Calculus I: Lesson 6: Finding the Equation of a Tangent LineDr. Karen Brucks
I tried looking for and solving this for an hour and I am stumped. the question I am trying to have answered is: find an equation of the tangent line to the curve at the given point y=4x-3x^2, (2,-4) 0 Comments Sign in to comment. ...
The normal to a curve is the line perpendicular to the tangent to the curve at a given point.mtangent×mnormal=−1mtangent×mnormal=−1Worked example 13: Finding the equation of a tangent to a curve Find the equation of the tangent to the curve y=3x2y=3x2 at the point (1;...
微积分求解,Calculus test.Find rhe equation of the line tangent to the graph of f(x)=ln(1-x^2+2x^4) at x=1. 答案 是求导吧...df(x)/dx=(d(1-x^2+2x^4)/dx)/(1-x^2+2x^4)=(-2x+8x^3)/(1-x^2+2x^4)x=1时,代入,得df(x)/dt=3相关推荐 1微积分求解,Calculus test.Fin...
Find an equation in x and y for the line tangent to the curve c(t) = 3t+2, y(t) = t4 at the point (5,1). a) 4x + 3y +1=0 b) -42 + 3y - 17 = 0 c) 42 + 3y - 17 = 0 d) 2-3y - 11 = 0 e) 4x - 3y - ...
Solution Share Answered by Calculus expert Step 1 Need to find the equation of the tangent and the normal line to the curve y=2x+1 at x=4. First, find the deriv...View the full answer Step 2 Unlock Step 3 Unlock Step 4 Unlock Answer U...
A tangent line is a straight line that touches only one point on a given curve. In order to determine its slope it is necessary to understand the basic differentiation rules of differential calculus in order to find the derivative function f '(x) of the initial function f(x). The value ...
Tangent lines intersect a curve at exactly one point, so that is the point we will use for the equation. In this case, the tangent line will intersect the curve at the point (5π/18, f(5π/18)). While we were given the x value of the point of intersection, we need to find ...
a circle is a closed curve that is drawn from the fixed point called the center, in which all the points on the curve are having the same distance from the center point of the center. the equation of a circle with (h, k) center and r radius is given by: (x-h) 2 + (y-k) ...
The equation of tangent and normal can be evaluated just like any other straight line. But to find the gradient of tangents and normals to a curve, students will need the derivative. If the slope of the tangent to a curve y = f(x) at a point a is f'(a) (derivative of f(x) ...