Section: 参数形式和微积分:差异 | 10.5 参数形式和微积分:差异

Section outline

Maria has her house key attached to a carabineer. As she studies for her upcoming Calculus test, she twirls it around her finger. A key becomes detached from the ring and flies off the carabineer and across the room. Maria’s floor is covered in papers and laundry. How can she narrow down her search and find her key?
::Maria的家用钥匙被绑在一名汽车司机身上。当她为即将到来的Calculus测试而学习时,她用手指环绕着她的手指。钥匙与戒指脱钩,从汽车司机身上飞走,然后穿过房间。 Maria的地板被纸张和洗衣机覆盖。她怎么能够缩小搜索范围,找到钥匙呢?

Derivatives of Parametric Equations
::参数等量的衍生物

By now, you’ve gotten used to finding the for functions in terms of $\begin{aligned} y \end{aligned}$ and $\begin{aligned} x \end{aligned}$ . You may recall that the derivative of a function at a point $\begin{aligned} (x, y) \end{aligned}$ gives the slope of the line that forms a tangent to the function at that point. You may also remember that the slope of a function is the change in the $\begin{aligned} y \end{aligned}$ value of the function divided by the change in the $\begin{aligned} x \end{aligned}$ value of the function.
::y 和 x 的函数已经习惯了。您可以记得, 函数在某个点( x,y) 的衍生函数会给该点函数带来正切线的斜度。您也可以记住, 函数的斜度是函数的y值变化, 函数的y值被函数 x 值的变化所分隔。

Since a parametric function is a function of the form $\begin{aligned} F (t) = (x (t), y (t)) \end{aligned}$ where $\begin{aligned} x (t) \end{aligned}$ and $\begin{aligned} y (t) \end{aligned}$ are themselves functions, you can express the derivative for a parametric function at point $\begin{aligned} t \end{aligned}$ as $\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} \end{aligned}$ . That is, you can find the derivatives for $\begin{aligned} y (t) \end{aligned}$ and $\begin{aligned} x (t) \end{aligned}$ , and then divide. When you differentiate a parametric function, you use the same identities and techniques that you’ve used to differentiate functions written in terms of $\begin{aligned} x \end{aligned}$ and $\begin{aligned} f (x) \end{aligned}$ .
::由于参数函数是表F(t) =(x(t),y(t)) 的函数, x(t) 和 y(t) 本身是函数, 您可以在 t 点表示参数函数的衍生物 y&(t) x(t) 。也就是说, 您可以找到 y(t) 和 x(t) 的衍生物, 然后进行分隔。当您区分参数函数时, 您使用的特性和技术与您用来区分以 x 和 f(x) 写成的函数相同。

Take the parametric function
::采取参数函数

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 4 t + 7 \\ y (t) & = 10 t^{2} - 3 t \end{aligned}$

::F( t) = (x( t) = (x( t) y( t) ) x( t) = 4t+7y( t) = 10t2- 3t

Let's find the slope of the line tangent to the curve when $\begin{aligned} t = 7 \end{aligned}$ .
::让我们在t=7时找到与曲线正切线的斜坡。

First, find the derivative of $\begin{aligned} x (t) \end{aligned}$ : $\begin{aligned} x^{'} (t) = 4 \end{aligned}$
::首先,找到 x( t): x}( t)=4的衍生物

Now, find the derivative of $\begin{aligned} y (t) \end{aligned}$ : $\begin{aligned} y^{'} (t) = 20 t - 3 \end{aligned}$
::现在,找到y( t) y` ( t) =20t- 3的衍生物

Divide to find the derivative of $\begin{aligned} F (t) \end{aligned}$ : $\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{20 t - 3}{4} \end{aligned}$
::要找到 F( t) 的衍生物的除号: y`( t)x_( t) =20t- 34

Evaluate when $\begin{aligned} t = 7 \end{aligned}$ :
::评估t=7:

