The vector equation for the curve is r(t)=(e^(-t)cos t,e^(-t)sin t,e^(-t)), so (split)r'(t)&=(e^(-t)(-sin t)+(cos t)(-e^(-t)), e^(-t)cos t+(sin t)(-e^(-t)), (-e^(-t))) &=(-e^(-t)(cos t+sin t), e^(-t)(cos t-sin t),-e^(-t))...
Given the parametric curve below, find an expression for \frac{dy}{dx} and the values of t where there is a horizontal tangent line. x(t) = t - cos(2t) y(t) = (t^3 - 3t)^5 Consider the parametric curve given by x (t) = cos (2 t) - 2 cos (t...
(1)Find dy/dx and d^2y/dx^2. (2)Find the equation of the tangent line at the point where θ=π/6. (3)Find all points of horizontal tangency. (4)Determine where the curve is concave upward or concave downward. (5)Find the length of one are of the curve. 相关知识点: 试题...
Given the parametric curve below, find an expression for \frac{dy}{dx} and the values of t where there is a horizontal tangent line. x(t) = t - cos(2t) y(t) = (t^3 - 3t)^5 (a) Assume a curve is given by the parametric equations x = g ( ...
Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen., , ; 相关知识点: 试题来源: 解析 ,. At , and . Thus, parametric equations of the tang...
1. 在模型树中,选择“基准点 ID 1690”(Datum Point id 1690) ,然后单击“通过 点的曲线”(Curve through Points) 。 2. 在操控板中单击“使用线”(Use Line) ,从样条切换为直线。 3. 单击“预览特征”(Preview Feature) 。 图 2 4. 单击“恢复特征”(Resume Feature) 。 5. 单击“倒圆角曲线”(...
1. 单击“基准”(Datum) 组下拉菜单并选择“曲线”(Curve) 。 2. 在模型树中,选择“基准点标识 1690”(Datum Point id 1690)。 3. 在操控板中单击“使用线”(Use Line) ,从样条切换为直线。 4. 单击“预览特征”(Preview Feature) 。 Enlarge Image 图 3 5. 单击“恢复特征”(Resume Feature) 。 6...
Given a parametric curve: x = t^2 + 1, y = t^3 - 3 t. Find (a) An equation of the tangent line at t = 3. (b) The points where the tangent is horizontal. Find parametric equations for the line that is tangent to the curve...
Answer to: Find the parametric equations for the tangent line to the curve x = t^{3} - 1, y = t^{2} + 1, z = t^{3}, at the point (26, 10, 27). Use...
In the second grid line, the vertical component is held constant, yielding a horizontal line through (ui,vj)(ui,vj). The corresponding grid curves are r(ui,v)r(ui,v) and r(u,vj)r(u,vj), and these curves intersect at point PijPij. Figure 9. Grid lines on a parameter domain ...