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Find the equations of the tangents to the curve x = 9t2 + 6, y = 6t3 + 3 that pass through the point (15, 9).

Respuesta :

LammettHash

Tangents to the curve have slope [tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}[/tex]. We have

[tex]\dfrac{\mathrm dy}{\mathrm dt}=18t^2[/tex]

[tex]\dfrac{\mathrm dx}{\mathrm dt}=18t[/tex]

[tex]\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{18t^2}{18t}=t[/tex]

The point (15, 9) occurs on the curve and corresponds to

[tex]\begin{cases}9t^2+6=15\\6t^3+3=9\end{cases}\implies\begin{cases}t=\pm1\\t=1\end{cases}\implies t=1[/tex]

so that the slope of the line tangent to the curve at (15, 9) is 1. Then the equation of this tangent line is

[tex]y-9=1(x-15)\implies y=x-6[/tex]

ernikhil

The equation of the tangent is [tex]y-9=1(x-15)[/tex].

The given functions are [tex]x=9t^2+6[/tex] and [tex]y=6t^3+3[/tex].

Differentiate the given function with respect to [tex]t[/tex] as-

[tex]\dfrac{dx}{dt}=\dfrac{d(9t^2+6)}{dt}\\\dfrac{dx}{dt}=18t[/tex]

Similarly,

[tex]\dfrac{dy}{dt}=\dfrac{d(6t^3+3)}{dt}\\\dfrac{dy}{dt}=18t^2[/tex]

So,

[tex]\dfrac{dy}{dx}=\dfrac{18t^2}{18t}\\\dfrac{dy}{dx}=t[/tex]

Now, the point [tex](15, 9)[/tex] occurs on the curve and corresponds to

[tex]\left \{ {{9t^2+6=15} \atop {6t^3+3=9}} \right. \rightarrow \left \{ {{t=\pm 1} \atop {t=1}} \right. \rightarrow t=1[/tex]

Hence, the slope of the line tangent to the curve at [tex](15, 9)[/tex] is 1.

And, the equation of the tangent is [tex]y-9=1(x-15)[/tex].

Learn more about tangents here:

https://brainly.com/question/19064965?referrer=searchResults

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