$

$f'(2)=\frac{2}{5}$

Remember that the derivative of $ln(u) = \frac{u'}{u}$ .

To find the adjustment pick a point (for example) in the original function. For simplicity, let's plug in , which gives us a of , so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

$3=\frac{2}{5}(0)+b$

The coefficient in front of the is the slope.

Now you simply solve for , which is .

Final equation of the line tangent to at is $y=\frac{2}{5}x +3$ .

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Question

Consider a function with first-derivative:

$\frac{dy}{dx}=\tan{x}$ .

Which integral can calculate the length along this curve between and $x=2\pi$ ?

Answer

To determine the length of a curve between two points, we evaluate the integral

$L = \int_0^{2\pi}{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$

There are many reasons this works, but we'll give an informal explanation here:

If we divide this curve into three line-segments, we can see that they become more and more similar to the original curve. By adding up all the little hypotenuses, we can arrive at the length of the curve. Note, that

$\sqrt{dx^2 + dy^2} = \sqrt{1+\left(\frac{dy}{dx} \right )^2}dx$

If we think of the integration symbol as a sum of infinitely small parts, this gets us the formula for length:

$L = \int_a^b{\sqrt{1+\left(\frac{dy}{dx} \right )^2}}dx$ .

Returning to the problem, we plug the derivative into the length formula:

$L=\int_{0}^{2\pi}{\sqrt{1+\tan^2{x}}}\quad dx$

substitute:

$L=\int_{0}^{2\pi}{\sqrt{\left( \frac{\cos{x}}{\cos{x}} \right)^2+\left(\frac{\sin{x}}{\cos{x}} \right )^2}}\quad dx$

Simplify:

$L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}}{\cos^2{x}}+\frac{\sin^2{x}}{\cos^2{x}}}}\quad dx$

$L=\int_{0}^{2\pi}{\sqrt{\frac{\cos^2{x}+\sin^2{x}}{\cos^2{x}}}}\quad dx$

By trigonometric identities we get:

$L=\int_{0}^{2\pi}{\sqrt{\frac{1}{\cos^2{x}}}}\quad dx$

$L=\int_{0}^{2\pi}{\frac{1}{\cos{x}}}\quad dx$

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Question

Find the length of the line segment between points A and B:

Answer

The distance between two points can be easily found using the Distance Formula:

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Applying the points we are given to this formula results in:

$D=\sqrt{((-5)-2)^2+(10-(-7))^2}=\sqrt{(-7)^2+(17)^2}=\sqrt{338}$

This is one of the answer choices.

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Question

What is the equation of the line tangent to f(x) = 4x3 – 2x2 + 4 at x = 5?

Answer

First, take the derivative of f(x). This is very easy:

f'(x) = 12x2 – 4x

Now, the slope of the tangent line at x = 5 is f'(5). Evaluated, this is: f'(5) = 12 * 5 * 5 – 4 * 5 = 300 – 20 = 280

Now, we must find the intersection point on the original line:

f(5) = 4 * 53 – 2 * 52 + 4 = 500 – 50 + 4 = 454

Therefore, the point of tangent intersection is (5,454).

Using the point-slope form of linear equations, we can find the line:

(y – 454) = 280(x – 5)

y – 454 = 280x – 1400

y = 280x – 946

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Question

Find the equation of the line tangent to at the point .

Answer

The equation of the tangent line will have the form $y-y_{0}=m(x-x_{0})$ , where is the slope of the line and $(x_{0},y_{0})=(1,3)$ .

To find the slope, we need to evaluate the derivative at :

$\frac{dy}{dx}=\frac{d(x^2-x+3)}{dx}=2x-1\Rightarrow m=f'(1)=2*1-1=1$

Now that we have the slope, we can determine the equation of the tangent line using the point-slope formula:

$y-y_{0}=m(x-x_{0})\Rightarrow y-3=1(x-1)\Rightarrow y=x+2.$

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Question

Find the equation of the line tangent to at .

$f(x)=e^{3x} +x^2$

Answer

To find the equation of the line at that point, you need two things: the slope at that point and the y adjustment of the function, in the form , where m is the slope and is the y adjustment. To get the slope, find the derivative of and plug in the desired point for , giving us an answer of for the slope.