$\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{20 (7) - 3}{4} = \frac{137}{4} \end{aligned}$

::y`(t)x_(t)=20(7)-34=1374

The slope of the line tangent to the curve when $\begin{aligned} t = 7 \end{aligned}$ is $\begin{aligned} \frac{137}{4} \end{aligned}$ .
::t=7为1374时,线的斜坡与曲线相切。

Examples
::实例

Example 1
::例1

Earlier, you were asked to help Maria narrow down her search for her keys. Assume the radius of Maria’s circle (carabineer and keys together), is 10 centimeters. The keys make about one revolution per second. That means that you can write a parametric equation for the keys’ path as they circle her finger. The following equation can model the keys’ movement.
::早些时候,有人要求你帮助玛丽亚缩小对钥匙的搜索范围。假设玛利亚圆圈的半径(carabineer和键合起来)为10厘米。键每秒发生一次革命。这意味着在键盘绕着她的手指时,你可以为键盘的路径写一个参数方程。以下方程可以模拟键的移动。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 10 \cos (2 π t) \\ y (t) & = 10 \sin (2 π t) \end{aligned}$

::F( t) = (x( t) ,y( t) ) x( t) = 10cos @ ( 2) t) y( t) = 10sin @ ( 2)

Multiplying $\begin{aligned} t \end{aligned}$ by $\begin{aligned} 2 π \end{aligned}$ means that when 1 second has passed, the keys have made one revolution. Assume that the keys fly off Marias finger about halfway through their spin, when $\begin{aligned} t \end{aligned}$ is 0.5.
::乘以 2 表示当 1 秒过后 , 键发生了一次革命。假设钥匙在 Marias 的旋转中、当 t 是 0. 5 时从 Marias 的手指中飞过半。

Now, find the derivative of the equation and its value when $\begin{aligned} t \end{aligned}$ is 0.5.
::现在,在 t为 0.5 时, 找到方程的衍生物及其值。

$\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{10 \cos (2 π t)}{- 10 \sin (2 π t)} \end{aligned}$

::y&( t) x_( t) = 10cos @ ( 2°t) - 10sin @ ( 2°t)

When $\begin{aligned} t = 0.5 \end{aligned}$ :
::当t=0.5:

$\begin{aligned} \frac{y^{'} (0.5)}{x^{'} (0.5)} = - \frac{\cos (π)}{\sin (π)} = u n d e f i n e d \end{aligned}$

::y( 0.5)x( 0. 5) sin =未定义

Notice that $\begin{aligned} \sin (π) \end{aligned}$ is 0 which makes the slope undefined. This means that the keys fly off along a vertical line that runs parallel to the $\begin{aligned} y \end{aligned}$ -axis of your coordinate plane. So the equation of the line will take the form:
::注意 sin 是 0 使斜坡没有定义。这意味着键沿着与您的坐标平面 Y 轴平行的垂直线飞走。因此线的方程式将采取下列形式:

$\begin{aligned} x = K \end{aligned}$ where $\begin{aligned} K \end{aligned}$ is a constant. To find the value of $\begin{aligned} K \end{aligned}$ , find the value of $\begin{aligned} x \end{aligned}$ when $\begin{aligned} t \end{aligned}$ is 0.5.
::x=K, K是常数。要找到 K 的值,请在 t 为 0. 5 时找到 x 的值。

$\begin{aligned} x (.5) & = 10 \cos (2 π (.5)) \\ x (.5) & = 10 \cos (π) \\ x (.5) & = 10 (- 1) \\ x (.5) & = - 10 \end{aligned}$

::x( 5) = 10cos @ (2) (5) x( 5) = 10cos @ x( 5) = 10( 1) x( 5) = 10( 1) x( 5) @% 10

The keys will fly away and land somewhere along the line $\begin{aligned} x = - 10 \end{aligned}$ .
::钥匙会飞走降落在10号线的某处

