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1 year ago

Find the gradient of the curve y=-x^-3-5x^2+7x at x=2

Mathematics

1 answer:

Harman [31]1 year ago

5 0

-8

Step-by-step explanation:

The gradient or slope of a curve y is simply the derivative of the curve at a point x. So we can proceed as follows

$\dfrac{dy}{dx} = \dfrac{d}{dx}(x^3 -5x^2 +7) = 3x^2 -10x$

Evaluated at x = 2, the gradient of the slope is

$\left.\dfrac{dy}{dx}\right|_{x=2} = 3(2)^2 - 10(2) = -8$

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Convert the Cartesian equation (x 2 + y 2)2 = 4(x 2 - y 2) to a polar equation.

SashulF [63]

<u>ANSWER</u>

${r}^{2} = 4 \cos2\theta$

<u>EXPLANATION</u>

The Cartesian equation is

${( {x}^{2} + {y}^{2} )}^{2} = 4( {x}^{2} - {y}^{2} )$

We substitute

$x = r \cos( \theta)$

$y = r \sin( \theta)$

and

${x}^{2} + {y}^{2} = {r}^{2}$

This implies that

${( {r}^{2} )}^{2} = 4(( { r \cos\theta) }^{2} - {(r \sin\theta) }^{2} )$

Let us evaluate the exponents to get:

${r}^{4} = 4({ {r}^{2} \cos^{2}\theta } - {r}^{2} \sin^{2}\theta)$

Factor the RHS to get:

${r}^{4} = 4{r}^{2} ({ \cos^{2}\theta } - \sin^{2}\theta)$

Divide through by r²

${r}^{2} = 4 ({ \cos^{2}\theta } - \sin^{2}\theta)$

Apply the double angle identity

$\cos^{2}\theta -\sin^{2}\theta= \cos(2 \theta)$

The polar equation then becomes:

${r}^{2} = 4 \cos2\theta$

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Step-by-step explanation:

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Find the gradient of the curve y=-x^-3-5x^2+7x at x=2​

Find the gradient of the curve y=-x^-3-5x^2+7x at x=2