已知椭圆参数方程$x= a cos \ t, y=b sin \ t$， 则椭圆在$t=\frac{\pi}{4}$点处的斜率为
A: $\frac{b}{a}$
B: $-\frac{b}{a}$
C: $\frac{a}{b}$
D: $-\frac{a}{b}$
A: $\frac{b}{a}$
B: $-\frac{b}{a}$
C: $\frac{a}{b}$
D: $-\frac{a}{b}$
举一反三
- 一平面简谐波以速度\(u\)沿\(x\)轴正方向传播，在\(t＝t＇\)时波形曲线如图所示．则坐标原点\(O\)的振动方程为 A: \(y=a\)cos[\(\frac{u}{b}\)\((t-t')\)\(+\frac{\pi}{2}\)] B: \(y=a\)cos[2\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)] C: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t+t')\)\(+\frac{\pi}{2}\)] D: \(y=a\)cos[\(\pi\)\(\frac{u}{b}\)\((t-t')\)\(-\frac{\pi}{2}\)]
- 设函数$$y=y(x)$$由$$\left\{ \begin{matrix} x=a(t-\sin t), \\ y=a(1-\cos t) \\ \end{matrix} \right.$$确定，则$${y}''(x)=$$()． A: $$-\frac{1}{a(1-\cos t)}$$ B: $$-\frac{1}{a{{(1-\cos t)}^{2}}}$$ C: $$\frac{1}{a(1-\cos t)}$$ D: $$\frac{1}{a{{(1-\cos t)}^{2}}}$$
- 函数$f(x)=\sin x + \cos x,x \in [0,2 \pi]$的上凸区间为 A: $[0,\frac{\pi}{4}] \cup [\frac{5}{4} \pi,2 \pi] $ B: $[\frac{\pi}{4},\frac{5}{4} \pi]$ C: $[0,\frac{3}{4}\pi] \cup [\frac{7}{4} \pi,2 \pi] $ D: $[\frac{3}{4} \pi,\frac{7}{4} \pi] $
- $\int_{0}^{\frac{\text{ }\!\!\pi\!\!\text{ }}{4}}{[\cos (2t)\mathbf{i}+\sin (2t)\mathbf{j}+t\sin t\mathbf{k}]}\operatorname{dt}=$( ) A: $(\frac{1}{2},\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ B: $(1,\frac{1}{2},\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ C: $(\frac{1}{2},1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$ D: $(1,1,\frac{4-\text{ }\!\!\pi\!\!\text{ }}{4\sqrt{2}})$
- 将函数\(f(x)=\sin^4 x\)展开成Fourier级数为 ____ . A: \(f(x) = \frac{3}{8}-\frac{1}{2}\cos 2x +\frac{1}{8}cos 4x\) B: \(f(x) = \frac{1}{4}-\frac{1}{2}\cos x +\frac{3}{8}cos 4x\) C: \(f(x) = \frac{1}{4}-\frac{1}{2}\sin 2x -\frac{3}{8}cos 4x\) D: \(f(x) = \frac{3}{8}-\frac{1}{2}\sin x -\frac{1}{8}cos 4x\)