Where m is the slope of the tangent line to the curve.
The key word is equation.
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
Well, this is the derivative I got:
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 .
Now, by convention, we let be the arbitrary point (for clarification and are not the same thing!) and substitute this into your derivative to see that
Now this value for the slope will change for any ; in other words, 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 , , or , then you would replace with 3, or respectively.
So now, in general, your tangent line at the arbitrary value will have the form . Thus the answer should be:
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.
This values acts as a temporary value for what you might want to make .
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 and an in the following equation.
The is were you will substitute in a co-ordinate where the 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. and finish question like normal. What do you get? Think about what it means.
It might help you bring the bigger picture together.