Video: Finding the Equation of the Tangent to a Curve Defined by Parametric Equation at a Given Value for the Parameter

Find an equation of the tangent to the curve π‘₯ = βˆšπ‘‘, 𝑦 = 𝑑² βˆ’ 2𝑑 at the point corresponding to the value 𝑑 = 4.

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Video Transcript

Find an equation of the tangent to the curve π‘₯ equals root 𝑑, 𝑦 equals 𝑑 squared minus two 𝑑 at the point corresponding to the value 𝑑 equals four.

So the first thing we want to do is actually find out what point on the curve is the tangent. And to do that, what we’re gonna do is actually substitute in 𝑑 equals four to each of our parametric equations.

So we’re gonna start with our π‘₯-coordinate. And to find our π‘₯-coordinate, what we’re gonna do is substitute 𝑑 equals four into π‘₯ equals root 𝑑. And when we do that, we get that π‘₯ is equal to the root four. So therefore, our π‘₯-value is gonna be equal to two.

Okay, great, so we’ve got our first coordinate. So now, let’s find the 𝑦-coordinate. Well, the 𝑦-coordinate is gonna be equal to four squared minus two multiplied by four, which is gonna give us 16 minus eight. So therefore, we’re gonna get a final answer of 𝑦 is equal to eight.

So therefore, we can say that the point on the curve that the tangent acts is actually two, eight. Okay, great, so now what? So what we need to do actually is actually find the equation of the tangent to the curve. So therefore, what we want to do first is find the slope at the point that we’ve got on the curve. And the reason we want to do that is because at that point the slope of the curve and the slope of the tangent will be equal to each other.

So what we actually have is a pair of parametric equations. And we actually have a special relationship that tells us what the slope function is for parametric equations. So our slope function d𝑦 dπ‘₯ is equal to d𝑦 d𝑑 divided by dπ‘₯ d𝑑. The reason it’s really useful is because it allows us to deal with each of parametric equations separately and then bring them together to find our slope.

So I’m gonna to start with π‘₯ equals root 𝑑. So what we want to do is actually differentiate this to find dπ‘₯ d𝑑. But in order to make this easier, what we want to do first is actually rewrite this in exponent form. So we got π‘₯ is equal to 𝑑 to the power of a half. So therefore, dπ‘₯ d𝑑 is gonna be equal to a half 𝑑 to the power of negative a half.

And to just remind us how we got that is cause when you differentiate, what you do is to multiply the coefficient which is one by the exponent which is a half. So we get a half and then it’s 𝑑 to the power of and then you subtract one from the exponent. So a half minus one gives us negative a half.

Okay, great, we found dπ‘₯ d𝑑. Now, let’s move on and find d𝑦 d𝑑. So what we have is that 𝑦 is equal to 𝑑 squared minus two 𝑑. So we’re gonna differentiate this. And when we differentiate this, we get two 𝑑 minus two. So great, what we’ve done is actually found dπ‘₯ d𝑑 and d𝑦 d𝑑. So we can actually put it back into our form to find d𝑦 dπ‘₯.

So therefore, what we do is when we substitute this back in to our formula for d𝑦 dπ‘₯, what we’re gonna get is that our slope function d𝑦 dπ‘₯ is equal to two 𝑑 minus two because that’s our d𝑦 d𝑑 divided by a half 𝑑 to the power of negative a half cause that’s our dπ‘₯ d𝑑. So therefore, this is gonna be equal to two 𝑑 minus two multiplied by two 𝑑 to the power of a half. And we’ve got that because actually if you divide by a fraction, that’s the same as multiplying by the reciprocal of that fraction.

So now, what we do is we actually substitute in 𝑑 equals four to find out the value of our slope. So therefore, we can say that d𝑦 dπ‘₯ is equal to two multiplied by four minus two multiplied by two root four, which is gonna be equal to six multiplied by four which gives us 24.

Okay, great, so we’ve now found the slope at the point two, eight. But why has this been useful? Well, it’s useful because actually we’re trying to find the equation of the tangent to the curve. So that’s gonna be a straight line. And the general form for a straight line is 𝑦 equals π‘šπ‘₯ plus 𝑐, where π‘š is our slope and 𝑐 is our 𝑦-intercept.

So therefore, we can say that 𝑦 is gonna be equal to 24π‘₯ plus 𝑐. And that’s because our slope was 24. But now, what we need to do is actually find out what 𝑐 is. And to find out 𝑐, we can actually substitute in our values from our points, so our π‘₯- and 𝑦-coordinates.

So therefore, when we do this, we get eight β€” because that’s what 𝑦 is equal to β€” is equal to 24 multiplied by two because π‘₯ is equal to two then plus 𝑐. So therefore, eight is equal to 48 plus 𝑐. So now, if we actually actually subtract 48 from each side of the equation, we get 𝑐 is equal to negative 40.

So therefore, if we actually substitute this into our 𝑦 equals π‘šπ‘₯ plus 𝑐 along with our 24 for our π‘š, we’re gonna get the equation of the tangent to the curve π‘₯ equals root 𝑑, 𝑦 equals 𝑑 squared minus two 𝑑 at the point corresponding to the value 𝑑 equals four, which is the point two, eight, is going to be 𝑦 equals 24π‘₯ minus 40.

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