Question Video: Finding the Second Derivative of a Function Defined by Parametric Equations Mathematics • Higher Education

Video Transcript

Given that 𝑥 is equal to three 𝑡 squared plus one and 𝑦 is equal to three 𝑡 squared plus five 𝑡, find d two 𝑦 by d𝑥 squared.

In this question, we’ve been given a pair of parametric equations. And we’ve been asked to find d two 𝑦 by d𝑥 squared, which is the second derivative of 𝑦 with respect to 𝑥. Now, we in fact have an equation to help us find the second derivative of parametric equations. The equation tells us that d two 𝑦 by d𝑥 squared is equal to d by d𝑡 of d𝑦 by d𝑥 over d𝑥 by d𝑡. In order to use this equation, we also need to find d𝑦 by d𝑥. And we also have an equation for finding d𝑦 by d𝑥 when given parametric equations. This equation tells us that d𝑦 by d𝑥 is equal to d𝑦 by d𝑡 over d𝑥 by d𝑡.

We can start our solution by finding d𝑦 by d𝑡 and d𝑥 by d𝑡. We have that 𝑥 is equal to three 𝑡 squared plus one. Using the power rule for differentiation, we multiply by the power and decrease the power by one. And so, differentiating the term three 𝑡 squared, we obtain six 𝑡. Since one is a constant and we differentiate it, it will go to zero. Therefore, we have that d𝑥 by d𝑡 is equal to six 𝑡. We’ve also been given that 𝑦 is equal to three 𝑡 squared plus five 𝑡. This can again be differentiated using the power rule. Differentiating the first term, we again get six 𝑡. And when we differentiate the five 𝑡, we simply get five. Therefore, we have that d𝑦 by d𝑡 is equal to six 𝑡 plus five.

We’ve now found all the components in order to find d𝑦 by d𝑥. We obtain that d𝑦 by d𝑥 is equal to six 𝑡 plus five over six 𝑡. When we look at our formula for the second derivative of 𝑦 with respect to 𝑥, we spot that we have to find d by d𝑡 of d𝑦 by d𝑥. So that’s d by d𝑡 of six 𝑡 plus five over six 𝑡, which is a quotient. Therefore, we can use the quotient rule to help us differentiate here. We find that the derivative of a quotient of some functions, 𝑢 over 𝑣, is equal to 𝑣 multiplied by d𝑢 by d𝑥 minus 𝑢 multiplied by d𝑣 by d𝑥 all over 𝑣 squared.

Now, in our case, our numerator is equal to six 𝑡 plus five. Therefore, it’s equal to 𝑢. And our denominator is six 𝑡. And so it’s equal to 𝑣. We can differentiate six 𝑡 plus five with respect to 𝑡 to find that d𝑢 by d𝑡 is equal to six. And differentiating six 𝑡 with respect to 𝑡, we find that d𝑣 by d𝑡 is also equal to six. Now, we’re ready to substitute 𝑢, 𝑣, d𝑢 by d𝑡, and d𝑣 by d𝑡 into the formula given to us by the quotient rule. What we get is six 𝑡 times six minus six 𝑡 plus five times six all over six 𝑡 squared. Multiplying through, we obtain 36𝑡 minus 36𝑡 minus 30 over 36𝑡 squared. Therefore, we can cancel the 36𝑡 with the minus 36𝑡. And we can cancel through our factor of five to obtain that d by d𝑡 of d𝑦 by d𝑥 is equal to negative five over six 𝑡 squared.

So if we look back at our formula for d two 𝑦 by d𝑥 squared, we have now found d by d𝑡 of d𝑦 by d𝑥. And it’s equal to negative five over six 𝑡 squared. We also found d𝑥 by d𝑡 earlier. And it’s equal to six 𝑡. Using our formula, we have that d two 𝑦 by d𝑥 squared is equal to negative five over six 𝑡 squared over six 𝑡, which simplifies to give us a solution that d two 𝑦 by d𝑥 squared is equal to negative five over 36𝑡 cubed.