Calculus II - Trig Substitutions

Show Mobile Notice Show All Notes Hide All Notes

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 7.3 : Trig Substitutions

Back to Problem List

4. Use a trig substitution to eliminate the root in \(\sqrt {{{\left( {w + 3} \right)}^2} - 100} \).

Show All Steps Hide All Steps

Start Solution

Okay, first off we need to acknowledge that this does look a little bit different from the first few problems in this section. However, it isn’t really all that different. We still have a difference between a squared term with a variable in it and a number. This looks similar to the following trig identity (ignoring the coefficients as usual).

\[{\sec ^2}\left( \theta \right) - 1 = {\tan ^2}\left( \theta \right)\]

So, secant is the trig function we’ll need to use for the substitution here and we now need to deal with the numbers on the terms and get the substitution set up.

Show Step 2

Before dealing with the coefficient on the trig function let’s notice that we’ll be substituting in for \(w + 3\) in this case since that is the quantity that is being squared in the first term.

So, to get the coefficient on the trig function notice that we need to turn the 1 (i.e. the coefficient of the squared term) into a 100 once we’ve done the substitution. With that in mind it looks like the substitution should be,

\[w + 3 = 10\sec \left( \theta \right)\]

Now, all we have to do is actually perform the substitution and eliminate the root.

Show Step 3

\[\begin{align*}\sqrt {{{\left( {w + 3} \right)}^2} - 100} & = \sqrt {{{\left( {10\sec \left( \theta \right)} \right)}^2} - 100} = \sqrt {100{{\sec }^2}\left( \theta \right) - 100} = 10\sqrt {{{\sec }^2}\left( \theta \right) - 1} \\ & = 10\sqrt {{{\tan }^2}\left( \theta \right)} = \require{bbox} \bbox[2pt,border:1px solid black]{{10\left| {\tan \left( \theta \right)} \right|}}\end{align*}\]

Note that because we don’t know the values of \(\theta \) we can’t determine if the tangent is positive or negative and so cannot get rid of the absolute value bars here.