Example 2
::例2

Find the equation of the line tangent to the following ellipse when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ .
::查找直线正切值的方程式, 以在 t4 时找到以下的椭圆。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 4 \cos (t) \\ y (t) & = - 6 \sin (t) \end{aligned}$

::F(t) = (x(t) y(t) ) x(t) = 4cos(t) y(t) 6sin(t)

First, use the derivatives for trigonometric functions to find $\begin{aligned} x^{'} (t) \end{aligned}$ and $\begin{aligned} y^{'} (t) \end{aligned}$ .
::首先,使用三角函数的衍生物来查找 x(t) 和 y(t) 。

$\begin{aligned} x^{'} (t) & = - 4 \sin (t) \\ y^{'} (t) & = - 6 \cos (t) \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

Divide to find the derivative of $\begin{aligned} F (t) \end{aligned}$ :
::要找到 F( t) 的衍生物而分隔开来:

$\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{- 6 \cos (t)}{- 4 \sin (t)} \end{aligned}$

::y"(t)x"(t)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Evaluate the derivative when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ :
::评估衍生物时, t4:

$\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{- 6 \cos \frac{π}{4}}{- 4 \sin \frac{π}{4}} = \frac{3}{2} \end{aligned}$

::y"(t)x"x"(t)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

So, the slope of the line tangent to the ellipse is $\begin{aligned} \frac{3}{2} \end{aligned}$ at time $\begin{aligned} \frac{π}{4} \end{aligned}$ .
::因此,与椭圆正切线的斜坡是32 时为4。

Now, you’ll need to find the values for $\begin{aligned} x (t) \end{aligned}$ and $\begin{aligned} y (t) \end{aligned}$ at time $\begin{aligned} \frac{π}{4} \end{aligned}$ .
::现在,您需要找到时间 4 的 x( t) 和 y( t) 值。

$\begin{aligned} x (\frac{π}{4}) & = 4 \cos (\frac{π}{4}) = 2 \sqrt{2} \\ y (\frac{π}{4}) & = - 6 \sin (\frac{π}{4}) = - 3 \sqrt{2} \end{aligned}$

:%4) = 4cos(%4) = 22y(%4) 6sin(%4) 32

So, $\begin{aligned} (x (t), y (t)) = (2 \sqrt{2}, - 3 \sqrt{2}) \end{aligned}$ when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ . Now that you have a point and a slope, you can write an equation in point-slope form:
:x(t),y(t))=(22,-32) t4。既然您有一个点和一个斜度,您可以用点-斜体形式写一个方程式 :

$\begin{aligned} y - (- 3 \sqrt{2}) = \frac{3}{2} (x - 2 \sqrt{2}) \end{aligned}$
::y-(-32)=32(x-22)

or
::或

$\begin{aligned} y + 3 \sqrt{2} = \frac{3}{2} (x - 2 \sqrt{2}) \end{aligned}$
::y+32=32(x-22)

Example 3
::例3

At Cheapskate Charlie’s Carnival of Fun, the swing ride is a chair attached to a rope swung by Strongman Steve as he spins in a circle. The motion of the chair can be described by the following :
::在Cheapskate Charlie的欢乐狂欢节上,摇摆座椅挂在Strongerman Steve绕圈旋转时的绳子上。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 6 \cos (π t) \\ y (t) & = 6 \sin (π t) \end{aligned}$
Unfortunately, on Tuesday, Strongman Steve had greasy carnival pizza for lunch and didn’t wash his hands before he returned to work. As a result, he lost his grip on the rope at $\begin{aligned} \frac{π}{6} \end{aligned}$ seconds into a spin. The carnival guest went flying off at a tangent to the circle. What equation describes her flight through the air?
::F(t) = (x(t),y(t) ) x(t) = 6cos(t) = 6sin) y(t) = 6sin(t) 不幸的是,周二,Strugman Steve有油腻的嘉年华披萨吃午餐,在上班前没有洗手。结果,他在 6 秒的旋转中失去了绳索。嘉年华嘉年华嘉宾飞向圆圈的正切方向飞去。什么方程式描述她在空中飞行?

