The question is to find a vector equation for the tangent line to the curve of intersection of the cylinders x2+y2=25x2+y2=25 and y2+z2=20y2+z2=20 at the point (3,4,2)(3,4,2). This was their solution: I'm confused about the first part where they...
Calculus I: Lesson 6: Finding the Equation of a Tangent LineDr. Karen Brucks
knowledgeable and patient tutor with a ph.d. in electrical eng. see tutors like this in order to get the slope of the line tangent to the curve, we take the first order derivative of f ( x ) which is given as follows: f '( x ) = (3 - 2 x - 4 x 2 )' = -2 - 8 ...
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. ...
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 - ...
Determine the coordinates of the indicated point by plugging the value of x into the function. For example, to find the tangent line where x = 2 of the function F(x) = -x^2 + 3x, plug x into the function to find F(2) = 2. Thus the coordinate would be (2, 2). ...
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...
the tangent line has slope =-2 y = -2x + b solve for b by plugging in the point (1,9) 9 =-2 + b b = 9+2 = 11 y=-2x + 11 is the equation of the line tangent to the curve at x=1 Upvote • 0 Downvote Add comment More Report Still...
The process for finding the equation of the tangent line involves finding the derivative of the function at the given point. This derivative will give us the slope of the tangent line. Then, we can plug in the x-value of the given point into the derivative to find the y-value. This po...
The equation for the curve is: {eq}y=(-1/2)x^2+1 {/eq} Which looks like this: This is a graph of the equation y=(-1/2)x^2+1 Now that the curve and its equation are known, it's possible to discover the tangent line. The tangent line is a straight line that touches ...