An equation of the tangent to the curve at the point corresponding to the given value of the parameter. x = t3 + 1, y = t10 + t; t = −1 is:
y = -3x
The equation of the tangent line can be estimated by applying the formula y – y₁ = m (x – x₁).
where;
From the information given:
x = t³ + 1 ; y = t¹⁰ + t ; t = -1
The first thing to do is to differentiate the given parameters;
∴
[tex]\mathbf {\dfrac{dx}{dt}= \dfrac{d}{dt}(t^3+1)}[/tex]
[tex]\mathbf {\dfrac{dx}{dt}= 3t^2+0}[/tex]
[tex]\mathbf {\dfrac{dx}{dt}= 3t^2}[/tex]
Also;
[tex]\mathbf {\dfrac{dy}{dt}= \dfrac{d}{dt}(t^{10}+t)}[/tex]
[tex]\mathbf {\dfrac{dy}{dt}= \dfrac{d}{dt}(10t^{9}+1)}[/tex]
Now, when t = -1
[tex]\mathbf{\dfrac{dx}{dt} = 3(-1)^2}[/tex]
[tex]\mathbf{\implies 3(1)} \\ \\ \mathbf{=3}[/tex]
[tex]\mathbf {\dfrac{dy}{dt}= 10(-1)^{9}+1)}[/tex]
[tex]\mathbf{\implies -10+1} \\ \\ \mathbf{=-9}[/tex]
The slope (m) can be computed as:
[tex]\mathbf{m = \dfrac{dy}{dx} }[/tex]
[tex]\mathbf{m = \dfrac{\dfrac{dy}{dt} }{\dfrac{dx}{dt} } }[/tex]
[tex]\mathbf{m = \dfrac{-9 }{3} }[/tex]
m = -3
The equation on a tangent line can be expressed as:
y - y₁ = m(x - x₁)
y - 0 = -3(x - 0)
y = -3x
Therefore, we can conclude that the equation of the tangent to the curve at the point corresponding to the given value of the parameter is -3x.
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