Find the slope of the line tangent to the circle at the point where $\begin{aligned} t = \frac{π}{6} \end{aligned}$ .
::在 t6 点找到与圆形相切线的斜坡。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 6 \cos (π t) \\ y (t) & = 6 \sin (π t) \\ x^{'} (t) & = - 6 \sin (π t) \\ y^{'} (t) & = - 6 \cos (π t) \\ \frac{y^{'} (t)}{x^{'} (t)} & = \frac{- 6 \cos (π t)}{- 6 \sin (π t)} \end{aligned}$

::F( t) = (x( t) ,y( t) x( t) = 6cos}( t) = 6sin ( tt) x_( t) 6sin ( tt) ( tt) ( tt) 6cos( tt) - 6sin( tt)

Now, evaluate the derivative at $\begin{aligned} \frac{π}{6} \end{aligned}$ .
::现在,在六点六分评价衍生物

$\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{- 6 \cos (π t)}{- 6 \sin (π t)} = \frac{- 6 \cos \frac{π^{2}}{6}}{- 6 \sin \frac{π^{2}}{6}} = - .074274 \end{aligned}$

:t)xxx(t)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\6\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Evaluate the original equation at $\begin{aligned} \frac{π}{6} \end{aligned}$ so that you can find values for $\begin{aligned} x (\frac{π}{6}) \end{aligned}$ and $\begin{aligned} y (\frac{π}{6}) \end{aligned}$ .
::将原始方程在 6 上进行估量,以便您能找到 x( 6) 和 y( 6) 的值。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 6 \cos (π t) \\ y (t) & = 6 \sin (π t) \\ x (\frac{π}{6}) & = 6 \cos (\frac{π^{2}}{6}) = - .4444 \\ y (\frac{π}{6}) & = 6 \sin (\frac{π^{2}}{6}) = 5.9835 \end{aligned}$

::F( t) = (x( t) ,y( t) ) x( t) = 6cos ( lt)y( t) = 6sin ( lt) x( 6) = 6cos ( }) ( 26) ( 444444y( 6) = 6sin ( }26) = 5. 9835

Now use point-slope form to write the equation for the line that describes the path of the unfortunate carnival guest.
::现在使用点窗体来为描述不幸嘉年华嘉年华嘉宾路径的行写方程式。

$\begin{aligned} y - 5.9835 = - .074274 (x - .4444) \end{aligned}$ describes the path of the guest flung by Strongman Steve.
::y-5.9835.074274(x-444444) 描述由Strongerman Steve 挂起的客人路径。

Example 4
::例4

Find the equation for the line tangent to the following ellipse when $\begin{aligned} t = \frac{π}{3} \end{aligned}$ . Assume an object makes one revolution around the ellipse every $\begin{aligned} 2 π \end{aligned}$ units of time.
::当 t3. 假设一个对象每2时间单位一次围绕椭圆发生革命时, 查找线切换到以下椭圆的方程式。

$\begin{aligned} \frac{x^{2}}{64} + \frac{y^{2}}{16} = 1 \end{aligned}$

::x264+y216=1

To find the tangent to $\begin{aligned} \frac{x^{2}}{64} + \frac{y^{2}}{16} = 1 \end{aligned}$ at $\begin{aligned} t = \frac{π}{3} \end{aligned}$ , you’ll first need to put the equation into parametric form .
::要在 t3 找到 x264+y216=1 的正切值, 您首先需要将方程式设置为参数形式。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 8 \cos (t) \\ y (t) & = 4 \sin (t) \end{aligned}$

::F(t) = (x(t) y(t) ) x(t) = 8cos(t) y(t) = 4sin(t)

Now, find $\begin{aligned} x^{'} (t) \end{aligned}$ and $\begin{aligned} y^{'} (t) \end{aligned}$ .
::现在,找到 x_( t) 和 y_( t) 。

