# Question Video: Differentiating Parametric Functions Involving Trigonometric Ratios at a Point Mathematics • Higher Education

Find dπ¦/dπ₯ at π = π/3, given π₯ = 5 cos π + 7 cos 2π and π¦ = 7 sin π + 4 sin 2π.

04:14

### Video Transcript

Find dπ¦ by dπ₯ at π equals π over three, given π₯ is equal to five cos π plus seven cos two π and π¦ is equal to seven sin π plus four sin two π.

In this question, we are given parametric equations for the coordinates π₯ and π¦ in terms of the variable π. We know that dπ¦ by dπ₯ is equal to dπ¦ by dπ divided by dπ₯ by dπ. This can also be written using the chain rule as dπ¦ by dπ₯ is equal to dπ¦ by dπ multiplied by dπ by dπ₯. Letβs begin by considering π₯, which is equal to five cos π plus seven cos two π. We can find an expression for dπ₯ by dπ by differentiating term by term.

We know that if π¦ is equal to cos π₯, dπ¦ by dπ₯ is equal to negative sin π₯. It also follows that if π¦ is equal to π multiplied by cos ππ₯, then dπ¦ by dπ₯ is equal to negative ππ multiplied by sin ππ₯. Differentiating five cos π gives us negative five sin π. And differentiating seven cos two π gives us negative 14 sin two π. We can repeat this for our π¦-coordinate, which is equal to seven sin π plus four sin two π. If π¦ is equal to sin π₯, dπ¦ by dπ₯ is cos π₯. Also, if π¦ is equal to π multiplied by sin ππ₯, then dπ¦ by dπ₯ is equal to ππ multiplied by cos ππ₯.

In our question, dπ¦ by dπ is equal to seven cos π plus eight cos two π. dπ¦ by dπ₯ is, therefore, equal to seven cos π plus eight cos two π divided by negative five sin π minus 14 sin two π. We need to calculate this when π is equal to π over three. At this point, we need to ensure that our calculator is in radian mode. cos of π over three is equal to one-half. cos of two π over three is equal to negative one-half. sin of π over three is equal to root three over two. And sin of two π over three is also equal to root three over two.

Substituting in these values, the numerator becomes seven multiplied by a half plus eight multiplied by negative a half. The denominator becomes negative five multiplied by root three over two minus 14 multiplied by root three over two. Seven multiplied by a half plus eight multiplied by negative a half is equal to negative one-half. Negative five multiplied by root three over two minus 14 multiplied by root three over two is equal to negative 19 root three over two.

Dividing a negative by a negative gives a positive answer. Therefore, this simplifies to one over 19 root three. We can then rationalize the denominator by multiplying the top and bottom of the fraction by root three. On the top, one multiplied by root three is root three. Multiplying root three by root three gives us three, and when we multiply this by 19, we get 57. So, the denominator is 57.

The value of dπ¦ by dπ₯ when π equals π over three is root three over 57.