Video Transcript
Find the equation of the tangent to
the curve 𝑥 equals 𝑡 cubed plus one, 𝑦 equals 𝑡 to the fourth power plus 𝑡 at
the point corresponding to the value 𝑡 equals negative one.
In this question, we have a
parametric equation where the coordinates of 𝑥 and 𝑦 are given in terms of 𝑡. We can work out the equation of the
tangent using the equation 𝑦 minus 𝑦 one is equal to 𝑚 multiplied by 𝑥 minus 𝑥
one, where 𝑥 one, 𝑦 one is a point that lies on the curve and 𝑚 is the slope or
gradient. The gradient of the tangent 𝑚 is
equal to d𝑦 by d𝑥. When dealing with parametric
equations, this in turn is equal to d𝑦 by d𝑡 divided by d𝑥 by d𝑡. We can also write this as d𝑦 by
d𝑡 multiplied by d𝑡 by d𝑥 using the chain rule and our knowledge of
reciprocals.
As 𝑥 is equal to 𝑡 cubed plus
one, we can find an expression for d𝑥 by d𝑡 by differentiating 𝑡 cubed plus one
with respect to 𝑡. We can do this term by term giving
us d𝑥 by d𝑡 is equal to three 𝑡 squared as differentiating any constant gives us
zero. We can repeat this process to find
an expression for d𝑦 by d𝑡. Differentiating 𝑡 to the fourth
power gives us four 𝑡 cubed and differentiating 𝑡 gives us one. d𝑦 by d𝑥 is
therefore equal to four 𝑡 cubed plus one divided by three 𝑡 squared.
We need to calculate the value of
this when 𝑡 is equal to negative one. We need to calculate four
multiplied by negative one cubed plus one divided by three multiplied by negative
one squared. Negative one cubed is negative
one. Multiplying this by four gives us
negative four and adding one gives us a numerator of negative three. As negative one squared is equal to
positive one, when we multiply this by three, the denominator becomes three. Finally, negative three divided by
three is equal to negative one. Our value of d𝑦 by d𝑥 when 𝑡 is
equal to negative one is negative one.
We also need to find the values of
𝑥 and 𝑦 when 𝑡 is equal to negative one. Our value of 𝑥 is equal to zero as
negative one cubed plus one equals zero. 𝑦 is also equal to zero as
negative one to the fourth power is equal to one and adding negative one to this
also gives us zero. We can now substitute our values of
𝑥, 𝑦, and d𝑦 by d𝑥 into the equation 𝑦 minus 𝑦 one is equal to 𝑚 multiplied
by 𝑥 minus 𝑥 one. We have a gradient 𝑚 equal to
negative one and a coordinate zero, zero. Substituting in these values gives
us 𝑦 minus zero is equal to negative one multiplied by 𝑥 minus zero.
This simplifies to the equation 𝑦
is equal to negative 𝑥. This is the equation of the tangent
to the curve at the point corresponding to the value 𝑡 equals negative one.
