Tangent line to a curve at a point and local linear approximation - AP Calculus AB

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Question

Find the minimum value of $f(x)=\frac{1}{\sqrt{9-x^2}}$

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Answer

In order to find the extreme value we need to take the derivative of the function.

$f'(x)=\frac{x}{(9-x^2)^{\frac{3}{2}}}$

After setting it equal to 0, we see that the only candidate is for . After setting into , we get the coordinate $(0,\frac{1}{3})$ as an extreme value. To confirm it is a minimum we can plot the function.

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Question

Find the equation of the line that is tangent to the graph of $f(x) = sin(2\pi x) + x^{3}$ when .

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Answer

First, evaluate $f(3)=sin(6\pi ) + 3^{3} = 27$ .

Then, to find the slope of the tangent line, find .

$f'(x) = 2\pi cos(2\pi x) + 3x^{2}$ , so $f'(3) = 2\pi + 27$ .

Therefore, the equation of the tangent line is

$y-27 = (2\pi + 27)(x-3)$ .

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Question

Differentiate,

$y = \sin(\sin(\sin\(3x)))$

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Answer

Differentiate,

$y = \sin(\sin(\sin\(3x)))$

Strategy

This one at first glance appears difficult even if we recognize that the chain rule is needed; we have a function within a function within a function within a function. To avoid making mistakes, it's best to start by defining variables to make the calculation easier to follow.

Let's start with the outermost function, we will write as a function of by setting,

$u = \sin(\sin(3x))$

$y = \sin(\underset{u}{ \underbrace{\sin(\sin(3x))}}) = \sin(u)$

$y = \sin(u)$                                                                                                                  Similarly, define to write as a function of

$u = \sin(\underset{v}{ \underbrace{\sin(3x)}}) = \sin(v)$

$u = \sin(v)$                                                                                                                  Write as a function of

$v = \sin(\underset{w}{ \underbrace{3x}} )=\sin(w)$

$v = \sin(w)$                                                                                                                  Finally, define the inner-most function, , as the function of

                                                                                                                   $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dw}\frac{dw}{dx}$

Since $\frac{dw}{dx}=3$ we will just substitute that in and move to the front.

$\frac{dy}{dx}=3\cos(u)\cos(v)\cos(w)$

That was easy enough, now just write everything in terms of by going back to the definitions of and .

$u = \sin(\sin(3x))$

$v = \sin(3x)$

$\frac{dy}{dx}=3\cos(\sin(\sin(3x)))\times \cos(\sin(3x))\times \cos(3x)$

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Question

Find the equation of the line tangent to $f(x)=x^{3}-3x^{2}+2x+4$ at the point .

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Answer

The first step is to find the derivative of the function given, which is $f'(x)=3x^{2}-6x+2$ . Next, find the slope at (1,4) by plugging in x=1 and solving for , which is the slope. You should get . This means the slope of the new line is also -1 because at the point where a slope and a line are tangent they have the same slope. Use the equation to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into to get . Now you can express the general equation of the line as .

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Question

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

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Answer

The first step is to find the derivative of the function given, which is $f'(x)=\frac{-3}{x^{2}}+2x$ .

Next, find the slope at by plugging in and solving for , which is the slope. You should get . This means the slope of the new line is also 3 because at the point where a slope and a line are tangent they have the same slope. Use the equation to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into to get . Now you can express the general equation of the line as .

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Question

Find the tangent line. Given the point (1,2)

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Answer

To find the tangent line at the given point, we need to first take the derivative of the given function.

Power Rule:

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent.

Therefore, becomes

From there we plug in the "1" from the point to get our m value of the equation . When we plug in "1" to y' we get m=-1. Then from there, we will plug our point into now that we have found m to find our b value. So,

Therefore, the tangent line is equal to

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Question

Find the line tangent at the point (0,1)

$y=e^{3x}$

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Answer

To find the tangent line at the given point, we need to first take the derivative of the given function. The rule for functions with "e" in it says that the derivative of However with this function there is also a 3 in the exponent so we will also use chain rule. Chain Rules states that we work from the outside to the inside. Meaning we will take the derivative of the outside of the equation and multiply it by the derivative of the inside of the equation.

