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Consider the following parametric equations.
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We have that x is equal to a t and that y is equal to 5t minus 5.
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We want to find the first and second partial derivatives of y with respect to x, then the slope, and the concavity at t equal to 1.
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So let's start with our first partial derivative, dx over dy.
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Via chain rule, we can show that dy over dx will write as dy over dt divided by dx over dt.
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Dy over dt is simply equal to 5, so the derivative of 5t minus 5 is equal to 5.
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And dx over dt, its derivative is simply equal to 8.
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So we find that our first derivative, dy over dx, is simply equal to 5 eighths.
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Now we want to find the second partial derivative, dy squared over dx squared.
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Now because our previous derivative was a constant, this is simply going to be equal to 0.
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And you can also verify this by calculating the derivative with respect to t of dy over dt, and to divide this again by dx over dt.
01:36
And it's this first derivative that is equal to 0...