$\begin{aligned} x (t) & = 8 \cos (t) \\ x^{'} (t) & = - 8 \sin (t) \\ y (t) & = 4 \sin (t) \\ y^{'} (t) & = 4 \cos (t) \\ \frac{y^{'} (t)}{x^{'} (t)} & = \frac{4 \cos (t)}{- 8 \sin (t)} \end{aligned}$

:t)=8cos(t)x_(t)_(t)_(t)_(t)_(t)__(t)=4cos(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_8sin}(t)

Evaluate at $\begin{aligned} t = \frac{π}{3} \end{aligned}$ .
::评估在t3。

$\begin{aligned} \frac{y^{'} (t)}{x^{'} (t)} = \frac{4 \cos (\frac{π}{3})}{- 8 \sin (\frac{π}{3})} = \frac{4 (\frac{1}{2})}{- 8 (\frac{\sqrt{3}}{2})} = \frac{- \sqrt{3}}{6} \end{aligned}$

::y"(t)x"x"(t)=4cos(3)-8sin(3)=4(12)-8(32)36

Now, find the coordinates of the point where $\begin{aligned} t = \frac{π}{3} \end{aligned}$ .
::现在,找到t3点的坐标。

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (\frac{π}{3}) & = 8 \cos (\frac{π}{3}) = 8 (.5) = 4 \\ y (\frac{π}{3}) & = 4 \sin (\frac{π}{3}) = 4 (\frac{\sqrt{3}}{2}) = 2 \sqrt{3} \end{aligned}$

::F( t) = (x( t),y( t) ) x 3) = 8cos = 8( 3) = 8( 5) = 4y( 3) = 4sin = 4( 32) = 23

Put the slope, the $\begin{aligned} x \end{aligned}$ coordinate, and the $\begin{aligned} y \end{aligned}$ coordinate into point-slope form.
::将斜坡、 x 坐标和 Y 坐标放入点窗体。

$\begin{aligned} y - 2 \sqrt{3} = - \frac{\sqrt{3}}{6} (x - 4) \end{aligned}$ is the tangent line to the ellipse at $\begin{aligned} t = \frac{π}{3} \end{aligned}$ .
::y- 2336(x- 4) 是 t3 的椭圆的正切线。

Example 5
::例5

Find the slope of the tangent line to the described by the equation below at $\begin{aligned} t = \frac{π}{6} \end{aligned}$ :
::查找正切线的斜坡到以下t6方程式所描述的直线的斜坡:

$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 5 \sec (t) \\ y (t) & = 4 \tan (t) \end{aligned}$

::F(t) = (x(t) ,y(t) ) x(t) = 5sec}(t) y(t) = 4tan(t)

First, see if the function can be differentiated at $\begin{aligned} \frac{π}{6} \end{aligned}$ . If $\begin{aligned} x (t) \end{aligned}$ is undefined at this point, $\begin{aligned} x^{'} (t) \end{aligned}$ will also be undefined. The function is defined at this point, and it’s continuous, so you can differentiate it.
::首先, 看看该函数是否可以在 6. 如果 x( t) 此时尚未定义, x_( t) 也将被定义。此函数在目前点被定义, 并且是连续的, 所以您可以区分它。

Find the derivatives of $\begin{aligned} x (t) \end{aligned}$ and $\begin{aligned} y (t) \end{aligned}$ to find the derivative for the parametric equations:
::查找 x(t) 和 y(t) 的衍生物以找到参数方程式的衍生物:

$\begin{aligned} x (t) & = 5 \sec (t) \\ x^{'} (t) & = 5 \sec (t) \tan (t) \\ y (t) & = 4 \tan (t) \\ y^{'} (t) & = 4 \sec^{2} (t) \\ \frac{y^{'} (t)}{x^{'} (t)} & = \frac{4 \sec^{2} (t)}{5 \sec (t) \tan (t)} = \frac{4 \sec (t)}{5 \tan (t)} \end{aligned}$

