【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)...
解析 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....
Find the equation of the tangent to the curve at the point (1,1) and the equation of the tangent to the curve at the point . Deduce that for . 答案 y=-号于小,boobah+图恩维r绦丝绿下垂条万2}O2uC$相关推荐 1Find the equation of the tangent to the curve at the point (1,1) a...
The equation of the tangent line to the curve y = (18 -2x)^1/3 at the point where x = 5 is? 答案 y = (18 -2x)^1/3x=5,y=³√(18 -10)=2;根据题意:The equation of the tangent line就是求此点的切线方程,先求过此点切线的斜率:求导得:y'=(1/3)(18 -2x)^(1/3-1)=(...
百度试题 结果1 题目Find the equation of the tangent to the curve y=(x^2-5)^3 at the point (2,-1) 相关知识点: 试题来源: 解析 求导y'=6x(x²-5)²,f'(2)=12,所以y=12(x-2)-1=12x-25 反馈 收藏
Find the equation of the tangent to the curve given parametrically byx=a(t+cos t), y=a(1-sin t), at the point where t= (π )4. 相关知识点: 试题来源: 解析 , ⇒ ⇒.When t= (π )4, x=a( (π )4+ (√ 2)2), y=a(1- (√ 2)2), y=a(1- (√ 2)2).Tangent:...
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Find the equation of the tangent to the curve 4x2 + 3y2 = 7 at the point (1,1) .5个回答 找到曲线的切线方程4×2 + 3Y2上面的点(1,1)= 7。2013-05-23 12:21:38 回答:匿名 发现正切的等式对曲线4x2 + 3y2 = 7在点(1,1)。 2013-05-23 12:23:18 回答:匿名...
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 ...