To put this into equation it will look like $3e^{3x}$

From there we plug in the "0" from the point to get our m value of the equation . When we plug in "0" to y' we get m=3. Then from there, we will plug our point into now that we have found m to find our m value. So,

then plug this all back into the equation once more and we are left with

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Question

Find the tangent line given the point (2,4) and the equation

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Answer

To find the tangent line at the given point, we need to first take the derivative of the given function using Power Rule

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. $(nx^{^{n-1}})$

From there we plug in the "2" from the point to get our m value of the equation . When we plug in "2" to y' we get m=8. Then from there, we will plug our point into now that we have found m to find our b value. So,

Plug this back into

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Question

Find the equation of the line tangent to the curve at the point where $x = \frac{\pi}{4}$

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

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Answer

Find the equation of the line tangent to the curve at the given point

$f(x)=\tan(x)+x$ $x = \frac{\pi}{4}$

The slope of the line tangent at the given point will be equal to the derivative of at that point. Compute the derivative and find the slope for our line:

$f'(x)=\sec^2(x)+1$

$f'\left(\frac{\pi}{4} \right )=\sec^2\left(\frac{\pi}{4}\right)+1$

Evaluate the secant term:

$\sec^2\left(\frac{\pi}{4}\right)=\frac{1}{\cos^2\left(\frac{\pi}{4}\right)}=\left(\frac{2}{\sqrt{2}} \right )^2=2$

Therefore slope of the tangent line is simply:

$f'\left(\frac{\pi}{4} \right )=3$

So now we know the slope of the tangent line and can write the equation then solve for

In order to solve for we need one point on the line. Use the point where the tangent line meets the curve. Use the original function to find the "y" coordinate at this point:

$f\left(\frac{\pi}{4} \right ): : =: : \tan\left(\frac{\pi}{4}\right) +\frac{\pi}{4}: :=: : 1+\frac{\pi}{4}: : =: : \frac{4+\pi}{4}$

We now have our point:

$\left(\frac{\pi}{4}: : ,: \frac{4+\pi}{4}\right )$

Use the point to find

$\frac{4+\pi}{4}=3\left(\frac{\pi}{4} \right )+b$

$b = \frac{2-\pi}{2}$

$y = 3x+\frac{2-\pi}{2}$

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Question

Find the slope of the line tangent to the curve of d(g) when g=6.

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Answer

Find the slope of the line tangent to the curve of d(g) when g=6.

All we need here is the power rule. This states that to find the derivative of a polynomial, simply subtract 1 from each exponent and then multiply each term by their original exponent.

Constant terms will drop out when we do this, and linear terms will become constants.

From here substitute in g=6.

$\d'(6)=42(6)^2+12(6)-2 \d'(6)=1582$

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Question

Give the equation of the line tangent to the graph of the equation

$f(x) = 4x+ e^{x-2}$

at the point .

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Answer

The tangent line to the graph of at point is the line with slope that passes through that point. Find the derivative :

$f(x) = 4x+ e^{x-2}$

$f(x) = 4x+ \frac{1}{e^{2}} \cdot e^{x}$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x} \left (4x+ \frac{1}{e^{2}} \cdot e^{x} \right )$

Apply the sum rule:

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x} (4x ) + \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{1}{e^{2}} \cdot e^{x} \right )$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x} (4x ) + \frac{1}{e^{2}} \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^{x} \right )$

$f'(x) =4 + \frac{1}{e^{2}} e^{x}$

$f'(x) =4 + e^{x-2}$

$f'(2)= 4 + e^{2-2} = 4 + e^{0} = 4 + 1 = 5$

The tangent line is therefore the line with slope 5 through .Apply the point-slope formula:

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Question

Give the equation of the line tangent to the graph of the equation

$f(x) =3 \sin \frac{x}{2} +2 \cos \frac{x}{3}$

at the point .

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Answer

The tangent line to the graph of at point is the line with slope that passes through that point. Find the derivative :

$f(x) =3 \sin \frac{x}{2} +2 \cos \frac{x}{3}$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left (3 \sin \frac{x}{2} +2 \cos \frac{x}{3} \right )$

Apply the constant multiple and sum rules:

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left (3 \sin \frac{x}{2} +2 \cos \frac{x}{3} \right )$

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left (3 \sin \frac{x}{2} \right )+ \frac{\mathrm{d} }{\mathrm{d} x}\left ( 2 \cos \frac{x}{3} \right )$

$f'(x) =3 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin \frac{x}{2} \right )+2 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos \frac{x}{3} \right )$

Set $u = \frac{x}{2}$ and $v= \frac{x}{3}$ and apply the chain rule.