$f'(x)=3e^{3x} + 2x$

To find the y adjustment pick a point 0 in the original function. For simplicity, let's plug in , which gives us a y of 1, so an easy point is . Next plug in those values into the equation of a line, . The new equation with all parameters plugged in is

Now you simply solve for , which is .

Final equation of the line tangent to at is

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Question

Find the equation of the line tangent to at $\frac{\pi }{2}$ .

Answer

The equation of the tangent line is To find , the slope, calculate the derivative and plug in the desired point.

$y'=2cos\left(2\cdot \frac{\pi}{2}\right)$

$y'=2cos(\pi)$

$y'\left(\frac{\pi}{2} \right )=-2$

The next step is to choose a coordinate on the original function. We can choose any value and calculate its value.

Let's choose $\frac{\pi }{2}$ .

$y=sin\left(2\cdot \frac{\pi}{2}\right)=sin(\pi)=0$

The value at this point is .

Plugging in those values we can solve for .

$0=(-2)\left(\frac{\pi }{2}\right)+b$

Solving for we get = $\pi$ .

$y=-2x+\pi$

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Question

A function, , is given by

.

Find the line tangent to at $x=\frac{\pi}{2}$ .

Answer

First we need to find the slope of at $x=\frac{\pi}{2}$ . To do this we need the derivative of . To take the derivative we need to use the power rule for the first term and recognize that the derivative of sine is cosine.

$\frac{df}{dx}=2x-cos(x)$

At,

$x=\frac{\pi}{2}, \frac{df}{dx}=\pi$

Now we need to know

$f\left(\frac{\pi}{2}\right)=\frac{\pi^2}{4}-1$ .

Now we have a slope, $m=\pi$ and a point $(x_1,y_1)=\left(\frac{\pi}{2},\frac{\pi^2}{4}-1\right)$

so we can use the point-slope formula to find the equation of the line.

Plugging in and rearranging we find

$y=\pi x-\frac{\pi^2}{4}-1$ .

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Question

What is the equation of the line tangent to $f(x)=\arctan (1-x^3)$ at ? Round to the nearest hundreth.

Answer

The tangent line to at must have the same slope as .

Applying the chain rule we get

$f'(x)=\frac{1}{1+(1-x^3)^2}(-3x^2)$ .

Therefore the slope of the line is,

$f'(-1)=-\frac{3}{5}=-0.6$ .

In addition, the tangent line touches the graph of at . Since , the point lies on the line.

Plugging in the slope and point we get .

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Question

Let .

Find the equation for a line tangent to when .

Answer

First, evaluate when .

$f(1)=2(1)^4+e^1-e(1)=2+0\textbf{=2}$

Thus, we need a line that contains the point

Next, find the derivative of and evaluate it at .

To find the derivative we will use the power rule,

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

This indicates that we need a line with a slope of 8.

In point-slope form, $y-y_{1}=m(x-x_{1})$ , a line with the point and a slope of 8 will be:

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Question

Give the general equation for the line tangent to at the point $(x_{0},y_{0})$ .

Answer

The equation of the tangent line has the form $y - y_{0} = m(x-x_{0})$ .

The slope can be determined by evaluating the derivative of the function at $x = x_{0}$ .

$f'(x_{0}) = 2x_{0} - 8$

Plugging this into the point slope equation, we get

$y - y_{0} = (2x_{0} - 8)(x - x_{0})$

$y_{0}$ can be determined by evaluating the original function at $x = x_{0}$ .

$y_{0} = x_{0}^2 - 8x_0 + 4$

Plugging this into the previous equation and simplifying gives us

$-x_{0}^2 + x_0(2x-8) + 4$

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Question

Find the equation for the line tangent to the curve $\ln(\frac{x}{2})+2x^2-2$ at .

Answer

The derivative of the function is $\frac{1}{x}+4x$ , and is found using the power rule

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

and the rule for the derivative of natural log which is,

$\frac{d}{dx}ln(x)=\frac{1}{x}$

so plugging in gives $\frac{17}{2}$ , which must be the slope of the line since the tangent line's slope is determined by the derivative.

Thus, the line is of the form $f(x)=\frac{17x}{2}+b$ , where b is unknown.

