Question

The tangent to the curve y=3x^2-5 at the point (2,7) makes an angle theta with positive x axis. Find value of tan theta

Answer 1

Answer:

$\large{\bold{tan\theta=12}}$

Step-by-step explanation:

$GIVEN\\\\y=3x^2-5\\\\diff\;w.r.t.\;\;x\\\\m=\frac{dy}{dx}=6x\\At ( 2,7)\\\\m=6\times 2=12\\\\WKT\\m=\frac{dy}{dx}=tan\theta\\\\=>tan\theta=12\\\\where\;m\;is\;slope\;of\;the\;curve$

HOPE IT HELPS

Answer 2

The value of tan θ = 12

Given :

The tangent to the curve y = 3x² - 5 at the point (2,7) makes an angle θ with positive x axis.

To find :

The value of tan θ

Solution :

Step 1 of 3 :

Write down the given equation of the curve

Here the given equation of the curve is

y = 3x² - 5

Step 2 of 3 :

Find the slope of the tangent to the curve at the point (2,7)

$\displaystyle \sf{y = 3 {x}^{2} - 5}$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2} - 5) }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = \frac{d}{dx} (3 {x}^{2}) - \frac{d}{dx} (5) }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = 3 \frac{d}{dx} ( {x}^{2}) - 0 }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = 3 \times 2x }$

$\displaystyle \sf{ \implies \frac{dy}{dx} = 6x }$

So the slope of the tangent to the curve at the point (2,7)

= m

$\displaystyle \sf{ = \frac{dy}{dx} \bigg|_{(2,7)} }$

$\displaystyle \sf{ = 6 \times 2 }$

$= 12$

Step 3 of 3 :

Find the value of tan θ

Now the tangent to the curve y = 3x² - 5 at the point (2,7) makes an angle θ with positive x axis

∴ m = tan θ

⇒ tan θ = m

⇒ tan θ = 12

Hence the value of tan θ = 12

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