I am trying to calculate the equation for the tangent line that passes through the point (0,1) of the curve y=lnx^3. When I differentiate the equation y=lnx^3, I get f'(x) = 3/x. When I use that equation to calculate the slope however it equals 3/0... which obviously is wrong. I'm not sure what mistakes I'm making, but I can't get it to work!
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equation of tangent line
- Thread starter lea.g
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I am trying to calculate the equation for the tangent line that passes through the point (0,1) of the curve y=lnx^3. When I differentiate the equation y=lnx^3, I get f'(x) = 3/x. When I use that equation to calculate the slope however it equals 3/0... which obviously is wrong. I'm not sure what mistakes I'm making, but I can't get it to work!
It is not wrong - at x = 0 the tangent to the curve is vertical (slope = DNE) also known as assymptote.
What is the equation of a vertical line at x = x1 ?
It is not wrong - at x = 0 the tangent to the curve is vertical (slope = DNE) also known as assymptote.
What is the equation of a vertical line at x = x1 ?
That part does make sense. I am still struggling to understand how to write a proper equation for other tangent lines to the curve though. How would I go about calculating a slope for other points?
In addition to the question, there is also a diagram showing the curve y=lnx^3, with a tangent line drawn at a point that is approx (4,2) (although the values are not given). This tangent line crosses the y axis at (0,1) which is stated, but no other information is given.
In addition to the question, there is also a diagram showing the curve y=lnx^3, with a tangent line drawn at a point that is approx (4,2) (although the values are not given). This tangent line crosses the y axis at (0,1) which is stated, but no other information is given.
Sorry for the confusion (I've been looking at this problem for far too long), but it is the equation of that line in particular that I have to solve for. I do not have the point of the tangent line that actually touches the curve, and that is where my confusion is coming from.
Sorry for the confusion (I've been looking at this problem for far too long), but it is the equation of that line in particular that I have to solve for. I do not have the point of the tangent line that actually touches the curve, and that is where my confusion is coming from.
Graph the equation. Look at it at x=0. Someone else already told you the tangent line is vertical at x=0. What is the equation for a vertical line at x=0?
calculate the equation for the tangent line that passes through the point (0,1) of the curve y=lnx^3
Hi Lea:
The phrase highlighted in red is poorly worded; the curve of ln(x^3) does not pass through the point (0,1).
The statement above is better worded as, "Find the equation for the line passing through (0,1) that is also tangent to the curve y=ln(x^3)".
You found everything that you need, except for the value of the x-coordinate at the point of tangency.
Let's call that value c
In other words, the point of tangency is ( c, ln(c^3) )
Now you can express the slope of the tangent line in terms of c.
Write out the Point-Slope formula for the tangent line, using the coordinates of the known point (0,1) and your expression for the slope.
Solve that equation for c, and you will have the equation for your tangent line.
If you need more help, please show us what you've done so far.
Cheers :cool:
Hi Lea:
The phrase highlighted in red is poorly worded; the curve of ln(x^3) does not pass through the point (0,1).
The statement above is better worded as, "Find the equation for the line passing through (0,1) that is also tangent to the curve y=ln(x^3)".
You found everything that you need, except for the value of the x-coordinate at the point of tangency.
Let's call that value c
In other words, the point of tangency is ( c, ln(c^3) )
Now you can express the slope of the tangent line in terms of c.
Write out the Point-Slope formula for the tangent line, using the coordinates of the known point (0,1) and your expression for the slope.
Solve that equation for c, and you will have the equation for your tangent line.
If you need more help, please show us what you've done so far.
Cheers :cool:
Got it! Thanks so much!