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Slope of the Tangent

Date: 9 Jan 1995 07:17:37 -0500
From: Secret Someone
Subject: (none)

I'm stumped:

        Given y=A sin x + B tan x. Find A and B if the slope of the 

tangent to the curve at x1=pi/4 is m1=4+root 2. and at x=0 ; m2= 4

I do not get this one either. For some reason I am having a hard time 
visualizing some of these problems. If ya'll can help on any of these 
it'd be much appreciated.

Date: 9 Jan 1995 23:33:45 -0500
From: Dr. Ken
Subject: Re: your mail

Hello there!

Well, here goes.  What's happened here is that someone has given you a
function f(x) = A sin[x] + B tan[x], and they didn't tell you what the
constants A and B are.  So then they told you some other information about
how to find those constants, by telling you what the slopes of certain
tangent lines to the curve are.

So let's translate what they said into mathematics.  "Slope of the tangent"
means "the derivative with respect to x."  So let's first find out what the
derivative with respect to x of this function is.

We have f(x) = A sin[x] + B tan[x], so f'(x) = A cos[x] + B Sec^2[x].  So
what have they told us about the tangent lines, i.e. the value of f' at
certain points?

They said that at Pi/4, f' = 4 + Sqrt{2}.  A couple of other ways of saying
this are f'(Pi/4) = 4 + Sqrt{2}, and
A cos[Pi/4] + B Sec^2[Pi/4] = 4 + Sqrt {2}.  Let's simplify to get
A/Sqrt{2} + 2B = 4 + Sqrt{2}.

The equation from the other tangent line is nicer.  We get
A + B = 4.  Nice stuff.  So you can solve for one of these variables (I'll
pick A) and substitute into the nasty equation.  A = 4 - B.

So we have (4-B)/Sqrt{2} + 2B = 4 + Sqrt{2}.  

You can probably handle it from here.  You'll solve for B, then plug back in
to find A.  Then you're done.

-Ken "Dr." Math
Associated Topics:
High School Calculus

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