Video Transcript
Find an equation of the tangent to the curve 𝑥 is equal to one plus the natural logarithm of 𝑡, 𝑦 is equal to 𝑡 squared plus two at the point one, three.
In this question, we have a parametric equation where the 𝑥- and 𝑦-coordinates are written in terms of 𝑡. We know that the equation of a tangent can be written in the form 𝑦 minus 𝑦 sub one is equal to 𝑚 multiplied by 𝑥 minus 𝑥 sub one. We have been given a point with coordinates one, three which we can substitute for 𝑥 one and 𝑦 one, respectively. Our value of 𝑚 will be the slope or gradient of the tangent. This can also be written as d𝑦 by d𝑥. And we know when dealing with parametric equations, d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 multiplied by d𝑡 by d𝑥. This is also written as d𝑦 by d𝑡 divided by d𝑥 by d𝑡.
We are given that 𝑥 is equal to one plus ln, the natural logarithm, of 𝑡. Our first step is to differentiate this to get an expression for d𝑥 by d𝑡. When differentiating a constant, we get zero. And differentiating the natural logarithm of 𝑡 gives us one over 𝑡. We are also told that 𝑦 is equal to 𝑡 squared plus two. Differentiating this will give us an expression for d𝑦 by d𝑡. Once again, we can do this term by term. Differentiating 𝑡 squared gives us two 𝑡, and differentiating the constant two gives us zero. Therefore, d𝑦 by d𝑡 is equal to two 𝑡.
We can now find an expression for d𝑦 by d𝑥 by multiplying d𝑦 by d𝑡 by the reciprocal of d𝑥 by d𝑡. We need to multiply two 𝑡 by 𝑡 over one. This is equal to two 𝑡 squared.
We now need to calculate the value of 𝑡 when 𝑥 equals one and 𝑦 equals three. This involves solving the equations one is equal to one plus the natural logarithm of 𝑡 and three is equal to 𝑡 squared plus two. Subtracting one from both sides of the left-hand equation gives us zero is equal to ln of 𝑡. We know that the natural logarithm of one is equal to zero. Therefore, 𝑡 equals one.
We can subtract two from both sides of the second equation, giving us one is equal to 𝑡 squared. Square rooting both sides of this equation gives us 𝑡 is equal to positive or negative one. As our value must satisfy both equations, we know that 𝑡 equals one. At the point one, three that lies on the curve, we know that 𝑡 is equal to one. d𝑦 by d𝑥, the gradient, is therefore equal to two multiplied by one squared. This is equal to two.
We can now substitute our values of 𝑚, 𝑥 one, and 𝑦 one into the equation 𝑦 minus 𝑦 one is equal to 𝑚 multiplied by 𝑥 minus 𝑥 one. This gives us 𝑦 minus three is equal to two multiplied by 𝑥 minus one. Distributing the parentheses gives us two 𝑥 minus two. We can then add three to both sides of this equation. This gives us the equation of the tangent to the curve at the point one, three, which is 𝑦 is equal to two 𝑥 plus one.