$f'(x) =3 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \sin u \right )+2 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \cos v \right )$

$f'(x) =3 \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \left ( \sin u \right )+2 \frac{\mathrm{d} v}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} v} \left ( \cos v \right )$

$f'(x) =3 \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \cos u +2 \frac{\mathrm{d} v}{\mathrm{d} x}\cdot (- \sin v)$

Substituting back:

$f'(x) =3 \frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{x}{2} \right )\cdot \cos \frac{x}{2} -2 \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{x}{3} \right )\cdot \sin \frac{x}{3}$

$f'(x) =3 \left ( \frac{1}{2} \right )\cdot \cos \frac{x}{2} -2 \left ( \frac{1}{3} \right )\cdot \sin \frac{x}{3}$

$f'(x) = \frac{3}{2} \cos \frac{x}{2} - \frac{2}{3} \sin \frac{x}{3}$

Evaluate using substitution:

$f'(0) = \frac{3}{2} \cos \frac{0}{2} - \frac{2}{3} \sin \frac{0}{3}$

$f'(0) = \frac{3}{2} \cos 0 - \frac{2}{3} \sin 0$

$f'(0) = \frac{3}{2}(1) - \frac{2}{3}( 0)$

$f'(0) = \frac{3}{2}$

The tangent line is therefore the line with slope $\frac{3}{2}$ through . is a -intercept, so apply the slope-intercept formula to get the equation

$y = \frac{3}{2}x+2$ .

This is not among the choices given.

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Question

Find the equation of the line parallel to the function at , and passes through the point

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Answer

We first start by finding the slope of the line in question, which we do by taking the derivative of and evaluate at .

,

We then use point slope form to get the equation of the line at the point

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Question

Find the equation of the line perpendicular to the line tangent to the following function at x=1, and passing through (0, 6):

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Answer

To find the equation of the line perpendicular to the tangent line to the function at a certain point, we must find the slope of the tangent line to the function, which is the derivative of the function at that point:

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}, \frac{\mathrm{d} }{\mathrm{d} x}ax=a$

Now, we evaluate the derivative at the given point:

We now know the slope of the tangent line, but because the line we are solving for is perpendicular to this line, its slope is the negative reciprocal, $-\frac{1}{6}$ .

With a slope and a point, we can now find the equation of the line:

$y-6=-\frac{1}{6}(x+0)$

$y=\frac{-x}{6}+6$

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Question

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

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Answer

To find the equation of a tangent line, we need two things: The tangent point, which is given as , and the slope of the tangent line at that point, which is the derivative at that point.

To find the derivative at the point, we will find , using derivative rules. Then we will plug the given point's x-coordinate into and that will give us the slope we need.

Finding the derivative of will require the power rule for each term. Recall that the power rule is $\frac{\mathrm{d} }{\mathrm{d} x}[x^n]= n x^{n-1}$ . For the , the power rule effectively removes the . Also, the derivative of a constant is , so the will drop when we get the derivative.

Applying these rules, we get

$f'(x)=(4)x^{4-1} + (3)2x^{3-1}+5 +0$

Now that we have the derivative, we effectively have a formula to find the slope of a tangent line at any point we choose. The question asks for the tangent line at . So we plug the x-coordinate of the point into the derivative function to find the slope.

The last step is to use the point-slope equation of a line to construct the equation we need. The point-slope equation is , where is the slope, and is the given point.

Plugging our slope into , and our original point in for and , we get

Now we can solve for to compare to the answer choices.

This is the equation of the tangent line at the given point. We have our answer.

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Question

Let $h(x)=\sin(x)-2x\cos(x)$ . Find the equation of the line tangent to at the point $(\pi, 2\pi)$ .

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Answer

To find the equation of a tangent line at a point, we will need the slope of the function at that point. To find this, we find the derivative of the function.

Finding this derivative will use trigonometric function derivative rules, and the product rule.

Recall that the derivative of $\sin(x)$ is $\cos(x)$ . This takes care of the first term.

To find the derivative of the next term, we need to be careful with our signs. We will use the product rule which results in two terms. The negative sign can mess us up if we aren't careful. The way we will handle this is to associate the minus sign with the term when doing the product rule. So we will find the derivative of $-2x\cos(x)$ . The product rule is as follows $\frac{\mathrm{d} }{\mathrm{d} x}[{\color{Red} f(x)}\cdot {\color{Blue} g(x)}]= {\color{Red} f'(x)}{\color{Blue} g(x)}+{\color{Blue} g'(x)}{\color{Red} f(x)}$ , where the red and blue are the two factors of the term we are differentiating. In our case, ${\color{Red} -2x}$ and ${\color{Blue} \cos(x)}$ .

