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How do you find the tangent of the line at the curve at the point 11 using the equation 3x3 7x2y 7xy2 2y319?
I'm pretty sure that what you're looking for is: Either the slope or the full equation
of the line that is tangent to the curve represented by that messy equation, at
the point (1, 1).
If that's so, then . . .
-- You'd have to start out by knowing where the plus, minus, and 'equals' signs
are in the equation.
From there, what I would do is:
-- Take the implicit first derivative of the equation, and solve it for y' .
That gives you the slope of the curve as a function of points (x, y).
-- Use it to find the slope of the curve at the point (1, 1).
-- You know that the slope of the line that's tangent at that point is the
negative reciprocal of the slope of the curve. Now you have the slope of
the tangent line. If that's all you need, then you're done. Retreat and declare
victory.
-- If you need the equation of the tangent line, that's no problem. You have
its slope AND a point on it ==> (1, 1). Using that info to find the equation
of the line is supposed to be a piece o' cake for you by now.
=============================
Wait. Don't go away.
It looks to me as if the equation is
3x3 + 7x2y + 7xy2 + 2y3 = 19
just because the point (1, 1) is on it.
Do not trust anything I say past this point. I have already wasted too much time on this one and I am not going to check my work. But I did differentiate that mess, and I got the following:
Slope of the equation at (1, 1) . . . -10/9
Slope of the line tangent to the graph at (1, 1) . . . +0.9
Equation of the line . . . Y = 0.9x + 0.1
Take it for what it's worth, and Good Luck.
What else can I help you with?
How do you find the slope of a graph if it is a curve?
You find the slope of the tangent to the curve at the point of interest.
What t-word is define in geometry as a straight line that touches a curve but continues on with crossing it?
A tangent line touches a curve or the circumference of a circle at just one point.
What is the slope of a curve line?
The slope of a curved line changes as you go along the curve and so you may have a different slope at each point. Any any particular point, the slope of the curve is the slope of the straight line which is tangent to the curve at that point. If you know differential calculus, the slope of a curved line at a point is the value of the first derivative of the equation of the curve at that point. (Actually, even if you don't know differential calculus, the slope is still the value of the function's first derivative at that point.)
What is the equation of the line tangent to the curve y equals cos x at the point x equals pi over 6?
y = 2(x) - (pi/3) + (sqrt(3)/2)
How many ways can an arc be tangent to one line?
Only once. A tangent line shares only one point with any single arc/curve.
How do you get the tangent line without the graph?
The answer will depend on the context. If the curve in question is a differentiable function then the gradient of the tangent is given by the derivative of the function. The gradient of the tangent at a given point can be evaluated by substituting the coordinate of the point and the equation of the tangent, though that point, is then given by the point-slope equation.
How the slope of a curved line at a point can be found?
The slope of a curved line at a point is the slope of the tangent to the curve at that point. If you know the equation of the curve and the curve is well behaved, you can find the derivative of the equation of the curve. The value of the derivative, at the point in question, is the slope of the curved line at that point.
How do you find the point of tangency?
A tangent is an object, like a line, which touches a curve. The tangent only touches the curve at one point. That point is called the point of tangency. The tangent does not intersect (pass through) the curve.
How do you find the slope of the tangent line to each curve at the given point?
By differentiating the answer and plugging in the x value along the curve, you are finding the exact slope of the curve at that point. In effect, this would be the slope of the tangent line, as a tangent line only intersects another at one point. To find the equation of a tangent line to a curve, use the point slope form (y-y1)=m(x-x1), m being the slope. Use the differential to find the slope and use the point on the curve to plug in for (x1, y1).
Why we draw tangent in newton raphson method?
A line tangent to a curve, at a point, is the closest linear approximation to how the curve is "behaving" near that point. The tangent line is used to estimate values of the curve, near that point.
How do you find the slope of an Indifference curve?
You find the tangent to the curve at the point of interest and then find the slope of the tangent.
What is a point of tangency?
It is the point at which a tangent touches a curve.
What is the definition of a tangent line?
A tangent is a line that just touches a curve at a single point and its gradient equals the rate of change of the curve at that point.
As a curve approaches a maximum point the slope will do what?
The slope of the tangent line at the maximum point of the curve is zero. So we say that as a curve point approaches to the maximum point, the slope of the tangent line at that point approaches to zero.
How can you write the equation of the tangent line to the curve?
Well if you have found the derivative (slope of the tangent line) of the curve at that point and you know the xy coordinates for that point in the curve then you set it up in y=mx+b format where y is your y-coordinate, x is your x-coordinate and m is your derivative and solve for b
In maths what is a tangent and what is a chord?
Tangent:In geometry, the tangent line (or simply the tangent) is a curve at a given point and is the straight line that "just touches" the curve at that point. As it passes through the point where the tangent line and the curve meet the tangent line is "going in the same direction" as the curve, and in this sense it is the best straight-line approximation to the curve at that point.Chord:A chord of a curve is a geometric line segment whose endpoints both lie on the outside of the circle.
Find an equation of the line tangent to the curve defined by x^2+3xy+y^4=32 at the point (2,2)?
55
