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Unfortunately, the browser used for posting questions is hopelessly inadequate for mathematics: it strips away most symbols. All that we can see is "tangent line Yex(1-cosx) xpi2". From that it is not at all clear what the missing symbols (operators) might be. It makes little sense for me to try and guess - I may as well make up my own questions and answer them!
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How to measure the point at which two tangents intersect each other?
To measure the point at which two tangents intersect each other, find an equation for each tangent line and compute the intersection. The tangent is the slope of a curve at a point. Knowing that slope and the coordinates of that point, you can determine the equation of the tangent line using one of the forms of a line such as point-slope, point-point, point-intercept, etc. Do the same for the other tangent. Solve the two equations as a system of two equations in two unknowns and you will have the point of intersection.
How do you find the tangent of a circle?
a tangent is a line that touches the circle at only ONE point
How do you find the tangent of an angle?
budosnp
How did Descartes find the equation of a tangent line?
According to Wikipedia.'Suppose that a curve is given as the graph of a function, y = f(x). To find the tangent line at the point p = (a, f(a)), consider another nearby point q = (a + h, f(a + h)) on the curve. The slope of the secant passing through p and q is equal to the difference quotientAs the point q approaches p, which corresponds to making h smaller and smaller, the difference quotient should approach a certain limiting value k, which is the slope of the tangent line at the point p. If k is known, the equation of the tangent line can be found in the point-slope form:'
How do you find the equation of a tangent line in calculus?
Suppose a curve is defined by the function y = f(x) and you want the equation of the tangent at the point A.Suppose the x-coordinate at A is p. Then use y = f(p) to find the y-coordinate of A and let that be q. So, the point A is (p, q).Next, find dy/dx, the derivative of the function f, with respect to q. This will be a constant or a functio of x. In the latter case, find its value for x = p. This is m, the gradient of the tangent.So now you have the tangent line with gradient, m which passes through the point (p, q) and so its equation isy - q = m(x - p)Simplify this into the required form: y = mx + c or ax + by + c = 0.