Solve for b by setting the equation equal to and plugging in for x since that is the given point.

$0=\frac{17(2)}{2}+b$ $\Rightarrow b=-11$ , which gives us $f(x)=\frac{17x}{2}-11$

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Question

Find the equation of the tangent line, where

, at .

Answer

In order to find the equation of the tangent line at , we first find the slope.

To do this we need to find .

Since we have found , now we simply plug in 1.

Now we need to plug in 1, into , to find a point that the tangent line touches.

Now we can use point-slope form to figure out what the equation of the tangent line is at .

Remember that point-slope for is

where and is the point where the tangent line touches , and is the slope of the tangent line.

In our case, , , and .

Thus our tangent line equation at is

.

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Question

Find the equation of the tangent line of

, at .

Answer

In order to find the equation of the tangent line at , we first find the slope.

To do this we need to find using the power rule $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Since we have found , now we simply plug in 1.

Now we need to plug in 1, into , to find a point that the tangent line touches.

Now we can use point-slope form to figure out what the equation of the tangent line is at .

Remember that point-slope for is

where and is the point where the tangent line touches , and is the slope of the tangent line.

In our case, , , and .

Thus our tangent line equation at is

.

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Question

Find the slope of the line tangent to the following function at $\frac{\pi}{4}$ .

$\cos(2x)+\sin (x)$

Answer

To find the slope of the line tangent you must take the derivative of the function. The derivative of cosine is negative sine and the derivative of sine is cosine.

This makes the derivative of the function

$-2\sin (2x)+\cos (x)$ .

Plug in the given x to get the slope.

$-2\sin \left(\frac{2\pi}{4}\right)+\cos \left(\frac{\pi}{4}\right)$

$=-2+\frac{\sqrt{2}}{2}$

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Question

Find the slope of the tangent line to the following function at .

Answer

To find the slope of the line tangent to the function at a point you must first find the derivative.

The power rule states that the derivative of is $nx^{n-1}$ .

The derivative of is $\frac{1}{x}$ .

The derivative of the function is

$6x-6-\frac{2x}{x^2}$ .

Plugging in 1 for x gives

$6(1)-6-\frac{2(1)}{1^2}=-2$ .

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Question

Given the differential function , we are told that , , and . Which of the following must be true?

Answer

" is decreasing at ." is incorrect. The function is increasing at because .

" is increasing over the interval ." is possibly true, but there is not enough information to conclude that it must be true.

" has a point of inflection at ." is possibly true. Although we know that , a requirement for an inflection point, we do not know that changes signs at .

" must have at least one relative maximum." is possibly true, but there is not enough information to conclude that it must be true.

"The line is tangent to ." must be true. Because , the function travels through the point . Because , the slope of the line tangent to the curve at is 5. Use point-slope form to determine the equation of the tangent line.

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Question

Find the slope of the tangent line through the given point of the following function.

at the point

Answer

In order to find the slope of the tangent line through a certain point, we must find the rate of change (derivative) of the function. The derivative of is written as . This tells us what the slope of the tangent line is through any point in our function . In other words, all we need to do is plug-in (because our point has an x-value of 1) into . This will give us our answer, .

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Question

Find the tangent line to the function at the point .

Answer

To find the tangent line one must first find the slope, this can be given by the derivative evaluated at a point.

To find the derivative of this function use the power rule which states,

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

The derviative of is $f'(x)=3x^{2}-8x$ .

Evaluated at our point ,

$\f'(x)=3x^{2}-8x \f'(3)=3(3)^2-8(3) \f'(3)=27-24=3$

we find that the slope, m is also 3.

Now we may use the point-slope equation of a line to find the tangent line.

The point slope equation is

$y=m(x-x_{1})+y_{_{1}}$ Where $(x_{1},y_{1})$ is the point at which the line is tangent.

Using this definition we find the tangent line to be defined by .

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Question

What is the slope of the tangent line of f(x) = 3x4 – 5x3 – 4x at x = 40?

Answer

The first derivative is easy:

f'(x) = 12x3 – 15x2 – 4

The slope of the tangent line is found by calculating f'(40) = 12 * 403 – 15 * 402 – 4 = 768,000 – 24,000 – 4 = 743,996

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