:t)=5sec(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)__(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(x)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_(t)_

Now, evaluate the derivative when $\begin{aligned} t = \frac{π}{6} \end{aligned}$ .
::现在,在 T6 时评价衍生物。

$\begin{aligned} \frac{4 \sec (\frac{π}{6})}{5 \tan (\frac{π}{6})} = \frac{4 (2)}{5 (\frac{\sqrt{3}}{3})} = \frac{24}{5 \sqrt{3}} = \frac{24 \sqrt{3}}{15} = \frac{8}{5} \sqrt{3} \end{aligned}$

::4sec(6)5tan(6)=4(2)5(33)=2453=24315=853

Review
::回顾

A function is defined by the following parametric equation. Find the derivative, then evaluate it when $\begin{aligned} t = 6 \end{aligned}$ .
$\begin{aligned} M (t) & = (x (t), y (t)) \\ x (t) & = t + 6 \\ y (t) & = 3 t + t^{3} \end{aligned}$

::A 函数由以下的参数方程定义。查找衍生物, 然后在 t=6. M( t) =( x( t), y( t) ) x( t) =t+6y( t) =3t+t3 时对其进行评估

A function is defined by the following parametric equation. Find the derivative, then evaluate it when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ .
$\begin{aligned} N (t) & = (x (t), y (t)) \\ x (t) & = 7 \cos (2 t) \\ y (t) & = \sin (2 t) \end{aligned}$

::A 函数由以下的参数方程定义。查找衍生物, 然后在 t4. N( t) = (x( t), y( t)) x( t) = 7cos @ ( 2t)y( t) =sin @ ( 2t) =sin @ ( 2t) 时对其进行评估

A function is defined by the following parametric equation. Find the derivative, then evaluate it when $\begin{aligned} t = 1 \end{aligned}$ .
$\begin{aligned} P (t) & = (x (t), y (t)) \\ x (t) & = t^{2} \\ y (t) & = 3 t^{4} + 2 t^{2} - 6 t \end{aligned}$

::A 函数由以下的参数方程定义。查找衍生物, 然后在 t=1. P(t) =( x( t),y( t)) x( t) =t2y( t) =3t4+2t2 - 6t 时对它进行评估

Find the slope of the line tangent to the following curve when $\begin{aligned} t = 0 \end{aligned}$ .
$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 4 t \\ y (t) & = 3 t + t^{2} \end{aligned}$

::在t=0.F(t)=(x(t),y(t))x(t)=4ty(t)=3t+t2时,查找线线正切的斜度到下曲线的斜度

Find the slope of the line tangent to the following curve when $\begin{aligned} t = 3 \end{aligned}$ .
$\begin{aligned} G (t) & = (x (t), y (t)) \\ x (t) & = t^{2} \\ y (t) & = t^{3} + 1 \end{aligned}$

::当 t=3. G(t) =(x(t),y(t)) x(t) =t2y(t) =t3+1时, 查找线线正切的斜度到以下曲线

Find the slope of the line tangent to the following curve when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ .
$\begin{aligned} H (t) & = (x (t), y (t)) \\ x (t) & = 2 \cos (t) \\ y (t) & = 2 \sin (t) \end{aligned}$

::当 t4.H( t) = (x( t),y( t)) x( t) = 2cos *( t) y( t) = 2sin}( t) = 2sin ( t) 时, 查找线线正切的斜度到以下曲线的斜度

Find the slope of the line tangent to the following curve when $\begin{aligned} t = \frac{π}{3} \end{aligned}$ .
$\begin{aligned} J (t) & = (x (t), y (t)) \\ x (t) & = 3 \cos (t) \\ y (t) & = 5 \sin (t) \end{aligned}$

::当 t3. J(t) = (x(t),y(t)) x(t) = 3cos}(t) y(t) = 5sin}(t) = 5sin}(t) 时,查找线条正切的斜度至下曲线

