Find the equation of the tangent to:(1)y=x-2x^2+3 at x=2(2)y=√x+1 at x=4(3)y=x^3-5x at x=1(4)y=4(√x) at (1,4) 相关知识点: 试题来源: 解析 (1)y=-7x+11(2)4y=x+8(3)y=-2x-2(4)y=-2x+6反馈 收藏 ...
百度试题 结果1 题目求翻译:Find the equation of the tangent and the normal respectively to the curve y=cos x at the point (Π/3,1/2) 相关知识点: 试题来源: 解析 找到方程的切线和正常的曲线分别y = cos x点(Π/ 3,1/2) 反馈 收藏 ...
【题目】1.Find the equation of the tangent to the curve y=x^2 ,which is parallel to the r-aris2.Find the equation of the tangent to the curve y=x^2-2x which is perpendicular to the line 2y=-13.Find the equation of the normal to the curvey=3x^2-2x-1 which is parallel to ...
【题目】Find the equation of the tangent to the curvegiven parametrically byx=a(t+cost) y=a(1-sint) atthe pot where atthepOsintuheret=π/(4) t=π/(4) 相关知识点: 试题来源: 解析 【解析】(|x|)/(|d|)=a(1-sint) (v_1)/d=-acost⇒(v_1)/(v_1x)=(-cost)/(1-sint)...
Find an equation of the tangent line to the graph of the function f(x) = 8/(√x^2+3x) at the point (1,4) 相关知识点: 试题来源: 解析 题的意思是:求f(x)=8/√(x^2+3x)在点(1,4)处的切线方程 重点是求函数在x=1处的f'(x) 直接求导很麻烦,可采用对数求导法, lnf(x)=ln8-(...
x=tcos t, y=tsin t; t=π t=π , t=π , and t=π . When t=π, (x,y)=(-π ,0) and (x,y)=(-π ,0), so an equation of the tangent to the curve at the point corresponding to t=π is y-0=π [x-(-π )], or y=π x+π ^2.反馈...
(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 ...
When t=1, (x,y)=(0,2) and ( dy)( dx)= 22=1, so an equation of the tangent to the curve at the point corresponding to t=1 is y-2=1(x-0) or y=x+2.结果一 题目 Find an equation of the tangent to the curve at the point corresponding to the given value of the ...
百度试题 结果1 题目【题目】【题目】【题目】Find the equation of the tangent to the curve 相关知识点: 试题来源: 解析 【解析】 【解析】 反馈 收藏
Find the equation of the tangent to y = x2+ x + I at the point P on the curve where x = r. 相关知识点: 试题来源: 解析 Differentiating(dy)/(dx)=2x+1 so at P,(dy)/(dx)=2t+1The coordinates f(Parc(t,t^2+t+1)s abc=(2t+1)(x-t)=(tany-(t^2+1)=(-1)^2y=(2t+1...