Differentiating functions given parametrically

In this article we shall explore how to find the gradient of a curve that has been given in parametric form. The important rule to remember when working with curves that have been described with parametric equations are to differentiate x and y with respect to parameter t and then using the chain rule as we shall see in the following examples.

Example

In this example we’re going to find the gradient at the point P where t=2 on the curve given by the following parametric equations;
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First we must differentiate x and y with respect to the parameter t.
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…and…
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Now we can use the chain rule;
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We need to rearrange the rule to give dy/dx, this should give us;
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We know what dy/dx is and we also know what dx/dt is, therefore we get;
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…so when t=2;
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So at the point P where t=2, the gradient is 4/13.

Example

Here is another example; A curve has been given by the parametric equations x = 3sin θ, y = 5cos θ. In this example we’re going to find the equation of the normal at the point P where θ = π/6.
Again we shall differentiate x and y with respect to the parameter θ, that is;
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Now we can use the chain rule as we did before; this will result in;
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So at the point P. where θ = π/6,
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You must know that the normal is perpendicular to the curve, so it’s gradient is -1/m where m is the gradient of the curve at that point.
To find the value of x and y we simply substitute in θ = π/6 in the original equations. We know that the gradient of the normal at P is 3√3 / 5 and x = 3/2 and y = 5√3 / 2.
To find the equation of the normal we use the equation of line in the form;
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…this gives us;
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…therefore;
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