Find the slope of the line tangent to the following curve when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ .
$\begin{aligned} K (t) & = (x (t), y (t)) \\ x (t) & = 2 \sec (t) \\ y (t) & = 4 \tan (t) \end{aligned}$

::当 t4. K( t) = (x( t),y( t)) x( t) = 2sec}( t) y( t) = 4tan( t) 时, 查找线直线正切的斜度到以下曲线的斜度

Find the equation of the line tangent to the following curve when $\begin{aligned} t = 0 \end{aligned}$ .
$\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 4 t \\ y (t) & = 3 t + t^{2} \end{aligned}$

::当 t=0. F(t) = (x(t),y(t)) x(t) = 4ty(t) = 3t+t2时, 查找线正切线的方程式到下曲线的方程

Find the equation of the line tangent to the following curve when $\begin{aligned} t = 3 \end{aligned}$ .
$\begin{aligned} G (t) & = (x (t), y (t)) \\ x (t) & = t^{2} \\ y (t) & = t^{3} + 1 \end{aligned}$

::当 t=3. G(t) = (x(t),y(t)) x(t) =t2y(t) =t3+1时, 查找线正切的方程式到下曲线的方程

Find the equation of the line tangent to the following curve when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ .
$\begin{aligned} H (t) & = (x (t), y (t)) \\ x (t) & = 2 \cos (t) \\ y (t) & = 2 \sin (t) \end{aligned}$

::当 t4.H( t) = (x( t),y( t)) x( t) = 2cos *( t) y( t) = 2sin}( t) = 2sin} (t) = 2sin} (t)

Find the equation of the line tangent to the following curve when $\begin{aligned} t = \frac{π}{3} \end{aligned}$ .
$\begin{aligned} J (t) & = (x (t), y (t)) \\ x (t) & = 3 \cos (t) \\ y (t) & = 5 \sin (t) \end{aligned}$

::当 t3. J(t) = (x(t),y(t)) x(t) = 3cos*(t) y(t) =5sin}(t) =5sin}(t) 时,查找线正切的方程到下曲线的方程。

Find the equation of the line tangent to the following curve when $\begin{aligned} t = \frac{π}{4} \end{aligned}$ .
$\begin{aligned} K (t) & = (x (t), y (t)) \\ x (t) & = 2 \sec (t) \\ y (t) & = 4 \tan (t) \end{aligned}$

::当 t4. K( t) = (x( t),y( t)) x( t) = 2sec}( t) y( t) = 4tan( t) 时, 查找线正切的方程到以下曲线的曲线。 @ k( t) = ( x( t) = 2sec}( t) y( t) = 4tan( t)

Find the equation for the line tangent to the following ellipse when $\begin{aligned} t = \frac{π}{6} \end{aligned}$ . Assume an object makes one revolution around the ellipse every $\begin{aligned} π \end{aligned}$ units of time.
$\begin{aligned} \frac{x^{2}}{16} + \frac{y^{2}}{49} = 1 \end{aligned}$

::当 t\\\ 6. 假设一个对象每时间单位 .x216+y249=1 围绕椭圆发生一次革命时, 查找线切换到以下椭圆的方程式。 x216+y249=1

Find the equation for the line tangent to the following circle when $\begin{aligned} t = \frac{2 π}{3} \end{aligned}$ . Assume an object makes one revolution around the circle every $\begin{aligned} 2 π \end{aligned}$ units of time.
$\begin{aligned} {(x - 1)}^{2} + {(y + 4)}^{2} = 9 \end{aligned}$

::当 t=23. 假设一个对象每2时间单位环圆一次突变时, 查找直线切换到下圆的方程。 (x- 1) 2+(y+4) 2=9

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Section outline

Derivatives of Parametric Equations ::参数等量的衍生物

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Derivatives of Parametric Equations
::参数等量的衍生物

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)