Graph the curve {eq}\displaystyle x = - 2 \cos t,\ y = \sin t + \sin 2 t {/eq} to discover where it crosses itself. Then find equations of both tangents at that point. Tangent to a curve in 2D: The slope of the ...
31.Find the equation of the tangent to the curve y=x2,which is parallel to the x-axis2.Find the equation of the tangent to the curve y=x2-2x which is perpendicular to the line 2y=x-13.Find the equation of the normal to the curve y=3x2-2x-1 which is parallel to the line ...
Find the equation of the tangent t$$ o y = x ^ { 2 } + x + 1 $$at the point P on the curve where x = 相关知识点: 试题来源: 解析 Differentiating.$$ \frac { d y } { d x } = 2 x + 1 $$ so at P.$$ \frac { d y } { d x } = 2 t + 1 $$ The...
Answer to: Find the equation of two tangent lines to curve y = x^2 that intersect at (2,-2). Need exact values. By signing up, you'll get thousands...
are both the slopes of the same line y = x^3 dy/dx = 3x^2 Let (h k) be a point on the curve such that k = h^3 dy/dx = 3h^2 Equation of tangent: y - k = 3h^2 (x - h) When the tangent passes through the point (0 2) the equation of tangent ...
Find the equation(s) to the tangent lines to the curve y = (x - 1)/(x + 1) that are parallel to the line x - 2y = 2. Find equations of the tangent lines to the curve y = \frac{x-1}{x+1} that are parallel to the line x-2y=5 F...
(6)dydx=-sinx.80x=π//3 gives gradient -sinπ/3=-(√3)/2 and y value y=1/2Equation tangent is y-1/2=-(√3)/2(x-π/(3))rearranges to y=-(√3)/2x+((π√3)/6+1/2)(7) Using d/(dx)(cosex)=-cosexcotx(dy)/(dx)=-cosexcost-2cosxsinx=π/(6) gives gradient ...
(dy)/(dx)=4(x-2)^3When x =0(dy)/(dx)=-32y=16y=-32x+16Differentiates to obtain a correct derivative either 4(x-2)^3OE or4x^3-24x^2+48x-32P1by -32obtained with no errors seen in evaluating (dy)/(dx)Substitutes x=0into their (dy)/(dx)to obtain a numerical value or ...
Find the equation of the tangent line to: f(x)=2sin(x)−4cos(x) at x=0. The Tangent Line to a Curve: The tangent of the curve f(x) is y=mx+b where m is the slope and m= dx dy. To solve this problem, we'll use the common derivative: d dx...
Find the equation of the tangent line at the point where x=2x=2. Step 1 Find the point of tangency. Since x=2x=2, we evaluate f(2)f(2). f(2)=23=8f(2)=23=8 The point is (2,8)(2,8). Step 2 Find the value of the derivative at x=2x=2. f′(x)=3x2⟶...