${\color{Red} f'(x)}= {\color{Red} -2}$ by the special case of the power rule. The drops.

${\color{Blue} g'(x)}={\color{Blue} -\sin(x)}$ by the trigonometric derivative rules.

Putting these together we get the derivative of $-2x\cos(x)$ to be

${\color{Red} -2}\cdot{\color{Blue} \cos(x)}+ {\color{Blue} (-\sin(x))}{\color{Red} (-2x)}$

Simplifying, we get

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

Assembling all the pieces of the derivative together, we have found

$h'(x)=\cos(x)-2\cos(x)+2x\sin(x)$

Combining like terms gives

$h'(x)=-\cos(x)+2x\sin(x)$

Now we have a formula for the tangent slope of at any point. The point we care about is $(\pi,2\pi)$ . To find this point's tangent slope, we will plug its x-coordinate into our derivative.

$h'(\pi)=-\cos(\pi)+2(\pi)\sin(\pi)$

Simplifying gives

$-(-1)+2(\pi)(0)$

So we have found the slope of the tangent line at our point, $(\pi,2\pi)$ , is .

The last step is the point-slope equation of a line, , where is the slope and is the given point. Plugging in for , $\pi$ for , and $2\pi$ for , we get

$(y - 2\pi)= 1(x-\pi)$

Solving for , we get

$y - 2\pi=x - \pi$

$y = x -\pi + 2\pi$

$y = x + \pi$

This is the equation of the tangent line at the given point. We have our answer.

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Question

Find the minimum value of $f(x)=\frac{1}{\sqrt{9-x^2}}$

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Answer

In order to find the extreme value we need to take the derivative of the function.

$f'(x)=\frac{x}{(9-x^2)^{\frac{3}{2}}}$

After setting it equal to 0, we see that the only candidate is for . After setting into , we get the coordinate $(0,\frac{1}{3})$ as an extreme value. To confirm it is a minimum we can plot the function.

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Question

Find the equation of the line that is tangent to the graph of $f(x) = sin(2\pi x) + x^{3}$ when .

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Answer

First, evaluate $f(3)=sin(6\pi ) + 3^{3} = 27$ .

Then, to find the slope of the tangent line, find .

$f'(x) = 2\pi cos(2\pi x) + 3x^{2}$ , so $f'(3) = 2\pi + 27$ .

Therefore, the equation of the tangent line is

$y-27 = (2\pi + 27)(x-3)$ .

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Question

Differentiate,

$y = \sin(\sin(\sin\(3x)))$

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Answer

Differentiate,

$y = \sin(\sin(\sin\(3x)))$

Strategy

This one at first glance appears difficult even if we recognize that the chain rule is needed; we have a function within a function within a function within a function. To avoid making mistakes, it's best to start by defining variables to make the calculation easier to follow.

Let's start with the outermost function, we will write as a function of by setting,

$u = \sin(\sin(3x))$

$y = \sin(\underset{u}{ \underbrace{\sin(\sin(3x))}}) = \sin(u)$

$y = \sin(u)$                                                                                                                  Similarly, define to write as a function of

$u = \sin(\underset{v}{ \underbrace{\sin(3x)}}) = \sin(v)$

$u = \sin(v)$                                                                                                                  Write as a function of

$v = \sin(\underset{w}{ \underbrace{3x}} )=\sin(w)$

$v = \sin(w)$                                                                                                                  Finally, define the inner-most function, , as the function of

                                                                                                                   $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dw}\frac{dw}{dx}$

Since $\frac{dw}{dx}=3$ we will just substitute that in and move to the front.

$\frac{dy}{dx}=3\cos(u)\cos(v)\cos(w)$

That was easy enough, now just write everything in terms of by going back to the definitions of and .

$u = \sin(\sin(3x))$

$v = \sin(3x)$

$\frac{dy}{dx}=3\cos(\sin(\sin(3x)))\times \cos(\sin(3x))\times \cos(3x)$

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Question

Find the equation of the line tangent to $f(x)=x^{3}-3x^{2}+2x+4$ at the point .

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Answer

The first step is to find the derivative of the function given, which is $f'(x)=3x^{2}-6x+2$ . Next, find the slope at (1,4) by plugging in x=1 and solving for , which is the slope. You should get . This means the slope of the new line is also -1 because at the point where a slope and a line are tangent they have the same slope. Use the equation to express your line. Y and x are variables and m is the slope, so the only thing you need to find is b. Plug in the point and slope into to get . Now you can express the general equation of the line as .

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