Find the tangent equation to the given curve that passes through the point (4, 3). Note that due to the t2 in the x equation and the 3 in the y equation, the equation in the parameter t has more than one solution. This means that there is a second tangent equation to the given curve that passes through a different point.
x = 3t2+1
y = 2t3 + 1
y = (tangent at smaller t)
y = (tangent at larger t)
Find the tangent equation to the given curve that passes through the point (4, 3)
Find the tangent equation to the given curve that passes through the point (4, 3). Note that due to the t2 in the x equation and the 3 in the y equation, the equation in the parameter t has more than one solution. This means that there is a second tangent equation to the given curve that passes through a different point. x = 3t2 + 1 y = 2t2 + 1
Find the tangent equation to the given curve that passes through the point (18,9). Note that due to the t2 in the x equation and the t3 in the y equation, the equation in the parameter t has more than one solution. This means that there is a second tangent equation to the given curve that passes through a different point. x = 9t2 + 9 y = 6t3 + 3
4. [-16 Points] DETAILS SCALCCC4 3.4.090. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Find the tangent equation to the given curve that passes through the point (10, 10). Note that due to the t2 in the x equation and the t3 in the y equation, the equation in the parameter t has more than one solution. This means that there is a second tangent equation to the given curve that passes through a different point. x = 6t2 + 4...
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = t4 + 3 y = t3 +t t = 1 y(x) = _______
Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 6 + In(t), y = +2 +6, (6, 7) — x
QUESTION 3 Find an equation for the line tangent to the curve at the point defined by the given value of t. x = 6t2 3y = t2= 1 o y = 1 / 8x + 1 / 2 Oy=}x+1 o y = 2/X 2/2 o y = 6x - 1 / 1
Find an equation for the line tangent to the curve at the point defined by the given value of t. x = sin t, y = 2 sin t, t = wa y = 2x - 213 y = 2x y = 2x + 13 Oy=-2x+ 2/3 Find an equation for the line tangent to the curve at the point defined by the given value of t. x=t, y= V2t, t = 18 y=- X-3 y=+x+3 O y = 1...
Find an equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = cos(O) + sin(40) y = sin(0) + cos(40) O = 0 y(x) = Need Help? Read It Talk to a Tutor 2. [-14 Points] DETAILS SESSCALCET1 9.2.010. Consider the following: x = t3 - 12t y = 2 - 1 (a) Find the following. dy dr = dy dr2 = (b) For which values of t is the...
Find an equation for the line tangent to the given curve at the point defined by the given value of t. Also, find d2y/dx2 at this point: x = cost, y = 1 + sint, t = 1/2
1. Find an equation of the tangent line to the curve at the given point. x = ++ +1, y=+*+t at the point (2,-2). 2. Find y" by implicit differentiation. (note that a is a constant) x² + y² = a² 3. A piece of wire 12 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total...