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

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

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Example Question #1 : Derivatives

Differentiate, 

 

Possible Answers:

Correct answer:

Explanation:

Differentiate, 

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, 

 

 

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Similarly, define  to write  as a function of  

  

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Write  as a function of 

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Finally, define the inner-most function, , as the function of 

 ________________________________________________________

 

 

Since  we will just substitute that in and move to the front. 

 

 

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

 

 

Example Question #2 : Ap Calculus Ab

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

 

Possible Answers:

Correct answer:

Explanation:

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

Example Question #3 : Ap Calculus Ab

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

Possible Answers:

Correct answer:

Explanation:

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 

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

Example Question #4 : Ap Calculus Ab

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

Possible Answers:

Correct answer:

Explanation:

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. 

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 

Example Question #5 : Derivative At A Point

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

 

 

 

Possible Answers:

Correct answer:

Explanation:

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

              

 

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: 

 

Evaluate the secant term: 

 

Therefore slope of the tangent line is simply: 

 

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: 

 

We now have our point:

 

 

Use the point to find 

 

 

 

Example Question #1 : Tangent Line To A Curve At A Point And Local Linear Approximation

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

Possible Answers:

Correct answer:

Explanation:

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.

Example Question #6 : Ap Calculus Ab

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

at the point .

Possible Answers:

Correct answer:

Explanation:

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

Apply the sum rule:

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

Example Question #8 : Derivative At A Point

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

at the point .

Possible Answers:

None of the other choices gives the correct response.

Correct answer:

None of the other choices gives the correct response.

Explanation:

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

Apply the constant multiple and sum rules:

Set  and  and apply the chain rule. 

Substituting back:

Evaluate  using substitution:

The tangent line is therefore the line with slope  through  is a -intercept, so apply the slope-intercept formula to get the equation

.

This is not among the choices given.

Example Question #2 : Tangent Line To A Curve At A Point And Local Linear Approximation

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

Possible Answers:

Correct answer:

Explanation:

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 

Example Question #8 : Ap Calculus Ab

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

Possible Answers:

 

 

 

 

 

Correct answer:

 

Explanation:

The first step is to find the derivative of the function given, which is . 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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