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Calculate the derivative d y / d x for the plane curve defined by the equations

x ( t ) = t 2 4 t , y ( t ) = 2 t 3 6 t , −2 t 3

and locate any critical points on its graph.

x ( t ) = 2 t 4 and y ( t ) = 6 t 2 6 , so d y d x = 6 t 2 6 2 t 4 = 3 t 2 3 t 2 .
This expression is undefined when t = 2 and equal to zero when t = ±1 .
A curve going from (12, −4) through the origin and (−4, 0) to (−3, 36) with arrows in that order. The point (12, −4) is marked t = −2 and the point (−3, 36) is marked t = 3. On the graph there are also written three equations: x(t) = t2 – 4t, y(t) = 2t3 – 6t, and −2 ≤ t ≤ 3.

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Finding a tangent line

Find the equation of the tangent line to the curve defined by the equations

x ( t ) = t 2 3 , y ( t ) = 2 t 1 , −3 t 4 when t = 2 .

First find the slope of the tangent line using [link] , which means calculating x ( t ) and y ( t ) :

x ( t ) = 2 t y ( t ) = 2.

Next substitute these into the equation:

d y d x = d y / d t d x / d t d y d x = 2 2 t d y d x = 1 t .

When t = 2 , d y d x = 1 2 , so this is the slope of the tangent line. Calculating x ( 2 ) and y ( 2 ) gives

x ( 2 ) = ( 2 ) 2 3 = 1 and y ( 2 ) = 2 ( 2 ) 1 = 3 ,

which corresponds to the point ( 1 , 3 ) on the graph ( [link] ). Now use the point-slope form of the equation of a line to find the equation of the tangent line:

y y 0 = m ( x x 0 ) y 3 = 1 2 ( x 1 ) y 3 = 1 2 x 1 2 y = 1 2 x + 5 2 .
A curved line going from (6, −7) through (−3, −1) to (13, 7) with arrow pointing in that order. The point (6, −7) is marked t = −3, the point (−3, −1) is marked t = 0, and the point (13, 7) is marked t = 4. On the graph there are also written three equations: x(t) = t2 − 3, y(t) = 2t − 1, and −3 ≤ t ≤ 4. At the point (1, 3), which is marked t = 2, there is a tangent line with equation y = x/2 + 5/2.
Tangent line to the parabola described by the given parametric equations when t = 2 .
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Find the equation of the tangent line to the curve defined by the equations

x ( t ) = t 2 4 t , y ( t ) = 2 t 3 6 t , −2 t 3 when t = 5 .

The equation of the tangent line is y = 24 x + 100 .

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Second-order derivatives

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 ( x ) is defined to be the derivative of the first derivative; that is,

d 2 y d x 2 = d d x [ d y d x ] .

Since d y d x = d y / d t d x / d t , we can replace the y on both sides of this equation with d y d x . This gives us

d 2 y d x 2 = d d x ( d y d x ) = ( d / d t ) ( d y / d x ) d x / d t .

If we know d y / d x as a function of t, then this formula is straightforward to apply.

Finding a second derivative

Calculate the second derivative d 2 y / d x 2 for the plane curve defined by the parametric equations x ( t ) = t 2 3 , y ( t ) = 2 t 1 , −3 t 4 .

From [link] we know that d y d x = 2 2 t = 1 t . Using [link] , we obtain

d 2 y d x 2 = ( d / d t ) ( d y / d x ) d x / d t = ( d / d t ) ( 1 / t ) 2 t = t −2 2 t = 1 2 t 3 .
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Calculate the second derivative d 2 y / d x 2 for the plane curve defined by the equations

x ( t ) = t 2 4 t , y ( t ) = 2 t 3 6 t , −2 t 3

and locate any critical points on its graph.

d 2 y d x 2 = 3 t 2 12 t + 3 2 ( t 2 ) 3 . Critical points ( 5 , 4 ) , ( −3 , −4 ) , and ( −4 , 6 ) .

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Integrals involving parametric equations

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 ( t ) = t sin t , y ( t ) = 1 cos t . Suppose we want to find the area of the shaded region in the following graph.

A series of half circles drawn above the x axis with x intercepts being multiples of 2π. The half circle between 0 and 2π is highlighted. On the graph there are also written two equations: x(t) = t – sin(t) and y(t) = 1 – cos(t).
Graph of a cycloid with the arch over [ 0 , 2 π ] highlighted.

To derive a formula for the area under the curve defined by the functions

x = x ( t ) , y = y ( t ) , a t b ,

we assume that x ( t ) is differentiable and start with an equal partition of the interval a t b . Suppose t 0 = a < t 1 < t 2 < < t n = b and consider the following graph.

A curved line is drawn in the first quadrant. Below it are a series of rectangles marked that begin at the x axis and reach up to the curved line; the rectangle’s height is determined by the location of the curved line at the leftmost point of the rectangle. These lines are noted as x(t0), x(t1), …, x(tn).
Approximating the area under a parametrically defined curve.

We use rectangles to approximate the area under the curve. The height of a typical rectangle in this parametrization is y ( x ( t i ) ) for some value t i in the i th subinterval, and the width can be calculated as x ( t i ) x ( t i 1 ) . Thus the area of the i th rectangle is given by

A i = y ( x ( t i ) ) ( x ( t i ) x ( t i 1 ) ) .

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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