Find the equation of the tangent line to the graph

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January 16th 2011, 06:45 PM
zhangxupage2

Find the equation of the tangent line to the graph

Hi everyone. I'm working on a calc problem, but I found it to be difficult. Maybe I'm missing the point.

Find the equation of the tangent line to the graph of $f(x)=3x tan^2(x) + x^2 cos(x)$

Well, this is the derivative I got:

$3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)$

Since it just says the equation of the line, but it doesn't specify an x-value, so I just plug in a random value. However, I found it extremely difficult to do it the regular way, because trig functions are involved.

Is there a trick to solve this problem?

Thanks a lot.
January 16th 2011, 06:47 PM
dwsmith

$m=3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)$

Where m is the slope of the tangent line to the curve.

The key word is equation.
January 16th 2011, 06:59 PM
pickslides

You can leave it as $\displaystyle x=x_0$ for a future value, then $\displaystyle y-f(x_0)=f'(x_0)(x-x_0)$
January 16th 2011, 07:03 PM
dwsmith

I would just write

Let $m=3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)$

and as pickslides said let $(x_0,y_0) \in f(x)$

$y-y_0=m(x-x_0)\Rightarrow y=m(x-x_0)+y_0$

This way you can save yourself the trouble of writing out m.
January 16th 2011, 07:06 PM
zhangxupage2

What value should I choose for x?
January 16th 2011, 07:07 PM
dwsmith

Quote:

Originally Posted by zhangxupage2

What value should I choose for x?

The arbitrary value $x_0$
January 16th 2011, 07:16 PM
zhangxupage2

If I made x zero, then the slope is going to be zero also. The equation would become a horizontal line.
January 16th 2011, 07:17 PM
dwsmith

Quote:

Originally Posted by zhangxupage2

If I made x zero, then the slope is going to be zero also. The equation would become a horizontal line.

No one said to make it zero. That is the slope at x = 0. Does your question ask for the slope at x = 0?
January 16th 2011, 07:24 PM
zhangxupage2

What should I make x of? I'm not sure what you meant by arbitrary x0?
January 16th 2011, 07:25 PM
dwsmith

Quote:

Originally Posted by zhangxupage2

What should I make x of? I'm not sure what you meant by arbitrary x0?

Call it z than.
January 16th 2011, 07:27 PM
zhangxupage2

Maybe I haven't learned it in your way. Can I please explain it in a simpler way? Thanks a lot.
January 16th 2011, 07:28 PM
dwsmith

Quote:

Originally Posted by zhangxupage2

Maybe I haven't learned it in your way. Can I please explain it in a simpler way? Thanks a lot.

I am sorry. I don't know how else to explain it.
January 16th 2011, 07:35 PM
zhangxupage2

Thanks a lot. I'm trying to find a value for x, but I don't know which one to use.
January 16th 2011, 07:43 PM
Chris L T521

Quote:

Originally Posted by zhangxupage2

Hi everyone. I'm working on a calc problem, but I found it to be difficult. Maybe I'm missing the point.

Find the equation of the tangent line to the graph of $f(x)=3x tan^2(x) + x^2 cos(x)$

Well, this is the derivative I got:

$3tan^2(x)+6tan(x)sec^2(x) + 2xcos(x)-x^2sin(x)$

Since it just says the equation of the line, but it doesn't specify an x-value, so I just plug in a random value. However, I found it extremely difficult to do it the regular way, because trig functions are involved.

Is there a trick to solve this problem?

Thanks a lot.

As it has been said already, you need to evaluate your derivative at an arbitrary point to get the slope of your line.

So far, you've correctly shown that $f^{\prime}(x)=3\tan^2x+6\tan x\sec^2 x+2x\cos x-x^2\sin x$ .

Now, by convention, we let $x=x_0$ be the arbitrary point (for clarification $x_0$ and $0$ are not the same thing!) and substitute this into your derivative to see that

$m=f^{\prime}(x_0)=3\tan^2x_0+6\tan x_0\sec^2 x_0+2x_0\cos x_0-x_0^2\sin x_0$

Now this value for the slope will change for any $x_0$ ; in other words, $x_0$ is a place holder for a specific x value when its defined in the problem. If the problem had said that you needed to find the equation of the tangent line at $x=3$ , $x=2\pi$ , or $x=\frac{1}{2}\pi$ , then you would replace $x_0$ with 3, $2\pi$ or $\frac{1}{2}\pi$ respectively.

So now, in general, your tangent line at the arbitrary value $x=x_0$ will have the form $y-f(x_0)=m(x-x_0)$ . Thus the answer should be:

$y=\left(3\tan^2x_0+6\tan x_0\sec^2 x_0+2x_0\cos x_0-x_0^2\sin x_0\right)(x-x_0)+3x_0\tan^2x_0+x_0^2\cos x_0$

Certainly, this equation can be simplified when an actual value is given. Since one wasn't given, we have to leave it in this form.

I hope this clarifies things.
January 16th 2011, 07:45 PM
pickslides

Quote:

Originally Posted by zhangxupage2

What should I make x of? I'm not sure what you meant by arbitrary x0?

This values acts as a temporary value for what you might want to make $\displaystyle x$ .

In your case you were not given an x-value to use in the question. That is O.K. Nothing to worry about.

I think the confusion here is the fact you have a $\displaystyle x$ and an $\displaystyle x_0$ in the following equation.

$\displaystyle y-f(x_0)=f'(x_0)(x-x_0)$

The $\displaystyle x_0$ is were you will substitute in a co-ordinate where the $x$ is just part of the equation of the straight line. No need to replace him.

I understand this is a little confusing.

For your own sake pick a value i.e. $\displaystyle x=1$ and finish question like normal. What do you get? Think about what it means.

It might help you bring the bigger picture together.

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