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Calculate the derivative $dy\text{/}dx$ for the plane curve defined by the equations
and locate any critical points on its graph.
${x}^{\prime}\left(t\right)=2t-4$ and
${y}^{\prime}\left(t\right)=6{t}^{2}-6,$ so
$\frac{dy}{dx}=\frac{6{t}^{2}-6}{2t-4}=\frac{3{t}^{2}-3}{t-2}.$
This expression is undefined when
$t=2$ and equal to zero when
$t=\mathrm{\pm 1}.$
Find the equation of the tangent line to the curve defined by the equations
First find the slope of the tangent line using [link] , which means calculating ${x}^{\prime}\left(t\right)$ and ${y}^{\prime}(t)\text{:}$
Next substitute these into the equation:
When $t=2,$ $\frac{dy}{dx}=\frac{1}{2},$ so this is the slope of the tangent line. Calculating $x\left(2\right)$ and $y\left(2\right)$ gives
which corresponds to the point $\left(1,3\right)$ on the graph ( [link] ). Now use the point-slope form of the equation of a line to find the equation of the tangent line:
Find the equation of the tangent line to the curve defined by the equations
The equation of the tangent line is $y=24x+100.$
Our next goal is to see how to take the second derivative of a function defined parametrically. The second derivative of a function $y=f\left(x\right)$ is defined to be the derivative of the first derivative; that is,
Since $\frac{dy}{dx}=\frac{dy\text{/}dt}{dx\text{/}dt},$ we can replace the $y$ on both sides of this equation with $\frac{dy}{dx}.$ This gives us
If we know $dy\text{/}dx$ as a function of t, then this formula is straightforward to apply.
Calculate the second derivative ${d}^{2}y\text{/}d{x}^{2}$ for the plane curve defined by the parametric equations $x\left(t\right)={t}^{2}-3,y\left(t\right)=2t-1,\mathrm{-3}\le t\le 4.$
From [link] we know that $\frac{dy}{dx}=\frac{2}{2t}=\frac{1}{t}.$ Using [link] , we obtain
Calculate the second derivative ${d}^{2}y\text{/}d{x}^{2}$ for the plane curve defined by the equations
and locate any critical points on its graph.
$\frac{{d}^{2}y}{d{x}^{2}}=\frac{3{t}^{2}-12t+3}{2{\left(t-2\right)}^{3}}.$ Critical points $\left(5,4\right),\left(\mathrm{-3},\mathrm{-4}\right),\text{and}\phantom{\rule{0.2em}{0ex}}\left(\mathrm{-4},6\right).$
Now that we have seen how to calculate the derivative of a plane curve, the next question is this: How do we find the area under a curve defined parametrically? Recall the cycloid defined by the equations $x\left(t\right)=t-\text{sin}\phantom{\rule{0.2em}{0ex}}t,\phantom{\rule{1em}{0ex}}y\left(t\right)=1-\text{cos}\phantom{\rule{0.2em}{0ex}}t.$ Suppose we want to find the area of the shaded region in the following graph.
To derive a formula for the area under the curve defined by the functions
we assume that $x\left(t\right)$ is differentiable and start with an equal partition of the interval $a\le t\le b.$ Suppose ${t}_{0}=a<{t}_{1}<{t}_{2}<\text{\cdots}<{t}_{n}=b$ and consider the following graph.
We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is $y\left(x\left({\stackrel{\u2013}{t}}_{i}\right)\right)$ for some value ${\stackrel{\u2013}{t}}_{i}$ in the i th subinterval, and the width can be calculated as $x\left({t}_{i}\right)-x\left({t}_{i-1}\right).$ Thus the area of the i th rectangle is given by
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